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]
return dist;
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=
Iind=
10
4
654
789
Finding Maximum Distance
9 7 3 1
I=9 7 3 1
Iind=0 1 2 3
J=
Iind=
The furthest i for J[0]
10
4
654
789
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
J=
Iind=
The furthest i for J[1]
10
4
654
789
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
J=
Iind=
The furthest i for J[2]
10
4
654
789
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
J=
Iind=
The furthest i for J[3]
10
4
654
789
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)