Advanced Algorithms COMS31900 Lowest Common Ancestor (with a bit - - PowerPoint PPT Presentation

Advanced Algorithms – COMS31900 Lowest Common Ancestor

(with a bit on on Range Minimum Queries)

Rapha¨ el Clifford

Slides by Benjamin Sach

Advanced Algorithms – COMS31900 Rapha¨ el Clifford Lowest Common Ancestor

(with a bit on on Range Minimum Queries)

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . .

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j After preprocessing,

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j After preprocessing,

?! !

? ?

!?

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

i

root

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

i

root

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

i

ancestors of node i root

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

i

ancestors of node i

• nodes on the path

from i to the root root

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

j

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

j

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

j

ancestors of node j

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

j

ancestors of node j

• nodes on the path

from j to the root

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

j

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

i j

root

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing,

i j

root common ancestors of i and j the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing,

i j

root common ancestors of i and j

• nodes which are

ancestors of both i and j the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing,

i j

root the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing,

i j

root the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

4 5 3 1 2

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing,

i j

root the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j depth =

4 5 3 1 2

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing,

i j

root the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j depth = lowest common ancestor of i and j

4 5 3 1 2

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing,

i j

root depth = lowest common ancestor of i and j

• the common ancestor of i

and j furthest from the root the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

4 5 3 1 2

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

4 5 3 1 2

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

i j

4 5 3 1 2

LCA(i, j)

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

4 5 3 1 2

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

4 5 3 1 2

i j

LCA(i, j)

Lowest common ancestor

Preprocess a tree T (with n nodes) to answer lowest common ancestor queries. . . After preprocessing, the output to a query LCA(i, j) is the lowest common ancestor of nodes i and j

4 5 3 1 2

i j

LCA(i, j)

• Ideally, we would like O(n) space, O(n) prep. time and O(1) query time

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

7 9 6

1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

the nodes are numbered between

0 and (n − 1)

7 9 6

1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

7 9 6

1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats) Write down every node you visit . . . and its depth

7 9 6

1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

(node) N

Write down every node you visit . . . and its depth

(depth) D 0

7 9 6

1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1

(node) N

Write down every node you visit . . . and its depth

(depth) D

1

7 9 6

1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5

(node) N

Write down every node you visit . . . and its depth

(depth) D

2 1

7 9 6

1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1

7 9 6

9 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2

7 9 6

9 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3

7 9 6

9 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2

7 9 6

9 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1

7 9 6

9 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2

7 9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1

7 9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1

7 9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1

7 9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0

7 9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1

7 9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2

7

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1

7

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2

7

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1

7

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0

7

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1

7

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

7

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

7

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

7

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

how do we find LCA(i,j)?

7 i j

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

how do we find LCA(i,j)?

7 i j

Find i and j in N

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

how do we find LCA(i,j)?

7 i j

Find i and j in N

7 i′ j′

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

how do we find LCA(i,j)?

7 i j

Find i and j in N

7 i′ j′

Compute RMQ(i′, j′) in D

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

how do we find LCA(i,j)?

7 i j

Find i and j in N

7 i′ j′

Compute RMQ(i′, j′) in D

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

how do we find LCA(i,j)?

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

how do we find LCA(i,j)?

j i 9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

how do we find LCA(i,j)?

j i 9 6

Find i and j in N

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

how do we find LCA(i,j)?

j i 9 6

Find i and j in N

i′ j′ 9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

how do we find LCA(i,j)?

j i 9 6

Find i and j in N

i′ j′

Compute RMQ(i′, j′) in D

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

how do we find LCA(i,j)?

j i 9 6

Find i and j in N

i′ j′

Compute RMQ(i′, j′) in D

9 6 1 1 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

how do we find LCA(i,j)?

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

how do we find LCA(i,j)?

9 6

9 6 1

j i

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

how do we find LCA(i,j)?

9 6

9 6 1

j i

Find i and j in N. . .

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

how do we find LCA(i,j)?

9 6

9 6 1

j i

Find i and j in N. . . which copy of i?

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10

Compute an Euler tour of T . . . (a depth first search with repeats)

1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N

How long is the tour? We follow each edge twice. . . and there are (n − 1) edges Write down every node you visit . . . and its depth

(depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

how do we find LCA(i,j)?

9 6

9 6 1

j i

Find i and j in N. . . which copy of i?

any copy is fine

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n)

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(n)

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(n) O(?)

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(1) O(n) O(?)

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(1) O(n) O(?) O(1)

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(1) O(n) O(?) O(1) O(?)

