Largest permutation after at most k swaps

Given a permutation of first n natural numbers as array and an integer k. Print the lexicographically largest permutation after at most k swaps

Examples:

Input: arr[] = {4, 5, 2, 1, 3}
       k = 3
Output: 5 4 3 2 1
Swap 1^st and 2^nd elements: 5 4 2 1 3 
Swap 3^rd and 5^th elements: 5 4 3 1 2 
Swap 4^th and 5^th elements: 5 4 3 2 1 

Input: arr[] = {2, 1, 3}
       k = 1
Output: 3 1 2
Swap 1^st and 3^re elements: 3 1 2

Naive approach: The idea is to generate one by one permutation in lexicographically decreasing order. Compare every generated permutation with original array and count the number of swaps required to convert. If count is less than or equal to k, print this permutation. The problem of this approach is that it would be difficult to implement and will definitely time out for the large value of N.

Algorithm:
To find the minimum swaps to convert one array to another read this article.
Copy the original array and sort that array in decreasing order. So the sorted array is the largest permutation of the original array.
Now generate all permutation in lexicographically decreasing order. Previous permutation is calculated using prev_permutation() function.
Find the minimum steps required to convert the new array (permutation in decreasing order) to original array, if the count is less than or equal to k. Then print the array and break.

Implementation:

C++14

#include <bits/stdc++.h>
using namespace std;
 
// Function returns the minimum number
// of swaps required to sort the array
// This method is taken from below post
// https:// www.geeksforgeeks.org/
// minimum-number-swaps-required-sort-array/
int minSwapsToSort(int arr[], int n)
{
    // Create an array of pairs where first
    // element is array element and second
    // element is position of first element
    pair<int, int> arrPos[n];
    for (int i = 0; i < n; i++) {
        arrPos[i].first = arr[i];
        arrPos[i].second = i;
    }
 
    // Sort the array by array element
    // values to get right position of
    // every element as second
    // element of pair.
    sort(arrPos, arrPos + n);
 
    // To keep track of visited elements.
    // Initialize all elements as not
    // visited or false.
    vector<bool> vis(n, false);
 
    // Initialize result
    int ans = 0;
 
    // Traverse array elements
    for (int i = 0; i < n; i++) {
        // Already swapped and corrected or
        // already present at correct pos
        if (vis[i] || arrPos[i].second == i)
            continue;
 
        // Find out the number of  node in
        // this cycle and add in ans
        int cycle_size = 0;
        int j = i;
        while (!vis[j]) {
            vis[j] = 1;
 
            // move to next node
            j = arrPos[j].second;
            cycle_size++;
        }
 
        // Update answer by adding current
        // cycle.
        ans += (cycle_size - 1);
    }
 
    // Return result
    return ans;
}
 
// method returns minimum number of
// swap to make array B same as array A
int minSwapToMakeArraySame(
    int a[], int b[], int n)
{
    // Map to store position of elements
    // in array B we basically store
    // element to index mapping.
    map<int, int> mp;
    for (int i = 0; i < n; i++)
        mp[b[i]] = i;
 
    // now we're storing position of array
    // A elements in array B.
    for (int i = 0; i < n; i++)
        b[i] = mp[a[i]];
 
    /* We can uncomment this section to
      print modified b array
    for (int i = 0; i < N; i++)
        cout << b[i] << " ";
    cout << endl; */
 
    // Returning minimum swap for sorting
    // in modified array B as final answer
    return minSwapsToSort(b, n);
}
 
// Function to calculate largest
// permutation after atmost K swaps
void KswapPermutation(
    int arr[], int n, int k)
{
    int a[n];
 
    // copy the array
    for (int i = 0; i < n; i++)
        a[i] = arr[i];
 
    // Sort the array in descending order
    sort(arr, arr + n, greater<int>());
 
    // generate permutation in lexicographically
    // decreasing order.
    do {
        // copy the array
        int a1[n], b1[n];
        for (int i = 0; i < n; i++) {
            a1[i] = arr[i];
            b1[i] = a[i];
        }
 
        // Check if it can be made same in k steps
        if (
            minSwapToMakeArraySame(
                a1, b1, n)
            <= k) {
            // Print the array
            for (int i = 0; i < n; i++)
                cout << arr[i] << " ";
            break;
        }
 
