Question

This is an interview question:

Find the largest possible difference in an array of integers, such that the smaller integer occurs earlier in the array.

Constraint:
Numbers are not unique.
The range is Integer range of java. (or any other language)

Example:

input 1: {1, 100, 2, 105, -10, 30, 100}

The largest difference is between -10 and 100 -> 110 (here -10 is at the 5th index and 100 is at 7th index)

input 2: {1, 100, 2, 105, -10, 30, 80}

The largest difference is between 1 and 105 -> 104 (here 1 is at the 1st index and 105 is at 4th index)

Possible Solution:

One approach is check for all possible differences and keep a track of the biggest difference found till now O(n^2) complexity.

can this be done in better than O(n^2) time?

Answer 1

Start from the last element and move backwards. keep in memory the largest element occurred till now.
for each element subtract from the max and store at the respective position.

Also, you can keep an element to store the max difference and give the output straight away.
O(n) time, O(1) space.

int max = INT_MIN;
int maxdiff = 0;

for (i = sizeof(arr) / sizeof(int) - 1; i >= 0; i--) {
  if (max < arr[i]) {
    max = arr[i];
  }
  int diff = max - arr[i];
  if (maxdiff < diff) {
    maxdiff = diff;
  }
}

print maxdiff;