Distance maximizing in array
Download
1 / 13

Distance Maximizing in Array - PowerPoint PPT Presentation


  • 71 Views
  • Uploaded on

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). Brute Force. For(dist=n-1; dist>=0; dist--) for( i =0; i <n-dist; i ++) if(A[ i+dist ]>A[ i ]

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Distance Maximizing in Array' - sovann


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Problem
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)


Brute force
Brute Force

For(dist=n-1; dist>=0; dist--)

for(i=0; i<n-dist; i++)

if(A[i+dist]>A[i]

return dist;


O n algorithm
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 i s
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 j s
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
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 distance1
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 distance2
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 distance3
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 distance4
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
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
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)


ad