        // move to previous permutation
    } while (prev_permutation(arr, arr + n));
}
 
int main()
{
    int arr[] = { 4, 5, 2, 1, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
 
    cout << "Largest permutation after "
         << k << " swaps:\n";
 
    KswapPermutation(arr, n, k);
    return 0;
}
// This code is contributed by karandeep1234

Java

// Java program for above approach
import java.util.*;
 
public class Solution {
  static class Pair {
    int first, second;
    Pair(int f, int s)
    {
      first = f;
      second = s;
    }
  }
 
  // Function returns the minimum number
  // of swaps required to sort the array
  static int minSwapsToSort(int arr[], int n)
  {
    // Create an array of pairs where first
    // element is array element and second
    // element is position of first element
    ArrayList<Pair> arrPos = new ArrayList<>();
    for (int i = 0; i < n; i++) {
      arrPos.add(new Pair(arr[i], i));
    }
 
    // Sort the array by array element
    // values to get right position of
    // every element as second
    // element of pair.
    Collections.sort(arrPos, (Pair A, Pair B) -> {
      return A.first - B.first;
    });
 
    // To keep track of visited elements.
    // Initialize all elements as not
    // visited or false.
    boolean[] vis = new boolean[n];
 
    // Initialize result
    int ans = 0;
 
    // Traverse array elements
    for (int i = 0; i < n; i++) {
      // Already swapped and corrected or
      // already present at correct pos
      if (vis[i] || arrPos.get(i).second == i)
        continue;
 
      // Find out the number of  node in
      // this cycle and add in ans
      int cycle_size = 0;
      int j = i;
      while (!vis[j]) {
        vis[j] = true;
 
        // move to next node
        j = arrPos.get(j).second;
        cycle_size++;
      }
 
      // Update answer by adding current
      // cycle.
      ans += (cycle_size - 1);
    }
 
    // Return result
    return ans;
  }
 
  // method returns minimum number of
  // swap to make array B same as array A
  static int minSwapToMakeArraySame(int a[], int b[],
                                    int n)
  {
    // Map to store position of elements
    // in array B we basically store
    // element to index mapping.
    HashMap<Integer, Integer> mp = new HashMap<>();
    for (int i = 0; i < n; i++)
      mp.put(b[i], i);
 
    // now we're storing position of array
    // A elements in array B.
    for (int i = 0; i < n; i++)
      b[i] = mp.get(a[i]);
 
    /* We can uncomment this section to
              print modified b array
        for (int i = 0; i < N; i++)
              System.out.print(b[i]);
        System.out.println();
        */
 
    // Returning minimum swap for sorting
    // in modified array B as final answer
    return minSwapsToSort(b, n);
  }
 
  // Function to calculate largest
  // permutation after atmost K swaps
  static void KswapPermutation(int[] arr, int n, int k)
  {
    int a[] = new int[n];
 
    // copy the array
    for (int i = 0; i < n; i++)
      a[i] = arr[i];
 
    // Sort the array in descending order
    // Arrays.sort(arr);
    Arrays.sort(arr);
    for (int idx = 0; idx < n / 2; idx++) {
      int t = arr[idx];
      arr[idx] = arr[n - idx - 1];
      arr[n - idx - 1] = t;
    }
 
    // generate permutation in lexicographically
    // decreasing order.
    do {
      // copy the array
      int[] a1 = new int[n];
      int[] b1 = new int[n];
      for (int i = 0; i < n; i++) {
        a1[i] = arr[i];
        b1[i] = a[i];
      }
 
      // Check if it can be made same in k steps
      if (minSwapToMakeArraySame(a1, b1, n) <= k) {
        // Print the array
        for (int i = 0; i < n; i++)
          System.out.print(arr[i] + " ");
        break;
      }
 
      // move to previous permutation
    } while (prev_permutation(arr, n));
  }
 
  static boolean prev_permutation(int[] s, int n)
  {
    // Find the largest index i such that s[i-1] is more
    // than s[i]
    int i = n - 1;
    while (i > 0 && s[i - 1] <= s[i]) {
      // Return false if i is at the first index of
      // the string.
      if (--i == 0) {
        return false;
      }
    }
 
    // Find the highest index j such that j >= i and
    // s[j] < s[i-1]
    int j = i;
    while (j < n && s[j] <= s[i - 1]) {
      j++;
    }
 
    j--;
 
    // Swap character at index i-1 with index j
    int t = s[i - 1];
    s[i - 1] = s[j];
    s[j] = t;
 
    // Reverse substring s[i to n-1] and return true
    int l = i;
    int r = n - 1;
    while (l < r) {
      t = s[l];
      s[l] = s[r];
      s[r] = t;
      l++;
      r--;
    }
 
    return true;
  }
  public static void main(String[] args)
  {
    int arr[] = { 2, 1, 3 };
    int n = arr.length;
    int k = 1;
    System.out.println("Largest permutation after " + k
                       + " swaps:");
    KswapPermutation(arr, n, k);
  }
}
 
// This code is contributed by karandeep1234.

