Minimum Replacements To Sort The Array - Leetcode Solution

Problem Statement:

Given an integer array, you need to find the minimum number of replacement operations required to sort the array in non-decreasing order.

A replacement operation is performed by selecting any element of the array and replacing it with some other value.

Solution:

Approach:

The problem can be solved using a Greedy approach. We need to replace the maximum number of values to achieve the sorted order. We traverse through the array, and for every element, we check if it is smaller than the previous element. If it is, we increment the count, indicating that we need to replace the current element. We then replace the current element with the previous one, as replacing the current element with any other value would either increase the number of replacements required or would not contribute any benefit.

Steps:

1. Initialize a variable to store the count of replacements required to 0.
2. Traverse the array and for every element, perform the following steps:
   1. Compare the current element with the previous element.
   2. If the current element is smaller than the previous element, increment the count of replacements required.
   3. Replace the current element with the previous element by assigning the previous element value to the current element.
3. Return the count of replacements required.

Code:

class Solution {
public:
    int minReplacements(vector<int>& nums) {
        int count = 0;
        for(int i=1; i<nums.size(); i++) {
            if(nums[i] < nums[i-1]) {
                count++;
                nums[i] = nums[i-1];
            }
        }
        return count;
    }
};

Time Complexity: O(n)

Space Complexity: O(1)

Explanation:

We traverse through the array only once, so the time complexity is O(n). We don't use any extra space, except for the loop variable, so the space complexity is O(1).

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