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# Replace Elements with Greatest Element on Right Side LeetCode Solution

## Problem – Replace Elements with Greatest Element on Right Side LeetCode Solution

Given an array `arr`, replace every element in that array with the greatest element among the elements to its right, and replace the last element with `-1`.

After doing so, return the array.

Example 1:

``````Input: arr = [17,18,5,4,6,1]
Output: [18,6,6,6,1,-1]
Explanation:
- index 0 --> the greatest element to the right of index 0 is index 1 (18).
- index 1 --> the greatest element to the right of index 1 is index 4 (6).
- index 2 --> the greatest element to the right of index 2 is index 4 (6).
- index 3 --> the greatest element to the right of index 3 is index 4 (6).
- index 4 --> the greatest element to the right of index 4 is index 5 (1).
- index 5 --> there are no elements to the right of index 5, so we put -1.
``````

Example 2:

``````Input: arr = [400]
Output: [-1]
Explanation: There are no elements to the right of index 0.
``````

Constraints:

• `1 <= arr.length <= 104`
• `1 <= arr[i] <= 105`

### Replace Elements with Greatest Element on Right Side LeetCode Solution in Java

``````    public int[] replaceElements(int[] A) {
for (int i = A.length - 1, mx = -1; i >= 0; --i)
mx = Math.max(A[i], A[i] = mx);
return A;
}
``````

### Replace Elements with Greatest Element on Right Side LeetCode Solution in C++

``````    vector<int> replaceElements(vector<int>& A, int mx = -1) {
for (int i = A.size() - 1; i >= 0; --i)
mx = max(mx, exchange(A[i], mx));
return A;
}
``````

### Replace Elements with Greatest Element on Right Side LeetCode Solution in Python

``````    def replaceElements(self, A, mx = -1):
for i in xrange(len(A) - 1, -1, -1):
A[i], mx = mx, max(mx, A[i])
return A
``````
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