Python3

#Function returns the minimum number of swaps required to sort the array
def min_swaps_to_sort(arr, n):
    # Create an array of pairs where first
    # element is array element and second
    # element is position of first element
    arr_pos = [(arr[i], i) for i in range(n)]
 
 
    # Sort the array by array element values to get right position of
    # every element as second element of pair.
    arr_pos.sort(key=lambda x: x[0])
 
    # To keep track of visited elements. Initialize all elements as not
    # visited or false.
    vis = [False] * n
 
    # Initialize result
    ans = 0
 
    # Traverse array elements
    for i in range(n):
        # Already swapped and corrected or already present at correct pos
        if vis[i] or arr_pos[i][1] == i:
            continue
 
        # Find out the number of nodes in this cycle and add in ans
        cycle_size = 0
        j = i
        while not vis[j]:
            vis[j] = True
            # move to next node
            j = arr_pos[j][1]
            cycle_size += 1
 
        # Update answer by adding current cycle.
        ans += (cycle_size - 1)
 
    # Return result
    return ans
 
 
#Method returns minimum number of swap to make array B same as array A
def min_swap_to_make_array_same(a, b, n):
    # Map to store position of elements in array B we basically store
    # element to index mapping.
    mp = {b[i]: i for i in range(n)}
 
    # Now store position of array A elements in array B.
    for i in range(n):
        b[i] = mp[a[i]]
 
    # Returning minimum swap for sorting in modified array B as final answer
    return min_swaps_to_sort(b, n)
 
 
# Function to calculate largest permutation after at most K swaps
def k_swap_permutation(arr, n, k):
    # Copy the array
    a = arr.copy()
 
    # Sort the array in descending order
    arr.sort(reverse=True)
 
    # Generate permutation in lexicographically decreasing order.
    while True:
        # Copy the array
        a1 = arr.copy()
        b1 = a.copy()
 
        # Check if it can be made same in k steps
        if min_swap_to_make_array_same(a1, b1, n) <= k:
            # Print the array
            return arr
            break
 
        # Move to previous permutation
        if not prev_permutation(arr):
            break
 
 
arr = [4, 5, 2, 1, 3]
n = len(arr)
k = 3
 
print("Largest permutation after", k, "swaps:")
print(k_swap_permutation(arr, n, k))
# contributed by akashish__

C#

// C# program for above approach
using System;
using System.Collections.Generic;
using System.Linq;
 
public class GFG
{
    class Pair
    {
        public int first, second;
        public Pair(int f, int s)
        {
            first = f;
            second = s;
        }
    }
 
    // Function returns the minimum number
    // of swaps required to sort the array
    static int minSwapsToSort(int[] arr, int n)
    {
        // Create an array of pairs where first
        // element is array element and second
        // element is position of first element
        List<Pair> arrPos = new List<Pair>();
        for (int i = 0; i < n; i++)
        {
            arrPos.Add(new Pair(arr[i], i));
        }
 
        // Sort the array by array element
        // values to get right position of
        // every element as second
        // element of pair.
        arrPos.Sort((Pair A, Pair B) => {
            return A.first - B.first;
        });
 
        // To keep track of visited elements.
        // Initialize all elements as not
        // visited or false.
        bool[] vis = new bool[n];
 
        // Initialize result
        int ans = 0;
 
        // Traverse array elements
        for (int i = 0; i < n; i++)
        {
            // Already swapped and corrected or
            // already present at correct pos
            if (vis[i] || arrPos[i].second == i)
                continue;
 
            // Find out the number of  node in
            // this cycle and add in ans
            int cycle_size = 0;
            int j = i;
            while (!vis[j])
            {
                vis[j] = true;
 
                // move to next node
                j = arrPos[j].second;
                cycle_size++;
            }
 
            // Update answer by adding current
            // cycle.
            ans += (cycle_size - 1);
        }
 
