## 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 <= 10`^{4}

`1 <= arr[i] <= 10`^{5}

### 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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