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Largest lexicographic array with at-most K consecutive swaps

Last Updated : 09 Aug, 2022
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Given an array arr[], find the lexicographically largest array that can be obtained by performing at-most k consecutive swaps. 

Examples : 

Input : arr[] = {3, 5, 4, 1, 2}
        k = 3
Output : 5, 4, 3, 2, 1
Explanation :     Array given : 3 5 4 1 2
                After swap 1 : 5 3 4 1 2
                After swap 2 : 5 4 3 1 2
                After swap 3 : 5 4 3 2 1
Input : arr[] = {3, 5, 1, 2, 1}
        k = 3
Output : 5, 3, 2, 1, 1

Brute Force Approach : Generate all permutation of the array and then pick the one which satisfies the condition of at most K swaps. The time complexity of this approach is O(n!).

Optimized Approach : In this greedy approach, first find the largest element present in the array which is greater than(if the 1st position element is not the greatest) the 1st position and which can be placed at the 1st position with at-most K swaps. After finding that element, note its index. Then, swap elements of the array and update K value. Apply this procedure for other positions till k is non-zero or array becomes lexicographically largest.

Below is the implementation of above approach : 

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

// Function which modifies the array
void KSwapMaximum(int arr[], int n, int k)
{
    for (int i = 0; i < n - 1 && k > 0; ++i) {

        // Here, indexPosition is set where we
        // want to put the current largest integer
        int indexPosition = i;
        for (int j = i + 1; j < n; ++j) {

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

            // Find the maximum value from i+1 to
            // max k or n which will replace
            // arr[indexPosition]
            if (arr[j] > arr[indexPosition])
                indexPosition = j;
        }

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

        // Updates k after swapping indexPosition-i
        // elements
        k -= indexPosition - i;
    }
}

// Driver code
int main()
{
    int arr[] = { 3, 5, 4, 1, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;

    KSwapMaximum(arr, n, k);

    // Print the final Array
    for (int i = 0; i < n; ++i)
        cout << arr[i] << " ";
}
Java 
// Java program to find 
// lexicographically 
// maximum value after
// k swaps.
import java.io.*;

class GFG
{ 
    static void SwapInts(int array[], 
                         int position1, 
                         int position2)
    {
        // Swaps elements
        // in an array.
        
        // Copy the first 
        // position's element
        int temp = array[position1]; 
        
        // Assign to the 
        // second element
        array[position1] = array[position2];
        
        // Assign to the 
        // first element
        array[position2] = temp; 
    }
    
    // Function which 
    // modifies the array
    static void KSwapMaximum(int []arr, 
                             int n, int k)
    {
        for (int i = 0; 
                 i < n - 1 && k > 0; ++i) 
        {
    
            // Here, indexPosition
            // is set where we want to
            // put the current largest 
            // integer
            int indexPosition = i;
            for (int j = i + 1; j < n; ++j) 
            {
    
                // If we exceed the 
                // Max swaps then 
                // break the loop
                if (k <= j - i)
                    break;
    
                // Find the maximum value 
                // from i+1 to max k or n 
                // which will replace 
                // arr[indexPosition]
                if (arr[j] > arr[indexPosition])
                    indexPosition = j;
            }
    
            // Swap the elements from 
            // Maximum indexPosition
            // we found till now to 
            // the ith index
            for (int j = indexPosition; j > i; --j)
                SwapInts(arr, j, j - 1);
    
            // Updates k after swapping
            // indexPosition-i elements
            k -= indexPosition - i;
        }
    }
    
    // Driver code
    public static void main(String args[])
    {
        int []arr = { 3, 5, 4, 1, 2 };
        int n = arr.length;
        int k = 3;
    
        KSwapMaximum(arr, n, k);
    
        // Print the final Array
        for (int i = 0; i < n; ++i)
            System.out.print(arr[i] + " ");
    }
}

// This code is contributed by 
// Manish Shaw(manishshaw1)
Python3 
# Python program to find 
# lexicographically 
# maximum value after
# k swaps.

arr = [3, 5, 4, 1, 2]

# Function which
# modifies the array
def KSwapMaximum(n, k) :
    
    global arr
    for i in range(0, n - 1) : 
        if (k > 0) : 
        
            # Here, indexPosition
            # is set where we want to
            # put the current largest 
            # integer
            indexPosition = i
            for j in range(i + 1, n) :         
    
                # If we exceed the Max swaps
                # then break the loop
                if (k <= j - i) :
                    break
    
                # Find the maximum value 
                # from i+1 to max k or n
                # which will replace
                # arr[indexPosition]
                if (arr[j] > arr[indexPosition]) :
                    indexPosition = j
            
            # Swap the elements from 
            # Maximum indexPosition
            # we found till now to 
            # the ith index
            for j in range(indexPosition, i, -1) :
                t = arr[j]
                arr[j] = arr[j - 1]
                arr[j - 1] = t
    
            # Updates k after swapping 
            # indexPosition-i elements
            k = k - indexPosition - i

# Driver code
n = len(arr)
k = 3

KSwapMaximum(n, k)

# Print the final Array
for i in range(0, n) :
    print ("{} " . 
            format(arr[i]), 
                 end = "")
    
# This code is contributed by 
# Manish Shaw(manishshaw1)
C# 
// C# program to find 
// lexicographically 
// maximum value after
// k swaps.
using System;

class GFG
{ 
    static void SwapInts(int[] array, 
                         int position1, 
                         int position2)
    {
        // Swaps elements in an array.
        
