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### Distance Maximizing in Array

Yang Liu

Problem

- Given an array of integers, find the maximum of j-i subject to that A[i]<A[j]

Example

Input: 5 2 3 1 7

Output: 4(=4-0)

O(n) Algorithm

- Observation

For input “5 2 3 1 7”,

- 3 never could be i which maximize j-I since 2 is before 3.
- 1 never could be j since 1 is before 7
- Idea:
- Use lArray to store possible is.
- Use rArray to store possible js.

Possible is

k=0

I[k]=A[0]

Iind[k]=0;

For(i=1;i<n;i++)

if(A[i]<I[k])

k=k+1;

I[k]=A[i]

Iind[k]=i;

A=9 7 3 1 10 2 8 6 5 4

I=9 7 3 1

Iind=0 1 2 3

Possible js

k=0

J[k]=A[n-1]

Jind[k]=n-1;

For(i=n-2;i>=0;i--)

if(A[i]>J[k])

k=k+1;

J[k]=A[i]

Jind[k]=i;

A=9 7 3 1 10 2 8 6 5 4

J= 10 6 5 4

Iind= 4 7 8 9

Finding Maximum Distance

9 7 3 1

I=9 7 3 1

Iind=0 1 2 3

The furthest i for J[0]

J= 10 6 5 4

Iind= 4 7 8 9

10 6 5 4

dist: 4(=4-0)

Finding Maximum Distance

9 7 3 1

I=9 7 3 1

Iind=0 1 2 3

The furthest i for J[1]

J= 10 6 5 4

Iind= 4 7 8 9

10 6 5 4

dist: 5(=7-2)

Finding Maximum Distance

9 7 3 1

I=9 7 3 1

Iind=0 1 2 3

The furthest i for J[2]

J= 10 6 5 4

Iind= 4 7 8 9

10 6 5 4

dist: 6(=8-2)

Finding Maximum Distance

9 7 3 1

I=9 7 3 1

Iind=0 1 2 3

The furthest i for J[3]

J= 10 6 5 4

Iind= 4 7 8 9

10 6 5 4

dist: 7(=9-2)

Finding Maximum Distance

- Queue Q to store Iind
- Stack S to store Jind

maxDist=0;

While(S is not empty)

jInd=S.pop();

while(Q is not empty)

iInd=Q.front();

if(A[jInd]>A[iInd])

maxDist=(jInd-iInd>maxDist)?jInd-iInd:maxDist;

break;

else

Q.deque();

Exercise 1

- You have an array A such that A[i] is the price of a stock on day i.

You are only permitted to buy one share of the stock and sell one share of the stock.

Design an algorithm to find the best times to buy and sell.

Exercise 2

Given an array A, find maximum A[i]-A[j] where i<j

Example

Input: A: 2 9 3 1 4

Output: 8(=9-1)

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