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(1) O(n) O(?) O(1) O(?) O(1)

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(1) O(n) O(?) O(1) O(?) O(1)

• Prep. time O(n + prepRMQ(n))

Query time O(1 + queryRMQ(n)) Space O(n + spaceRMQ(n)) depends on the RMQ structure used

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(1) O(n) O(?) O(1) O(?) O(1)

• Prep. time O(n log log n)

Query time O(1) Space O(n log log n) using the best result from last lecture

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

why does this work?

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(1) O(n) O(?) O(1) O(?) O(1)

• Prep. time O(n log log n)

Query time O(1) Space O(n log log n) using the best result from last lecture

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

depth > d + 1

S2

x

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . . children

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

depth > d + 1

S2

x

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

subtrees

children

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

depth > d + 1

S2

x

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

subtrees

children

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1

S2

x

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1

S2

x

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1

y y y y

d + 1 d + 1 d + 1 d + 1

x y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1

y y y y

d + 1 d + 1 d + 1 d + 1

x y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

x y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

i j x y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

i j

imagine LCA(i, j) is not y

x y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

i j

imagine LCA(i, j) is not y

i′ and j′ are in here so RMQ does not return the location of a y x y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

i j

imagine LCA(i, j) is not y

i′ and j′ are in here so RMQ does not return the location of a y x y

(all of the ys are out of range)

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

x y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

i j x y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

i j

again, imagine LCA(i, j) is not y

x y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

i j

again, imagine LCA(i, j) is not y

i′ and j′ cross an x (which has smaller depth than y)

so the RMQ location isn’t a y

x y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

x y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

x i j y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

x i j y

now imagine LCA(i, j) is y

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

x i j y

now imagine LCA(i, j) is y

i′ and j′ cross a y (which is the smallest in the range)

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

x i j y

now imagine LCA(i, j) is y

i′ and j′ cross a y (which is the smallest in the range)

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

x i j y

now imagine LCA(i, j) is y

i′ and j′ cross a y (which is the smallest in the range)

Solving LCA using RMQ - correctness

We can also define a Euler tour of T recursively. . .

S1 S3 Sk

depth = d depth = d + 1

x x x x

(node) N (depth) D

x x d d d d d d

tour of S1 tour of S2 tour of S3 tour of Sk depth > d + 1 Claim the RMQ reports the location of some y in N iff LCA(i, j) = y

y y y y

d + 1 d + 1 d + 1 d + 1

x i j y

now imagine LCA(i, j) is y

i′ and j′ cross a y (which is the smallest in the range)

so the RMQ reports a y

Ongoing Summary

We have seen an O(n log log n) space, O(n log log n) prep. time and O(1) query time solution

for the Lowest Common Ancestor problem

which uses solution 3 for RMQ from last lecture

Ongoing Summary Can we do better?

We have seen an O(n log log n) space, O(n log log n) prep. time and O(1) query time solution

for the Lowest Common Ancestor problem

which uses solution 3 for RMQ from last lecture

Solving LCAs using RMQs - efficiency

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

Solving LCAs using RMQs - efficiency

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

Notice anything interesting about D?

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

Solving LCAs using RMQs - efficiency

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

Notice anything interesting about D?

D[i + 1] = D[i] ± 1

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

±1 Range minimum query

A

Preprocess an integer array A (length n) to answer range minimum queries. . .

n

After preprocessing, a range minimum query is given by RMQ(i, j) the output is the location of the smallest element in A[i, j] e.g. RMQ(3, 7) = 5, which is the location of the smallest element in A[3, 7]

i = 3 j = 7

RMQ(3, 7) = 5

• Can we exploit this ±1 property to get a more efficient RMQ data structure?
• Ideally we would like O(n) space, O(n) prep. time and O(1) query time

where for all k, we have A[k + 1] = A[k] ± 1

16 17 15 16 15 14 15 16 17 18 19 20 21 20 19 20

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

(in a tie, report the leftmost)

Low-resolution RMQ (again) A

Key Idea replace A with a smaller, ‘low resolution’ array H

H ˜ n

i j

and many small arrays L0, L1, L2 . . . ‘for the details’

n

i′ j′

Preprocess the array H

• which has length ˜

n =

2n log n

• to answer RMQs. . .

in O(n) space/prep time Preprocess each array Li (which has length (log n)/2) to answer RMQs. . . in O(log n log log n) space/prep time How do we answer a query in A in O(1) time? Do one query in H and one query in two different Li and return the smallest

˜ n =

2n log n

log n 2

min of these as there are O(n/ log n) Li arrays, we have O(n log log n) total space/prep time

