# C Exercises: Find minimum number of swaps required to gather all elements less than or equals to k

## C Array: Exercise-105 with Solution

Write a program in C to find the minimum number of swaps required to gather all elements less than or equal to k.

**Sample Solution:**

**C Code:**

```
#include<stdio.h>
// Function to find the minimum number of swaps required to bring elements less than or equal to 'k' together
int minSwap(int *arr1, int n, int k)
{
int ctr = 0;
// Count elements less than or equal to 'k' in the array
for (int i = 0; i < n; ++i)
if (arr1[i] <= k)
++ctr;
int gelem = 0;
// Count elements greater than 'k' within the first 'ctr' elements
for (int i = 0; i < ctr; ++i)
if (arr1[i] > k)
++gelem;
int noswp = gelem;
// Move the window of size 'ctr' through the array and count the number of elements greater than 'k' to minimize swaps
for (int i = 0, j = ctr; j < n; ++i, ++j)
{
if (arr1[i] > k)
{
--gelem;
}
if (arr1[j] > k)
{
++gelem;
}
}
// Update the minimum swaps required
if(noswp > gelem)
{
noswp = gelem;
}
if(noswp < gelem)
{
noswp = noswp;
}
return noswp;
}
// Main function
int main()
{
int arr1[] = {2, 7, 9, 5, 8, 7, 4}; // Given array
int n = sizeof(arr1) / sizeof(arr1[0]); // Calculate array size
int k = 7; // Given 'k' value for comparison
int i = 0;
// Display the given array
printf("The given array is: \n");
for(i = 0; i < n; i++)
{
printf("%d ", arr1[i]);
}
printf("\n");
// Calculate and display the minimum swaps required to bring elements less than or equal to 'k' together
printf("The minimum swap required is: %d", minSwap(arr1, n, k));
return 0;
}
```

Sample Output:

The given array is: 2 7 9 5 8 7 4 The minimum swap required is: 2

**Visual Presentation:**

**Flowchart 1:**

**Flowchart 2:**

**C Programming Code Editor:**

**Previous:** Write a program in C to rearrange an array such that even index elements are smaller and odd index elements are greater than their next.

**Next:** Write a program in C to convert the array in such a way that double its value and replace the next number with 0 if current and next element are same and rearrange the array such that all 0's shifted to the end.

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