        // Return result
        return ans;
    }
 
    // method returns minimum number of
    // swap to make array B same as array A
    static int minSwapToMakeArraySame(int[] a, int[] b,
                                      int n)
    {
        // Map to store position of elements
        // in array B we basically store
        // element to index mapping.
        Dictionary<int, int> mp = new Dictionary<int, int>();
        for (int i = 0; i < n; i++)
            mp.Add(b[i], i);
 
        // now we're storing position of array
        // A elements in array B.
        for (int i = 0; i < n; i++)
            b[i] = mp[a[i]];
 
        /* We can uncomment this section to
                  print modified b array
            for (int i = 0; i < N; i++)
                  System.out.print(b[i]);
            System.out.println();
            */
 
        // Returning minimum swap for sorting
        // in modified array B as final answer
        return minSwapsToSort(b, n);
    }
 
    // Function to calculate largest
    // permutation after atmost K swaps
    static void KswapPermutation(int[] arr, int n, int k)
    {
        int[] a = new int[n];
 
        // copy the array
        for (int i = 0; i < n; i++)
            a[i] = arr[i];
 
        // Sort the array in descending order
        // Arrays.sort(arr);
        Array.Sort(arr);
        for (int idx = 0; idx < n / 2; idx++)
        {
            int t = arr[idx];
            arr[idx] = arr[n - idx - 1];
            arr[n - idx - 1] = t;
        }
 
        // generate permutation in lexicographically
        // decreasing order.
        do
        {
            // copy the array
            int[] a1 = new int[n];
            int[] b1 = new int[n];
            for (int i = 0; i < n; i++)
            {
                a1[i] = arr[i];
                b1[i] = a[i];
            }
 
            // Check if it can be made same in k steps
            if (minSwapToMakeArraySame(a1, b1, n) <= k)
            {
                // Print the array
                for (int i = 0; i < n; i++)
                    Console.Write(arr[i] + " ");
                break;
            }
 
            // move to previous permutation
        } while (prev_permutation(arr, n));
    }
 
    static bool prev_permutation(int[] s, int n)
    {
        // Find the largest index i such that s[i-1] is more
        // than s[i]
        int i = n - 1;
        while (i > 0 && s[i - 1] <= s[i])
        {
            // Return false if i is at the first index of
            // the string.
            if (--i == 0)
            {
                return false;
            }
        }
 
        // Find the highest index j such that j >= i and
        // s[j] < s[i-1]
        int j = i;
        while (j < n && s[j] <= s[i - 1])
        {
            j++;
        }
 
        j--;
 
        // Swap character at index i-1 with index j
        int t = s[i - 1];
        s[i - 1] = s[j];
        s[j] = t;
 
        // Reverse substring s[i to n-1] and return true
        int l = i;
        int r = n - 1;
        while (l < r)
        {
            t = s[l];
            s[l] = s[r];
            s[r] = t;
            l++;
            r--;
        }
 
        return true;
    }
    static public void Main()
    {
        int[] arr = { 2, 1, 3 };
        int n = arr.Length;
        int k = 1;
        Console.WriteLine("Largest permutation after " + k
                           + " swaps:");
        KswapPermutation(arr, n, k);
    }
}
 
// This code is contributed by akashish__

Javascript

// Function returns the minimum number
// of swaps required to sort the array
// This method is taken from below post
// https:// www.geeksforgeeks.org/
// minimum-number-swaps-required-sort-array/
function minSwapsToSort(arr, n)
{
 
    // Create an array of pairs where first
    // element is array element and second
    // element is position of first element
    let arrPos = new Array(n);
    for (let i = 0; i < n; i++) {
        arrPos[i] = {first: arr[i], second: i};
    }
 
    // Sort the array by array element
    // values to get right position of
    // every element as second
    // element of pair.
    arrPos.sort((a, b) => a.first - b.first);
 
    // To keep track of visited elements.
    // Initialize all elements as not
    // visited or false.
    let vis = new Array(n).fill(false);
 
    // Initialize result
    let ans = 0;
 
    // Traverse array elements
    for (let i = 0; i < n; i++) {
        // Already swapped and corrected or
        // already present at correct pos
        if (vis[i] || arrPos[i].second === i)
            continue;
 
        // Find out the number of  node in
        // this cycle and add in ans
        let cycle_size = 0;
        let j = i;
        while (!vis[j]) {
            vis[j] = true;
 