L0 L1 L2 L3 L4 L5 L˜

n

goes in here all of these go in here

Low-resolution RMQ (again) A

Key Idea replace A with a smaller, ‘low resolution’ array H

H ˜ n

i j

and many small arrays L0, L1, L2 . . . ‘for the details’

n

i′ j′

Preprocess the array H

• which has length ˜

n =

2n log n

• to answer RMQs. . .

in O(n) space/prep time Preprocess each array Li (which has length (log n)/2) to answer RMQs. . . in O(log n log log n) space/prep time How do we answer a query in A in O(1) time? Do one query in H and one query in two different Li and return the smallest

˜ n =

2n log n

log n 2

min of these as there are O(n/ log n) Li arrays, we have O(n log log n) total space/prep time

too big and slow!

L0 L1 L2 L3 L4 L5 L˜

n

goes in here all of these go in here

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to (remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

RMQx(0, 2) = RMQy(0, 2) = 2 RMQx(3, 4) = RMQy(3, 4) = 4 RMQx(0, 4) = RMQy(0, 4) = 2 RMQx(0, 1) = RMQy(0, 1) = 1 (remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

1 1

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

1 1

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

1 1

dx = dy = = 2 = 2

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

Fact Lx is equivalent to Ly iff dx = dy

1 1

dx = dy = = 2 = 2

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

Fact Lx is equivalent to Ly

• We can precompute dx for each Lx in O(|Lx|) = O(log n) time.

iff dx = dy

1 1

dx = dy = = 2 = 2

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

Fact Lx is equivalent to Ly

• We can precompute dx for each Lx in O(|Lx|) = O(log n) time.
• How many different values of d are there?

iff dx = dy

1 1

dx = dy = = 2 = 2

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

Fact Lx is equivalent to Ly

• We can precompute dx for each Lx in O(|Lx|) = O(log n) time.
• How many different values of d are there?

iff dx = dy

1 1

dx = dy = = 2 = 2 d contains (log n)/2 − 1 bits so . . .

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

Fact Lx is equivalent to Ly

• We can precompute dx for each Lx in O(|Lx|) = O(log n) time.
• How many different values of d are there?

iff dx = dy

1 1

dx = dy = = 2 = 2 d contains (log n)/2 − 1 bits so . . .

at most 2(log n)/2 (remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

Fact Lx is equivalent to Ly

• We can precompute dx for each Lx in O(|Lx|) = O(log n) time.
• How many different values of d are there?

iff dx = dy

1 1

dx = dy = = 2 = 2 d contains (log n)/2 − 1 bits so . . .

at most 2(log n)/2

=

• 2log n1/2

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

Fact Lx is equivalent to Ly

• We can precompute dx for each Lx in O(|Lx|) = O(log n) time.
• How many different values of d are there?

iff dx = dy

1 1

dx = dy = = 2 = 2 d contains (log n)/2 − 1 bits so . . .

at most 2(log n)/2

=

• 2log n1/2 √n

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

Fact Lx is equivalent to Ly

• We can precompute dx for each Lx in O(|Lx|) = O(log n) time.
• How many different values of d are there?
• For each value of d we store RMQ(i, j) for all i, j

iff dx = dy

1 1

dx = dy = = 2 = 2 d contains (log n)/2 − 1 bits so . . .

at most 2(log n)/2

=

• 2log n1/2 √n

(remember these are the locations of the minimum)

Counting ±1 RMQ arrays

L

log n 2

How many different ±1 RMQ arrays like this. . . are there?

Ly

log n 2

We say that iff for all (i, j): RMQx(i, j) = RMQy(i, j)

Lx

log n 2

is equivalent to

16 15 14 15 14 10 9 8 9 8

is equivalent to

Lx Ly

1 2 3 4 1 2 3 4

16 15 14 15 14 10 9 8 9 8

1 2 3 4 1 2 3 4

Fact Lx is equivalent to Ly

• We can precompute dx for each Lx in O(|Lx|) = O(log n) time.
• How many different values of d are there?
• For each value of d we store RMQ(i, j) for all i, j

iff dx = dy . . . this requires O(√n log2 n) = O(n) total space and prep. time

1 1

dx = dy = = 2 = 2 d contains (log n)/2 − 1 bits so . . .

at most 2(log n)/2

=

• 2log n1/2 √n

(remember these are the locations of the minimum)

