Related Articles
Lexicographically smallest array after at-most K consecutive swaps
• Difficulty Level : Medium
• Last Updated : 17 Aug, 2018

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…(ak or an) [We consider ak 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,…(ak or an) 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.#includeusing 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; i0; ++i)    {        // Set the position where we want        // to put the smallest integer        int pos = i;        for (int j = i+1; j 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 codeint 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

## Java

 // Java program to find lexicographically minimum// value after k swaps.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

## Python

 # 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 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 Coden, k = 5, 3arr = [7, 6, 9, 2, 1]minimizeWithKSwaps(arr, n, k)  # Print the final Arrayfor 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

## php

 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?>
Output: 2 7 6 9 1

Time complexity: O(N2)
Auxiliary space: O(1)

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.