            // move to next node
            j = arrPos[j].second;
            cycle_size++;
        }
 
        // Update answer by adding current
        // cycle.
        ans += (cycle_size - 1);
    }
 
    // Return result
    return ans;
}
 
// method returns minimum number of
// swap to make array B same as array A
function minSwapToMakeArraySame(a, b, n) {
    // Map to store position of elements
    // in array B we basically store
    // element to index mapping.
    let mp = new Map();
    for (let i = 0; i < n; i++)
        mp.set(b[i], i);
 
    // now we're storing position of array
    // A elements in array B.
    for (let i = 0; i < n; i++)
        b[i] = mp.get(a[i]);
 
    /* We can uncomment this section to
      print modified b array
    for (let i = 0; i < N; i++)
        console.log(b[i]);
    console.log(); */
 
    // Returning minimum swap for sorting
    // in modified array B as final answer
    return minSwapsToSort(b, n);
}
 
// Function to calculate largest
// permutation after atmost K swaps
function KswapPermutation(arr, n, k) {
    // copy the array
    let a = arr.slice();
 
    // Sort the array in descending order
    arr.sort();
    arr.reverse();
 
    // generate permutation in lexicographically
    // decreasing order.
    do {
        // copy the array
        let a1 = [0]*n;
        let b1 = [0]*n;
        for (let i = 0; i < n; i++) {
            a1[i] = arr[i];
            b1[i] = a[i];
        }
 
        // Check if it can be made same in k steps
        if (
            minSwapToMakeArraySame(
                a1, b1, n)
            <= k) {
            // Print the array
            console.log(arr);
            break;
        }
 
        // move to previous permutation
    } while (prev_permutation(arr, arr + n));
}
 
let arr = [ 4, 5, 2, 1, 3 ];
let n = arr.length;
let k = 3;
 
console.log("Largest permutation after",k,"swaps:");
 
KswapPermutation(arr, n, k);
 
// This code is contributed by akashish__

Output

Largest permutation after 3 swaps:
5 4 3 2 1

Complexity Analysis:

Time Complexity: O(N!).
To generate all permutation O(N!) time complexity is required.
Space Complexity: O(n).
to store the new array O(n) space is required.

Another approach worth considering :

This problem can be considered as an instance of a “controlled” selection sort . By controlled , we mean that we are not performing the selection sort operation on the entire array. Instead, we are constructing ourselves to only the total number of swaps K that we are allowed to perform.

So in the following approach , all we need to do is simply let the selection sort go on for k times , not more than that . Also we need to check if the position of the maximum number we’re about to swap with the current position i is equal to the number already present in that position , and we need to jump over this particular situation , nevertheless we shouldn’t waste our limited number of swaps .

Implementation:

C++14

#include <bits/stdc++.h>
using namespace std;
void KswapPermutation(
    int arr[], int n, int k)
{
    for(int i=0;i<n-1;i++)
    {
        if( k>0)
        {
            int max = i;
            for(int j=i+1;j<n;j++)
            if(arr[j]>arr[max])
            max = j;
           
          // this condition checks whether the max value has changed since when
          // we started , or is it the same.
          // We need to ignore the swap if the value is same.
          // It means that the number ought to be present at the ith position , is already
          // there.
            if(max!=i)
            {
            int temp = arr[max];
            arr[max] = arr[i];
            arr[i] = temp;
            k--;
            }
        }
        else
        break;
    }
}
  
// Driver code
int main()
{
    int arr[] = { 4, 5, 2, 1, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    KswapPermutation(arr, n, k);
    cout << "Largest permutation after "
         << k << " swaps:"<<endl;
    for (int i = 0; i < n; ++i)
        cout<<arr[i]<<" ";
    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
class GFG {
 
  static void KswapPermutation(int arr[], int n, int k)
  {
    for (int i = 0; i < n - 1; i++) {
      if (k > 0) {
        int max = i;
        for (int j = i + 1; j < n; j++)
          if (arr[j] > arr[max])
            max = j;
 
        // this condition checks whether the max
        // value has changed since when we started ,
        // or is it the same. We need to ignore the
        // swap if the value is same. It means that
        // the number ought to be present at the ith
        // position , is already there.
        if (max != i) {
          int temp = arr[max];
          arr[max] = arr[i];
          arr[i] = temp;
          k--;
        }
      }
      else
        break;
    }
  }
  public static void main(String[] args)
  {
    int arr[] = { 4, 5, 2, 1, 3 };
    int n = arr.length;
    int k = 3;
    KswapPermutation(arr, n, k);
    System.out.println("Largest permutation after " + k
                       + " swaps:");
    for (int i = 0; i < n; ++i)
      System.out.print(arr[i] + " ");
  }
}
 
// This code is contributed by aadityaburujwale.

