Lexicographically smallest array after at-most K consecutive swaps

Given an array arr[], find the lexicographically smallest array that can be obtained after performing at maximum of k consecutive swaps. 
Examples :

Input: arr[] = {7, 6, 9, 2, 1}
        k = 3
Output: arr[] = {2, 7, 6, 9, 1}
Explanation: Array is: 7, 6, 9, 2, 1
Swap 1:   7, 6, 2, 9, 1
Swap 2:   7, 2, 6, 9, 1
Swap 3:   2, 7, 6, 9, 1
So Our final array after k = 3 swaps : 
2, 7, 6, 9, 1

Input: arr[] = {7, 6, 9, 2, 1}
        k = 1
Output: arr[] = {6, 7, 9, 2, 1}

We strongly recommend that you click here and practice it, before moving on to the solution.

Naive approach is to generate all the permutation of array and pick the smallest one which satisfy the condition of at-most k swaps. Time complexity of this approach is Ω(n!), which will definitely time out for large value of n.
An Efficient approach is to think greedily. We first pick the smallest element from array a₁, a₂, a₃…(a_k or a_n) [We consider a_k when k is smaller, else n]. We place the smallest element to the a₀ position after shifting all these elements by 1 position right. We subtract number of swaps (number of swaps is number of shifts minus 1) from k. If still we are left with k > 0 then we apply the same procedure from the very next starting position i.e., a₂, a₃,…(a_k or a_n) and then place it to the a₁ position. So we keep applying the same process until k becomes 0. 

C++

// C++ program to find lexicographically minimum
// value after k swaps.
#include<bits/stdc++.h>
using namespace std ;

// Modifies arr[0..n-1] to lexicographically smallest
// with k swaps.
void minimizeWithKSwaps(int arr[], int n, int k)
{
    for (int i = 0; i<n-1 && k>0; ++i)
    {
        // Set the position where we want
        // to put the smallest integer
        int pos = i;
        for (int j = i+1; j<n ; ++j)
        {
            // If we exceed the Max swaps
            // then terminate the loop
            if (j-i > k)
                break;

            // Find the minimum value from i+1 to
            // max k or n
            if (arr[j] < arr[pos])
                pos = j;
        }

        // Swap the elements from Minimum position
        // we found till now to the i index
        for (int j = pos; j>i; --j)
            swap(arr[j], arr[j-1]);

        // Set the final value after swapping pos-i
        // elements
        k -=  pos-i;
    }
}

// Driver code
int main()
{
    int arr[] = {7, 6, 9, 2, 1};
    int n = sizeof(arr)/sizeof(arr[0]);
    int k = 3;

    minimizeWithKSwaps(arr, n, k);

    //Print the final Array
    for (int i=0; i<n; ++i)
        cout << arr[i] <<" ";
}

Java

// Java program to find lexicographically minimum
// value after k swaps.
import java.io.*;
class GFG {

    // Modifies arr[0..n-1] to lexicographically
    // smallest with k swaps.
    static void minimizeWithKSwaps(int arr[], int n, int k)
    {
        for (int i = 0; i < n-1 && k > 0; ++i)
        {

            // Set the position where we want
            // to put the smallest integer
            int pos = i;
            for (int j = i+1; j < n ; ++j)
            {

                // If we exceed the Max swaps
                // then terminate the loop
                if (j - i > k)
                    break;

                // Find the minimum value from i+1 to
                // max k or n
                if (arr[j] < arr[pos])
                    pos = j;
            }

            // Swap the elements from Minimum position
            // we found till now to the i index
            int temp;

            for (int j = pos; j>i; --j)
            {
                temp=arr[j];
                arr[j]=arr[j-1];
                arr[j-1]=temp;
            }

            // Set the final value after swapping pos-i
            // elements
            k -= pos-i;
        }
    }

    // Driver method
    public static void main(String[] args)
    {

        int arr[] = {7, 6, 9, 2, 1};
        int n = arr.length;
        int k = 3;

        minimizeWithKSwaps(arr, n, k);

        //Print the final Array
        for (int i=0; i<n; ++i)
            System.out.print(arr[i] +" ");
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python program to find lexicographically minimum
# value after k swaps.
def minimizeWithKSwaps(arr, n, k):

    for i in range(n-1):

        # Set the position where we want
    # to put the smallest integer
        pos = i
        for j in range(i+1, n):

            # If we exceed the Max swaps
        # then terminate the loop
            if (j-i > k):
                break

            # Find the minimum value from i+1 to
            # max (k or n)
            if (arr[j] < arr[pos]):
                pos = j

        # Swap the elements from Minimum position
        # we found till now to the i index
        for j in range(pos, i, -1):
            arr[j],arr[j-1] = arr[j-1], arr[j]

