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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
#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 −
Output
Minimum swaps = 1
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