Python3

def KswapPermutation(arr, n, k):
    for i in range(0, n-1):
        if(k > 0):
            max = i
            for j in range(i+1, n):
                if(arr[j] > arr[max]):
                    max = j
 
            # this condition checks whether the max value has changed since when
            # we started , or is it the same.
            # We need to ignore the swap if the value is same.
            # It means that the number ought to be present at the ith position , is already
            # there.
            if(max != i):
                temp = arr[max]
                arr[max] = arr[i]
                arr[i] = temp
                k = k-1
 
        else:
            break
 
# Driver code
arr = [4, 5, 2, 1, 3]
n = len(arr)
k = 3
KswapPermutation(arr, n, k)
print("Largest permutation after "+str(k) + " swaps:")
for i in range(0, n):
    print(arr[i], end=' ')
 
# This code is contributed by Harsh Khatri

C#

using System;
using System.Collections.Generic;
class GFG {
  static void KswapPermutation(int[] arr, int n, int k)
  {
    for (int i = 0; i < n - 1; i++) {
      if (k > 0) {
        int max = i;
        for (int j = i + 1; j < n; j++)
          if (arr[j] > arr[max]) {
            max = j;
          }
 
        // this condition checks whether the max
        // value has changed since when we started ,
        // or is it the same. We need to ignore the
        // swap if the value is same. It means that
        // the number ought to be present at the ith
        // position , is already there.
        if (max != i) {
          int temp = arr[max];
          arr[max] = arr[i];
          arr[i] = temp;
          k--;
        }
      }
      else
        break;
    }
  }
 
  // Driver code
  public static void Main(string[] args)
  {
    int[] arr = { 4, 5, 2, 1, 3 };
    int n = 5;
    int k = 3;
    KswapPermutation(arr, n, k);
    Console.WriteLine("Largest permutation after " + k
                      + " swaps:");
    for (int i = 0; i < n; i++)
      Console.Write(arr[i] + " ");
  }
}
 
// This code is contributed by garg28harsh.

Javascript

<script>
function KswapPermutation(arr,  n,  k)
{
    for(let i=0;i<n-1;i++)
    {
        if( k>0)
        {
            let max = i;
            for(let j=i+1;j<n;j++)
            if(arr[j]>arr[max])
            max = j;
           
          // this condition checks whether the max value has changed since when
          // we started , or is it the same.
          // We need to ignore the swap if the value is same.
          // It means that the number ought to be present at the ith position , is already
          // there.
            if(max!=i)
            {
            let temp = arr[max];
            arr[max] = arr[i];
            arr[i] = temp;
            k--;
            }
        }
        else
        break;
    }
}
  
// Driver code
 
    let arr = [ 4, 5, 2, 1, 3 ];
    let n = 5;
    let k = 3;
    KswapPermutation(arr, n, k);
    document.write( "Largest permutation after "+ k + " swaps:");
    document.write(arr);
    
 </srcipt>

Output

Largest permutation after 3 swaps:
5 4 3 2 1

Complexity analysis :

Time complexity : O(n^2) , because this approach utilizes selection sort
Space complexity : O(1) , because the sort is in place and no extra space is needed

Efficient approach:

This is a greedy approach. The largest permutation is found when the largest elements are at the front of the array, i.e. the largest elements are sorted in decreasing order. There are at most k swaps so put the 1^st, 2^nd, 3^rd, …, k^th largest element at their respective position.

Note: If the number of swaps allowed is equal to the size of the array, then there is no need to iterate over the whole array. The answer will simply be the reverse sorted array.

Algorithm:
Create a HashMap or an array of length n to store element-index pair or map element to its index.
Now run a loop k times.
In each iteration swap the ith element with the element n – i. where i is the index or count of the loop. Also swap their position, i.e. update the hashmap or array. So in this step the largest element in remaining element is swapped to the front.
Print the output array.

Implementation 1: This uses simple arrays to arrive at the solution.