        // Copy the first position's element
        int temp = array[position1]; 
        
        // Assign to the second element
        array[position1] = array[position2];
        
        // Assign to the first element
        array[position2] = temp; 
    }
    
    // Function which 
    // modifies the array
    static void KSwapMaximum(int []arr, 
                             int n, int k)
    {
        for (int i = 0; 
                 i < n - 1 && k > 0; ++i) 
        {
    
            // Here, indexPosition
            // is set where we want to
            // put the current largest 
            // integer
            int indexPosition = i;
            for (int j = i + 1; j < n; ++j) 
            {
    
                // If we exceed the 
                // Max swaps then 
                // break the loop
                if (k <= j - i)
                    break;
    
                // Find the maximum value 
                // from i+1 to max k or n 
                // which will replace 
                // arr[indexPosition]
                if (arr[j] > arr[indexPosition])
                    indexPosition = j;
            }
    
            // Swap the elements from 
            // Maximum indexPosition
            // we found till now to 
            // the ith index
            for (int j = indexPosition; j > i; --j)
                SwapInts(arr, j, j - 1);
    
            // Updates k after swapping
            // indexPosition-i elements
            k -= indexPosition - i;
        }
    }
    
    // Driver code
    static void Main()
    {
        int []arr = new int[]{ 3, 5, 4, 1, 2 };
        int n = arr.Length;
        int k = 3;
    
        KSwapMaximum(arr, n, k);
    
        // Print the final Array
        for (int i = 0; i < n; ++i)
            Console.Write(arr[i] + " ");
    }
}
// This code is contributed by 
// Manish Shaw(manishshaw1)
PHP 
<?php
// PHP program to find 
// lexicographically 
// maximum value after
// k swaps.

function swap(&$x, &$y)
{
    $x ^= $y ^= $x ^= $y;
}

// Function which
// modifies the array
function KSwapMaximum(&$arr, $n, $k)
{
    for ($i = 0; 
         $i < $n - 1 && 
         $k > 0; $i++) 
    {
        
        // Here, indexPosition
        // is set where we want to
        // put the current largest 
        // integer
        $indexPosition = $i;
        for ($j = $i + 1; 
             $j < $n; $j++)
        {

            // If we exceed the Max swaps
            // then break the loop
            if ($k <= $j - $i)
                break;

            // Find the maximum value 
            // from i+1 to max k or n
            // which will replace
            // arr[indexPosition]
            if ($arr[$j] > $arr[$indexPosition])
                $indexPosition = $j;
        } 
        
        // Swap the elements from 
        // Maximum indexPosition
        // we found till now to 
        // the ith index
        for ($j = $indexPosition; 
             $j > $i; $j--)
            swap($arr[$j], $arr[$j - 1]);

        // Updates k after swapping 
        // indexPosition-i elements
        $k -= $indexPosition - $i;
    }
}

// Driver code
$arr = array( 3, 5, 4, 1, 2 );
$n = count($arr);
$k = 3;

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

// Print the final Array
for ($i = 0; $i < $n; $i++)
    echo ($arr[$i]." ");
    
// This code is contributed by 
// Manish Shaw(manishshaw1)
?>
JavaScript 
<script>

// JavaScript program to find 
// lexicographically 
// maximum value after
// k swaps.

    function SwapLets(array, position1, position2)
    {
        // Swaps elements
        // in an array.
          
        // Copy the first 
        // position's element
        let temp = array[position1]; 
          
        // Assign to the 
        // second element
        array[position1] = array[position2];
          
        // Assign to the 
        // first element
        array[position2] = temp; 
    }
      
    // Function which 
    // modifies the array
    function KSwapMaximum(arr, n, k)
    {
        for (let i = 0; 
                 i < n - 1 && k > 0; ++i) 
        {
      
            // Here, indexPosition
            // is set where we want to
            // put the current largest 
            // integer
            let indexPosition = i;
            for (let j = i + 1; j < n; ++j) 
            {
      
                // If we exceed the 
                // Max swaps then 
                // break the loop
                if (k <= j - i)
                    break;
      
                // Find the maximum value 
                // from i+1 to max k or n 
                // which will replace 
                // arr[indexPosition]
                if (arr[j] > arr[indexPosition])
                    indexPosition = j;
            }
      
            // Swap the elements from 
            // Maximum indexPosition
            // we found till now to 
            // the ith index
            for (let j = indexPosition; j > i; --j)
                SwapLets(arr, j, j - 1);
      
            // Updates k after swapping
            // indexPosition-i elements
            k -= indexPosition - i;
        }
    }

// Driver code
    
        let arr = [ 3, 5, 4, 1, 2 ];
        let n = arr.length;
        let k = 3;
      
        KSwapMaximum(arr, n, k);
      
        // Print the final Array
        for (let i = 0; i < n; ++i)
            document.write(arr[i] + " ");
            
    // This code is contributed by coode_hunt.
</script>

Output
5 4 3 1 2

Time Complexity: O(N*N) 
Auxiliary Space: O(1)


Next Article
Partition into two subsets of lengths K and (N - k) such that the difference of sums is maximum

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