Minimum swaps required to bring all elements less than or equal to k together in C++

Problem statement

Given an array of n positive integers and a number k. Find the minimum number of swaps required to bring all the numbers less than or equal to k together.

Example

If input array is = {1, 5, 4, 7, 2, 10} and k = 6 then 1 swap is required i.e. swap element 7 with 2.

Algorithm

Count all elements which are less than or equals to ‘k’. Let’s say the count is ‘cnt’
Using two pointer technique for window of length ‘cnt’,each time keep track of how many elements in this range are greater than ‘k’. Let’s say the total count is ‘outOfRange’.
Repeat step 2, for every window of length ‘cnt’ and take minimum of count ‘outOfRange’ among them. This will be the final answer.

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int getMinSwaps(int *arr, int n, int k) {
   int cnt = 0;
   for (int i = 0; i < n; ++i) {
      if (arr[i] <= k) {
         ++cnt;
      }
   }
   int outOfRange = 0;
   for (int i = 0; i < cnt; ++i) {
      if (arr[i] > k) {
         ++outOfRange;
      }
   }
   int result = outOfRange;
   for (int i = 0, j = cnt; j < n; ++i, ++j) {
      if (arr[i] > k) {
         --outOfRange;
      }
      if (arr[j] > k) {
         ++outOfRange;
      }
      result = min(result, outOfRange);
   }
   return result;
}
int main() {
   int arr[] = {1, 5, 4, 7, 2, 10};
   int n = sizeof(arr) / sizeof(arr[0]);
   int k = 6;
   cout << "Minimum swaps = " << getMinSwaps(arr, n, k) <<
   endl;
   return 0;
}

When you compile and execute above program. It generates following output −