C++

// Below is C++ code to print largest
// permutation after at most K swaps
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate largest
// permutation after atmost K swaps
void KswapPermutation(
    int arr[], int n, int k)
{
    // Auxiliary dictionary of
    // storing the position of elements
    int pos[n + 1];
 
    for (int i = 0; i < n; ++i)
        pos[arr[i]] = i;
 
    for (int i = 0; i < n && k; ++i) {
        // If element is already i'th largest,
        // then no need to swap
        if (arr[i] == n - i)
            continue;
 
        // Find position of i'th
        // largest value, n-i
        int temp = pos[n - i];
 
        // Swap the elements position
        pos[arr[i]] = pos[n - i];
        pos[n - i] = i;
 
        // Swap the ith largest value with the
        // current value at ith place
        swap(arr[temp], arr[i]);
 
        // decrement number of swaps
        --k;
    }
}
 
// Driver code
int main()
{
    int arr[] = { 4, 5, 2, 1, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
 
    KswapPermutation(arr, n, k);
    cout << "Largest permutation after "
         << k << " swaps:n";
    for (int i = 0; i < n; ++i)
        printf("%d ", arr[i]);
    return 0;
}

Java

// Below is Java code to print
// largest permutation after
// atmost K swaps
class GFG {
 
    // Function to calculate largest
    // permutation after atmost K swaps
    static void KswapPermutation(
        int arr[], int n, int k)
    {
 
        // Auxiliary dictionary of storing
        // the position of elements
        int pos[] = new int[n + 1];
 
        for (int i = 0; i < n; ++i)
            pos[arr[i]] = i;
 
        for (int i = 0; i < n && k > 0; ++i) {
 
            // If element is already i'th
            // largest, then no need to swap
            if (arr[i] == n - i)
                continue;
 
            // Find position of i'th largest
            // value, n-i
            int temp = pos[n - i];
 
            // Swap the elements position
            pos[arr[i]] = pos[n - i];
            pos[n - i] = i;
 
            // Swap the ith largest value with the
            // current value at ith place
            int tmp1 = arr[temp];
            arr[temp] = arr[i];
            arr[i] = tmp1;
 
            // decrement number of swaps
            --k;
        }
    }
 
    // Driver method
    public static void main(String[] args)
    {
 
        int arr[] = { 4, 5, 2, 1, 3 };
        int n = arr.length;
        int k = 3;
 
        KswapPermutation(arr, n, k);
 
        System.out.print(
            "Largest permutation "
            + "after " + k + " swaps:\n");
 
        for (int i = 0; i < n; ++i)
            System.out.print(arr[i] + " ");
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python code to print largest permutation after K swaps
 
def KswapPermutation(arr, n, k):
 
    # Auxiliary array of storing the position of elements
    pos = {}
    for i in range(n):
        pos[arr[i]] = i
 
    for i in range(n):
 
        # If K is exhausted then break the loop
        if k == 0:
            break
 
        # If element is already largest then no need to swap
        if (arr[i] == n-i):
            continue
 
        # Find position of i'th largest value, n-i
        temp = pos[n-i]
 
        # Swap the elements position
        pos[arr[i]] = pos[n-i]
        pos[n-i] = i
 
        # Swap the ith largest value with the value at
        # ith place
        arr[temp], arr[i] = arr[i], arr[temp]
 
        # Decrement K after swap
        k = k-1
 
# Driver code
arr = [4, 5, 2, 1, 3]
n = len(arr)
k = 3
KswapPermutation(arr, n, k)
 
# Print the answer
print ("Largest permutation after", k, "swaps: ")
print (" ".join(map(str, arr)))

C#

// Below is C# code to print largest
// permutation after atmost K swaps.
using System;
 
class GFG {
 
    // Function to calculate largest
    // permutation after atmost K
    // swaps
    static void KswapPermutation(int[] arr,
                                 int n, int k)
    {
 
        // Auxiliary dictionary of storing
        // the position of elements
        int[] pos = new int[n + 1];
 
        for (int i = 0; i < n; ++i)
            pos[arr[i]] = i;
 
        for (int i = 0; i < n && k > 0; ++i) {
 
            // If element is already i'th
            // largest, then no need to swap
            if (arr[i] == n - i)
                continue;
 
            // Find position of i'th largest
            // value, n-i
            int temp = pos[n - i];
 