RMQ on the L arrays in linear space A

Key Idea replace A with a smaller, ‘low resolution’ array H

n ˜ n =

2n log n

log n 2

L0 L1 L2 L3 L4 L5 L˜

n

d0 d1 d2 d3 d4 d5 d˜

n

precompute the value of dx for each Lx in O(n) total space and prep. time Precompute all the RMQ answers for every value 0 d √n in O(n) total space and prep. time To perform a query within some Lx

• Look up dx
• Find the row dx in the table
• Find the entry giving RMQx(i, j)

row d3

This takes O(1) time

Optimal ±1 RMQ

Key Idea replace A with a smaller, ‘low resolution’ array H and many small arrays L0, L1, L2 . . . ‘for the details’ Preprocess the array H to answer RMQs. . . in O(n) space/prep time Preprocess each array Li (which has length (log n)/2) to answer RMQs. . . build a complete table of answers How do we answer a query in A in O(1) time? Do one query in H and one query in two different Li and return the smallest

˜ n =

2n log n O(n) total space/prep time

A H ˜ n

i j

n

i′ j′

log n 2

min of these

L0 L1 L2 L3 L4 L5 L˜

n

goes in here all of these go in here

Ongoing Summary for the Lowest Common Ancestor problem

We have seen an O(n) space, O(n) prep. time and O(1) query time solution

for the ±1 Range Minimum Query problem

which improves solution 3 for RMQ from last lecture (but only for ±1 inputs) which uses solution 3 for RMQ from last lecture

We have seen an O(n log log n) space, O(n log log n) prep. time and O(1) query time solution

Ongoing Summary

How does this affect our LCA solution?

for the Lowest Common Ancestor problem

We have seen an O(n) space, O(n) prep. time and O(1) query time solution

for the ±1 Range Minimum Query problem

which improves solution 3 for RMQ from last lecture (but only for ±1 inputs) which uses solution 3 for RMQ from last lecture

We have seen an O(n log log n) space, O(n log log n) prep. time and O(1) query time solution

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(1) O(n) O(n) O(1) O(1) O(1)

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(1) O(n) O(n) O(1) O(1) O(1)

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(1) O(n) O(n) O(1) O(1) O(1)

This gives us O(n) space, O(n) prep. time and O(1) query time for the LCA problem

Solving LCAs using RMQs

3 1 2 2 3

4

5

8

10 1 5 5 10 5 1 1 0 2 0 3 3 8 3 0 4 0

(node) N (depth) D

3 2 1 2 3 2 1 2 1 1 0 1 2 1 2 1 0 1 0

2n−1

7

7

9 6

9 6 1

Preprocessing Summary

• 1. Construct N and D from T
• 2. Add a pointer from each
• 3. Preprocess D for RMQs

node i to some N[i′] = i Query Summary - LCA(i,j)

• 1. Find (any) i′ st. N[i′] = i
• 2. Find (any) j′ st. N[j′] = j
• 3. Compute RMQ(i′, j′) in D
• 4. LCA(i, j) = N[RMQ(i′, j′)]

O(n) O(1) O(n) O(n) O(1) O(1) O(1)

This gives us O(n) space, O(n) prep. time and O(1) query time for the LCA problem

by using the solution to ±1RMQ

Ongoing Summary

We have seen an O(n) space, O(n) prep. time and O(1) query time solution

for the Lowest Common Ancestor problem

which uses the solution to ±1RMQ

We have seen an O(n) space, O(n) prep. time and O(1) query time solution

for the ±1 Range Minimum Query problem

which improves solution 3 for RMQ from last lecture (but only for ±1 inputs)

Ongoing Summary

What about the general Range Minimum Query problem?

We have seen an O(n) space, O(n) prep. time and O(1) query time solution

for the Lowest Common Ancestor problem

which uses the solution to ±1RMQ

We have seen an O(n) space, O(n) prep. time and O(1) query time solution

for the ±1 Range Minimum Query problem

which improves solution 3 for RMQ from last lecture (but only for ±1 inputs) (when the inputs aren’t ±1)

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

Build the Cartesian tree, TA of the array A:

6 13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

Build the Cartesian tree, TA of the array A:

• The root is the smallest value

6 13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value

6 13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

partitions the array in two

6 13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

17 8 23 73 51 82 19 32 67 91 14 46 9 21 54

partitions the array in two

6 13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

17 8 23 73 51 82 19 32 67 91 14 46 9 21 54

partitions the array in two

• The rest of the tree is given by

recursing left and right. . .

6 13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

17 8 23 73 51 82 19 32 67 91 14 46 9 21 54 8 9

partitions the array in two

• The rest of the tree is given by

recursing left and right. . .

6 13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

8 9 17 23 73 51 82 19 32 67 91 14 46 21 54

partitions the array in two

• The rest of the tree is given by

recursing left and right. . .

