1 / 13

Distance Maximizing in Array

Explore an efficient algorithm to find the maximum of j - i for an integer array where A[i] < A[j]. For instance, given the input array [5, 2, 3, 1, 7], the goal is to compute the maximum distance, which equals 4. Using an O(n) approach, the solution leverages two auxiliary arrays to track potential indices for j and i, along with a queue and a stack to efficiently calculate the maximum distance. This method simplifies the brute-force approach while ensuring optimal performance.

sovann
Télécharger la présentation

Distance Maximizing in Array

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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Distance Maximizing in Array Yang Liu

  2. 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)

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

  4. 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.

  5. 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

  6. 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

  7. 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)

  8. 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)

  9. 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)

  10. 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)

  11. 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();

  12. 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.

  13. 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)

More Related