Maximum difference between two elements where larger element appears after the smaller element

Objective: Given an array A[], write an algorithm to find Maximum difference between two elements where larger element appears after the smaller element or in other words find A[i] and A[j] such that A[j]-A[i] is maximum where j > i.

Example:

int [] A = { 2, 5, 1, 7, 3, 9, 5};
Maximum Difference between two elements A[i] and A[j] and where j > i: 8

int [] A = { 22,2, 12, 5, 4, 7, 3, 19, 5};
Maximum Difference between two elements A[i] and A[j] and where j > i: 17

Approach 1:

Naive: This problem can be easily solved using two nested loops. Take each element at a time and compare it with all the other elements and keep the track of the maximum difference elements where larger element appears after the smaller element.

Time complexity is O(N^2).

Code:

Approach 2:

Divide and Conquer

1. We need to find the two elements A[i] and A[j] so that A[j] – A[i] is maximum and j > i
2. Divide the input array into 2 parts, left Half and right half.
3. We have divided the problem in 2 parts. Now we can conclude that for answer-
   1. Both indexes i and j are in the left half of the array.
   2. Both indexes i and j are in the right half of the array.
   3. Index i is in left half and index j is in right half of the array.
4. Solve the above 3 sub problems recursively and final answer will the maximum of these 3 sub problems.
5. This solution is very much similar to Merge sort

Time complexity is O(nlogn).

Code:

Approach 3:

Dynamic Programming:

1. Traverse the array from right to left.
2. Maintain 2 variables – maxDifference (this will be our final answer), and max_so_far(maximum element we encounter while traversing the array)
3. During iteration, every element has 2 choices
   1. Either current element is the maximum element among elements which are iterated, if yes then max_so_far = current element.
   2. OR current element is less than the max_so_far. If yes then update the maxDifference = max_so_far – current element.
4. Once the iteration is done, return maxDifference.

Time complexity: O(n), Space complexity: O(1)

Code:

Output:

Maximum Difference between two elements A[i] and A[j] and where j > i: 8