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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