        # Set the final value after swapping pos-i
        # elements
        k -= pos - i


# Driver Code
n, k = 5, 3
arr = [7, 6, 9, 2, 1]
minimizeWithKSwaps(arr, n, k)

# Print the final Array
for i in range(n):
    print(arr[i], end = " ")

C#

// C# program to find lexicographically
// minimum value after k swaps.
using System;

class GFG {

    // Modifies arr[0..n-1] to lexicographically
    // smallest with k swaps.
    static void minimizeWithKSwaps(int []arr, int n,
                                              int k)
    {
        for (int i = 0; i < n-1 && k > 0; ++i)
        {
            // Set the position where we want
            // to put the smallest integer
            int pos = i;
            for (int j = i+1; j < n ; ++j)
            {
                // If we exceed the Max swaps
                // then terminate the loop
                if (j - i > k)
                    break;

                // Find the minimum value from
                // i + 1 to max k or n
                if (arr[j] < arr[pos])
                    pos = j;
            }

            // Swap the elements from Minimum position
            // we found till now to the i index
            int temp;

            for (int j = pos; j>i; --j)
            {
                temp=arr[j];
                arr[j]=arr[j-1];
                arr[j-1]=temp;
            }

            // Set the final value after
            // swapping pos-i elements
            k -= pos-i;
        }
    }

    // Driver method
    public static void Main()
    {
        int []arr = {7, 6, 9, 2, 1};
        int n = arr.Length;
        int k = 3;

        // Function calling
        minimizeWithKSwaps(arr, n, k);

        // Print the final Array
        for (int i=0; i<n; ++i)
          Console.Write(arr[i] +" ");
    }
}

// This code is contributed by nitin mittal.

php

<?php
// php program to find lexicographically minimum
// value after k swaps.

// Modifies arr[0..n-1] to lexicographically
// smallest with k swaps.
function minimizeWithKSwaps($arr, $n, $k)
{
    for ($i = 0; $i < $n-1 && $k > 0; ++$i)
    {
        // Set the position where we want
        // to put the smallest integer
        $pos = $i;
        for ($j = $i+1; $j < $n ; ++$j)
        {
            // If we exceed the Max swaps
            // then terminate the loop
            if ($j-$i > $k)
                break;

            // Find the minimum value from
            // i+1 to max k or n
            if ($arr[$j] < $arr[$pos])
                $pos = $j;
        }

        // Swap the elements from Minimum
        // position we found till now to
        // the i index
        for ($j = $pos; $j > $i; --$j)
        {
            $temp = $arr[$j];
            $arr[$j] = $arr[$j-1];
            $arr[$j-1] = $temp;
        }

        // Set the final value after
        // swapping pos-i elements
        $k -= $pos-$i;
    }

    //Print the final Array
    for ($i = 0; $i < $n; ++$i)
        echo $arr[$i] . " ";
}

// Driver code
$arr = array(7, 6, 9, 2, 1);
$n = count($arr);
$k = 3;

minimizeWithKSwaps($arr, $n, $k);

// This code is contributed by Sam007
?>

Javascript

<script>
    // Javascript program to find lexicographically
    // minimum value after k swaps.

    // Modifies arr[0..n-1] to lexicographically
    // smallest with k swaps.
    function minimizeWithKSwaps(arr, n, k)
    {
        for (let i = 0; i < n - 1 && k > 0; ++i)
        {

            // Set the position where we want
            // to put the smallest integer
            let pos = i;
            for (let j = i+1; j < n ; ++j)
            {
                // If we exceed the Max swaps
                // then terminate the loop
                if (j - i > k)
                    break;

                // Find the minimum value from
                // i + 1 to max k or n
                if (arr[j] < arr[pos])
                    pos = j;
            }

            // Swap the elements from Minimum position
            // we found till now to the i index
            let temp;

            for (let j = pos; j > i; --j)
            {
                temp = arr[j];
                arr[j] = arr[j - 1];
                arr[j - 1] = temp;
            }

            // Set the final value after
            // swapping pos-i elements
            k -= pos - i;
        }
    }

    let arr = [7, 6, 9, 2, 1];
    let n = arr.length;
    let k = 3;

    // Function calling
    minimizeWithKSwaps(arr, n, k);

    // Print the final Array
    document.write("Output: ");
    for (let i = 0; i < n; ++i)
      document.write(arr[i] + " ");

    // This code is contributed by divyesh072019.
</script>

Output

2 7 6 9 1 

Time complexity: O(N²) 
Auxiliary space: O(1)
Reference:
http://stackoverflow.com/questions/25539423/finding-minimal-lexicographical-array
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