LeetCode 3507: Minimum Pair Removal to Sort Array I Solution

Master LeetCode problem 3507 (Minimum Pair Removal to Sort Array I), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation

To solve this problem, we can iterate through the array and keep track of the number of operations needed to make the array non-decreasing. At each step, we compare the current element with the next element. If the next element is smaller than the current element, we increment the number of operations by the difference between the current element and the next element. We then update the next element to be the sum of the current and next elements.

We repeat this process until we have iterated through the entire array, and the final count of operations will be the minimum number of operations needed to make the array non-decreasing.

Algorithm:

  1. Initialize a variable operations to 0.
  2. Iterate through the array starting from the second element.
  3. If the current element is greater than the next element, increment operations by the difference between the current and next elements. Update the next element to be the sum of the current and next elements.
  4. Repeat step 3 for all elements in the array.
  5. Return the final value of operations.

Time Complexity: O(n) where n is the number of elements in the array. Space Complexity: O(1)

Solution Code

class Solution {
    public int minOperations(int[] nums) {
        int operations = 0;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] < nums[i - 1]) {
                operations += nums[i - 1] - nums[i];
                nums[i] = nums[i - 1];
            }
        }
        return operations;
    }
}

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