6 13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

8 9 17 23 73 51 82 19 32 67 91 14 46 21 54

17 14 21

partitions the array in two

• The rest of the tree is given by

recursing left and right. . .

6

19

13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

8 9

17 14 21

23 73 51 82 32 67 91 46 54

partitions the array in two

• The rest of the tree is given by

recursing left and right. . .

6

19

13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

8 9

17 14 21

23 73 51 82 32 67 91 46 54

partitions the array in two

• The rest of the tree is given by

recursing left and right. . .

6

23 51 67 46 54 32 19

13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

8 9

17 14 21

73 82 91

partitions the array in two

• The rest of the tree is given by

recursing left and right. . .

6

23 51 67 46 54 32 19

13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

8 9

17 14 21 73 82 91

partitions the array in two

• The rest of the tree is given by

recursing left and right. . .

6

23 51 67 46 54 32 19

13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

This process isn’t very efficient. . . a better one takes O(n) time

8 9

17 14 21 73 82 91

partitions the array in two

• The rest of the tree is given by

recursing left and right. . .

6

23 51 67 46 54 32 19

13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

This process isn’t very efficient. . . a better one takes O(n) time

8 9

17 14 21 73 82 91

partitions the array in two

• The rest of the tree is given by

recursing left and right. . . it’s not tricky but we don’t have time to cover it

6

23 51 67 46 54 32 19

13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

This process isn’t very efficient. . . a better one takes O(n) time

8 9

17 14 21 73 82 91

partitions the array in two

• The rest of the tree is given by

recursing left and right. . . it’s not tricky but we don’t have time to cover it Key Fact: The LCA in TA equals the RMQ in A

6

23 51 67 46 54 32 19

13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 i = 3 j = 7

RMQ(3, 7) = 6

9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

This process isn’t very efficient. . . a better one takes O(n) time

8 9

17 14 21 73 82 91

partitions the array in two

• The rest of the tree is given by

recursing left and right. . . it’s not tricky but we don’t have time to cover it Key Fact: The LCA in TA equals the RMQ in A

6

23 51 32 67 46 54 73 32 19

13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

This process isn’t very efficient. . . a better one takes O(n) time

8 9

17 14 21 73 82 91

partitions the array in two

• The rest of the tree is given by

recursing left and right. . . it’s not tricky but we don’t have time to cover it Key Fact: The LCA in TA equals the RMQ in A

6

23 51 67 46 54 32 19

13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

This process isn’t very efficient. . . a better one takes O(n) time

8 9

17 14 21 73 82 91

partitions the array in two

• The rest of the tree is given by

recursing left and right. . . it’s not tricky but we don’t have time to cover it Key Fact: The LCA in TA equals the RMQ in A

6

23 51 67 46 54 32 19

9

91 54

i = 10 j = 15

13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

This process isn’t very efficient. . . a better one takes O(n) time

8 9

17 14 21 73 82 91

partitions the array in two

• The rest of the tree is given by

recursing left and right. . . it’s not tricky but we don’t have time to cover it Key Fact: The LCA in TA equals the RMQ in A This gives us O(n) space, O(n) prep. time and O(1) query time for the RMQ problem

6

23 51 67 46 54 32 19

9

91 54

i = 10 j = 15

13

Solving RMQs using LCAs A n

17 8 23 73 51 82 19 32 5 67 91 14 46 9 21 54

1 2 3 4 5 7 8 9 10 11 12 14 15

5

Build the Cartesian tree, TA of the array A:

• The root is the smallest value
• The selected location

This process isn’t very efficient. . . a better one takes O(n) time

8 9

17 14 21 73 82 91

partitions the array in two

• The rest of the tree is given by

recursing left and right. . . it’s not tricky but we don’t have time to cover it Key Fact: The LCA in TA equals the RMQ in A This gives us O(n) space, O(n) prep. time and O(1) query time for the RMQ problem

6

23 51 67 46 54 32 19

9

91 54

i = 10 j = 15

13

by using the solution to LCA :)

Summary

We have seen an O(n) space, O(n) prep. time and O(1) query time solution

for the Range Minimum Query problem

which uses the solution to LCA

We have seen an O(n) space, O(n) prep. time and O(1) query time solution

for the Lowest Common Ancestor problem

which uses the solution to ±1RMQ

We have seen an O(n) space, O(n) prep. time and O(1) query time solution

for the ±1 Range Minimum Query problem

which improves solution 3 for RMQ from last lecture (but only for ±1 inputs) (which works for all inputs)