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

Difficulty Level : Medium
Last Updated : 19 Jul, 2018

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.

Input:  arr[] = {2, 1, 5, 6, 3}, k = 3
Output: 1

Explanation: 
To bring elements 2, 1, 3 together, swap 
element '5' with '3' such that final array
will be-
arr[] = {2, 1, 3, 6, 5}

Input:  arr[] = {2, 7, 9, 5, 8, 7, 4}, k = 5
Output: 2

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A simple solution is to first count all elements less than or equals to k(say ‘good’). Now traverse for every sub-array and swap those elements whose value is greater than k. Time complexity of this approach is O(n²)

A simple approach is to use two pointer technique and sliding window.

Find count of 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 ‘bad’.

Repeat step 2, for every window of length ‘cnt’ and take minimum of count ‘bad’ among them. This will be the final answer.

C++

// C++ program to find minimum swaps required
// to club all elements less than or equals
// to k together
#include <iostream>
using namespace std;
  
// Utility function to find minimum swaps
// required to club all elements less than
// or equals to k together
int minSwap(int *arr, int n, int k) {
      
    // Find count of elements which are
    // less than equals to k
    int count = 0;
    for (int i = 0; i < n; ++i)
        if (arr[i] <= k)
            ++count;
      
    // Find unwanted elements in current
    // window of size 'count'
    int bad = 0;
    for (int i = 0; i < count; ++i)
        if (arr[i] > k)
            ++bad;
      
    // Initialize answer with 'bad' value of
    // current window
    int ans = bad;
    for (int i = 0, j = count; j < n; ++i, ++j) {
          
        // Decrement count of previous window
        if (arr[i] > k)
            --bad;
          
        // Increment count of current window
        if (arr[j] > k)
            ++bad;
          
        // Update ans if count of 'bad'
        // is less in current window
        ans = min(ans, bad);
    }
    return ans;
}
  
// Driver code
int main() {
      
    int arr[] = {2, 1, 5, 6, 3};
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    cout << minSwap(arr, n, k) << "\n";
      
    int arr1[] = {2, 7, 9, 5, 8, 7, 4};
    n = sizeof(arr1) / sizeof(arr1[0]);
    k = 5;
    cout << minSwap(arr1, n, k);
    return 0;
}

Java

// Java program to find minimum 
// swaps required to club all
// elements less than or equals
// to k together
import java.lang.*;
  
class GFG {
      
// Utility function to find minimum swaps
// required to club all elements less than
// or equals to k together
static int minSwap(int arr[], int n, int k) {
  
    // Find count of elements which are
    // less than equals to k
    int count = 0;
    for (int i = 0; i < n; ++i)
    if (arr[i] <= k)
        ++count;
  
    // Find unwanted elements in current
    // window of size 'count'
    int bad = 0;
    for (int i = 0; i < count; ++i)
    if (arr[i] > k)
        ++bad;
  
    // Initialize answer with 'bad' value of
    // current window
    int ans = bad;
    for (int i = 0, j = count; j < n; ++i, ++j) {
  
    // Decrement count of previous window
    if (arr[i] > k)
        --bad;
  
    // Increment count of current window
    if (arr[j] > k)
        ++bad;
  
    // Update ans if count of 'bad'
    // is less in current window
    ans = Math.min(ans, bad);
    }
    return ans;
}
  
// Driver code
public static void main(String[] args) 
{
    int arr[] = {2, 1, 5, 6, 3};
    int n = arr.length;
    int k = 3;
    System.out.print(minSwap(arr, n, k) + "\n");
  
    int arr1[] = {2, 7, 9, 5, 8, 7, 4};
    n = arr1.length;
    k = 5;
    System.out.print(minSwap(arr1, n, k));
}
}
  
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find 
# minimum swaps required
# to club all elements less
# than or equals to k together
  
# Utility function to find 
# minimum swaps required to 
# club all elements less than
# or equals to k together
def minSwap(arr, n, k) :
      
    # Find count of elements
    # which are less than 
    # equals to k
    count = 0
    for i in range(0, n) :
        if (arr[i] <= k) :
            count = count + 1
      
