LeetCode 2366: Minimum Replacements to Sort the Array

Explanation:

To solve this problem, we need to find the minimum number of operations required to sort the given array in non-decreasing order. The key idea is to iterate through the array and check if the current element is greater than the next element. If it is, we need to replace the current element with two smaller elements that sum up to the current element. This will ensure that the array becomes sorted in non-decreasing order.

We can achieve this by keeping track of the number of operations required to sort the array. Whenever we encounter an element that needs to be replaced, we increment the operation count by 1. Finally, the total number of operations needed will be our answer.

Algorithm:

1. Initialize a variable operations to 0.
2. Iterate through the array from index 0 to n-2.
3. For each element, if it is greater than the next element, calculate the difference diff.
4. Increment the operations count by diff/2 + diff%2.
5. Return the total number of operations as the answer.

Time Complexity: The time complexity of this algorithm is O(n) where n is the number of elements in the input array.

Space Complexity: The space complexity of this algorithm is O(1) as we are using a constant amount of extra space.

:

class Solution {
    public int minReplacements(int[] nums) {
        int operations = 0;
        for (int i = 0; i < nums.length - 1; i++) {
            if (nums[i] > nums[i + 1]) {
                int diff = nums[i] - nums[i + 1];
                operations += diff / 2 + diff % 2;
            }
        }
        return operations;
    }
}