# 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:

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 public class MaxDifferenceBruteForce { public static int maxDifference(int [] A){ int maxDiff = –1; for (int i = 0; i A[i] && (A[j]–A[i]>maxDiff)) maxDiff = A[j]–A[i]; } } return maxDiff; } public static void main(String[] args) { int [] A = { 2, 5, 1, 7, 3, 4, 9, 4, 5}; System.out.println("Maximum Difference between two elements A[i] and A[j] and where j > i: " + maxDifference(A)); } }

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:

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 public class MaxDifferenceDandC { public int maxDifference(int [] A, int start, int end){ if(start>=end){ return –1; } int mid = start + (end–start)/2; int leftDiff = maxDifference(A,start,mid); int rightDiff = maxDifference(A,mid+1,end); int minLeft = getMin(A, start, mid); int maxRight = getMax(A, mid, end); int centerDiff = maxRight – minLeft; return Math.max(centerDiff, Math.max(leftDiff,rightDiff)); } public int getMin(int [] A, int i, int j){ int min = A[i]; for (int k = i+1; k <=j ; k++) { if(A[k]max) max = A[k]; } return max; } public static void main(String[] args) { MaxDifferenceDandC m = new MaxDifferenceDandC(); int [] A = { 2, 5, 1, 7, 3, 4, 9, 4, 5}; int start = 0; int end = A.length–1; System.out.println("Maximum Difference between two elements A[i] and A[j] and where j > i: " + m.maxDifference(A, start, end)); } }

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:

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 public class MaxDifferenceDynamicProgramming { public static int maxDifference(int [] A){ int size = A.length; int maxDiff = –1; int max_so_far = A[size–1]; //assign the last element for (int i = size – 2 ; i >0 ; i—) { if(max_so_farA[i]) maxDiff = Math.max(maxDiff,max_so_far–A[i]); } return maxDiff; } public static void main(String[] args) { int [] A = { 12, 5, 1, 7, 3, 9, 5}; System.out.println("Maximum Difference between two elements A[i] and A[j] and where j > i: " + maxDifference(A)); } }

Output:

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