    # Find unwanted elements 
    # in current window of 
    # size 'count'
    bad = 0
    for i in range(0, count) :
        if (arr[i] > k) :
            bad = bad + 1
      
    # Initialize answer with 
    # 'bad' value of current
    # window
    ans = bad
    j = count
    for i in range(0, n) :
          
        if(j == n) :
            break
              
        # Decrement count of
        # previous window
        if (arr[i] > k) :
            bad = bad - 1
          
        # Increment count of 
        # current window
        if (arr[j] > k) :
            bad = bad + 1
          
        # Update ans if count 
        # of 'bad' is less in
        # current window
        ans = min(ans, bad)
  
        j = j + 1
  
    return ans
  
# Driver code
arr = [2, 1, 5, 6, 3]
n = len(arr)
k = 3
print (minSwap(arr, n, k))
  
arr1 = [2, 7, 9, 5, 8, 7, 4]
n = len(arr1)
k = 5
print (minSwap(arr1, n, k))
  
# This code is contributed by 
# Manish Shaw(manishshaw1)

C#

// C# program to find minimum 
// swaps required to club all
// elements less than or equals
// to k together
using System;
  
class GFG {
      
    // Utility function to find minimum swaps
    // required to club all elements less than
    // or equals to k together
    static int minSwap(int []arr, int n, int k) {
      
        // Find count of elements which are
        // less than equals to k
        int count = 0;
        for (int i = 0; i < n; ++i)
        if (arr[i] <= k)
            ++count;
      
        // Find unwanted elements in current
        // window of size 'count'
        int bad = 0;
        for (int i = 0; i < count; ++i)
        if (arr[i] > k)
            ++bad;
      
        // Initialize answer with 'bad' value of
        // current window
        int ans = bad;
        for (int i = 0, j = count; j < n; ++i, ++j) {
      
            // Decrement count of previous window
            if (arr[i] > k)
                --bad;
          
            // Increment count of current window
            if (arr[j] > k)
                ++bad;
          
            // Update ans if count of 'bad'
            // is less in current window
            ans = Math.Min(ans, bad);
        }
        return ans;
    }
      
    // Driver code
    public static void Main() 
    {
        int []arr = {2, 1, 5, 6, 3};
        int n = arr.Length;
        int k = 3;
        Console.WriteLine(minSwap(arr, n, k));
      
        int []arr1 = {2, 7, 9, 5, 8, 7, 4};
        n = arr1.Length;
        k = 5;
        Console.WriteLine(minSwap(arr1, n, k));
    }
}
  
// This code is contributed by vt_m.

PHP

<?php
// PHP program to find
// minimum swaps required
// to club all elements 
// less than or equals
// to k together
  
// Utility function to 
// find minimum swaps
// required to club all 
// elements less than
// or equals to k together
function minSwap($arr, $n, $k)
{
      
    // Find count of elements
    // which are less than
    // equals to k
    $count = 0;
    for ($i = 0; $i < $n; ++$i)
        if ($arr[$i] <= $k)
            ++$count;
      
    // Find unwanted elements in current
    // window of size 'count'
    $bad = 0;
    for ($i = 0; $i < $count; ++$i)
        if ($arr[$i] > $k)
            ++$bad;
      
    // Initialize answer 
    // with 'bad' value of
    // current window
    $ans = $bad;
    for ($i = 0, $j = $count; $j < $n; 
                           ++$i, ++$j) 
    {
          
        // Decrement count of 
        // previous window
        if ($arr[$i] > $k)
            --$bad;
          
        // Increment count of
        // current window
        if ($arr[$j] > $k)
            ++$bad;
          
        // Update ans if count of 'bad'
        // is less in current window
        $ans = min($ans, $bad);
    }
    return $ans;
}
  
// Driver code
$arr = array(2, 1, 5, 6, 3);
$n = sizeof($arr);
$k = 3;
echo(minSwap($arr, $n, $k) . "\n");
      
$arr1 = array(2, 7, 9, 5, 8, 7, 4);
$n = sizeof($arr1);
$k = 5;
echo(minSwap($arr1, $n, $k));
  
// This code is contributed by Ajit.
?>

Output :

1
2

Time complexity: O(n)
Auxiliary space: O(1)

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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP