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LeetCode 2422: Merge Operations to Turn Array Into a Palindrome

ArrayTwo PointersGreedy

LeetCode 2422 Solution Explanation

Explanation:

To solve this problem, we can use a two-pointer approach. We start by comparing elements from both ends of the array. If the elements at both ends are equal, we move the pointers towards the center. If the elements are not equal, we need to perform a merge operation to make them equal. We keep track of the number of merge operations needed to turn the array into a palindrome.

Here are the steps for the algorithm:

  1. Initialize two pointers, left pointing to the start of the array, and right pointing to the end of the array.
  2. While left pointer is less than or equal to the right pointer:
    • If arr[left] == arr[right], increment left and decrement right.
    • If arr[left] < arr[right], merge arr[left] with arr[left+1] by incrementing left and increment the merge operations count.
    • If arr[left] > arr[right], merge arr[right] with arr[right-1] by decrementing right and increment the merge operations count.
  3. Return the total number of merge operations.

The time complexity of this algorithm is O(n) where n is the length of the input array. The space complexity is O(1) as we are not using any extra space.

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LeetCode 2422 Solutions in Java, C++, Python

class Solution {
    public int minMergeOperations(int[] arr) {
        int left = 0, right = arr.length - 1;
        int merges = 0;
        
        while (left <= right) {
            if (arr[left] == arr[right]) {
                left++;
                right--;
            } else if (arr[left] < arr[right]) {
                arr[left + 1] += arr[left];
                left++;
                merges++;
            } else {
                arr[right - 1] += arr[right];
                right--;
                merges++;
            }
        }
        
        return merges;
    }
}

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