// The result will always include the smallest value of A[i] where i is
less than j,
// so just keep track of that minimum value as you scan through the array.
int findPair(int *A, int N)
{
min = A[0];
max = 0;
for(j = 1; j < N; ++j)
{
if ((A[j] - min) > max) max = A[j] - min;
if (A[j] < min) A[j] = min;
}
return max;
}
On Friday, April 18, 2014 4:49:26 PM UTC-4, TUSHAR wrote:
>
> Hi All,
>
> Can anyone help me solving this problem ?
>
> Given a non-empty array A consisting of N integers.
>
> Find Max(j-i), s.t 0<=i<=j<N and A[i]<=A[j]
>
> Here (i, j) is a Monotonic Pair, satisfying the above condition.
>
> Space Complexity : O(n) Time Complexity: O(n)
>
> --
> Regards...
>
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