**Question:**There is an integer array consisting positive and negative integers. Find maximum positive difference S defined as:

S = a[i] - a[j] where i>j

and

S > 0

**Answer:**

An inefficient solution to this problem is that we can traverse the array from the start run a loop for each element and then check for all other number which are coming later in array traversal. But this will be a very inefficient solution and will have runtime O(n^2).

A better solution is that we can calculate the max difference (glbdiff) for the subarray traversed till now and whenever we find a better difference value we can replace glbdiff by this value (locdiff). This is a dynamic approach and complete in O(1) space and O(n) time.

**Code:**

#include<iostream>

using namespace std;

int main()

{

int locdiff, loci, locj, glbdiff, glbi, glbj, i,n;

int in[100]={0,};

glbi= glbj=glbdiff= 0;

cin >> n;

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

cin >> in[i];

loci = 0; locj=1;

locdiff = in[locj] - in[loci];

for (i=2; i<n; i++)

{

if (in[i] < in[loci])

{

loci = i;

locj= i+1;

locdiff = in[locj] - in[loci];

if(locdiff > glbdiff)

{

glbi = loci;

glbj = locj;

glbdiff = locdiff;

}

}

else

{

locj = i;

locdiff = in[locj] - in[loci];

if(locdiff > glbdiff)

{

glbi = loci;

glbj = locj;

glbdiff = locdiff;

}

}

}

cout << "i = " << glbi << " j = " << glbj << " diff = " << glbdiff << endl;

}

**UPDATE:**A cleaner version of above code is as below:

void MaxDiff(int in[], int sz, int &start, int &end) {

int min = 0;

int maxDiff = 0;

start = end = 0;

for (int i = 0; i < sz; i++) {

if (in[i] < in[min])

min = i;

int diff = in[i] - in[min];

if (diff > maxDiff) {

start = min;

end = i;

maxDiff = diff;

}

}

}

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