            // Swap the elements position
            pos[arr[i]] = pos[n - i];
            pos[n - i] = i;
 
            // Swap the ith largest value with
            // the current value at ith place
            int tmp1 = arr[temp];
            arr[temp] = arr[i];
            arr[i] = tmp1;
 
            // decrement number of swaps
            --k;
        }
    }
 
    // Driver method
    public static void Main()
    {
 
        int[] arr = { 4, 5, 2, 1, 3 };
        int n = arr.Length;
        int k = 3;
 
        KswapPermutation(arr, n, k);
 
        Console.Write("Largest permutation "
                      + "after " + k + " swaps:\n");
 
        for (int i = 0; i < n; ++i)
            Console.Write(arr[i] + " ");
    }
}
 
// This code is contributed nitin mittal.

PHP

<?php
// PHP code to print largest permutation
// after atmost K swaps
 
// Function to calculate largest
// permutation after atmost K swaps
function KswapPermutation(&$arr, $n, $k)
{
 
    for ($i = 0; $i < $n; ++$i)
        $pos[$arr[$i]] = $i;
 
    for ($i = 0; $i < $n && $k; ++$i)
    {
        // If element is already i'th largest,
        // then no need to swap
        if ($arr[$i] == $n - $i)
            continue;
 
        // Find position of i'th largest
        // value, n-i
        $temp = $pos[$n - $i];
 
        // Swap the elements position
        $pos[$arr[$i]] = $pos[$n - $i];
        $pos[$n - $i] = $i;
 
        // Swap the ith largest value with the
        // current value at ith place
        $t = $arr[$temp];
        $arr[$temp] = $arr[$i];
        $arr[$i] = $t;
         
        // decrement number of swaps
        --$k;
    }
}
 
// Driver code
$arr = array(4, 5, 2, 1, 3);
$n = sizeof($arr);
$k = 3;
 
KswapPermutation($arr, $n, $k);
echo ("Largest permutation after ");
echo ($k);
echo (" swaps:\n");
for ($i = 0; $i < $n; ++$i)
{
    echo ($arr[$i] );
    echo (" ");
}
 
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

// Below is Javascript code to print largest
// permutation after at most K swaps
 
// Function to calculate largest
// permutation after atmost K swaps
function KswapPermutation(arr, n, k)
{
    // Auxiliary dictionary of
    // storing the position of elements
    let pos = new Array(n + 1);
 
    for (let i = 0; i < n; ++i)
        pos[arr[i]] = i;
 
    for (let i = 0; i < n && k; ++i) {
        // If element is already i'th largest,
        // then no need to swap
        if (arr[i] == n - i)
            continue;
 
        // Find position of i'th
        // largest value, n-i
        let temp = pos[n - i];
 
        // Swap the elements position
        pos[arr[i]] = pos[n - i];
        pos[n - i] = i;
 
        // Swap the ith largest value with the
        // current value at ith place
        let tmp1 = arr[temp];
        arr[temp] = arr[i];
        arr[i] = tmp1;
 
        // decrement number of swaps
        --k;
    }
}
 
 let arr = [ 4, 5, 2, 1, 3 ];
let n = arr.length;
let k = 3;
 
KswapPermutation(arr, n, k);
console.log("Largest permutation after " + k + " swaps:" + "</br>");
for (let i = 0; i < n; ++i)
    console.log(arr[i] + " ");
         
        // This code is contributed by garg28harsh.

Output

Largest permutation after 3 swaps:n5 4 3 2 1

Time complexity :- O(nlogn)

Space complexity :- O(N)

Implementation 2: This uses a hashmap to arrive at the solution.

C++

// C++ Program to find the
// largest permutation after
// at most k swaps using unordered_map.
#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
 
// Function to find the largest
// permutation after k swaps
void bestpermutation(
    int arr[], int k, int n)
{
    // Storing the elements and
    // their index in map
    unordered_map<int, int> h;
    for (int i = 0; i < n; i++) {
        h.insert(make_pair(arr[i], i));
    }
 
    // If number of swaps allowed
    // are equal to number of elements
    // then the resulting permutation
    // will be descending order of
    // given permutation.
    if (n <= k) {
        sort(arr, arr + n, greater<int>());
    }
    else {
 
        for (int j = n; j >= 1; j--) {
            if (k > 0) {
 
                int initial_index = h[j];
                int best_index = n - j;
 
                // if j is not at it's best index
                if (initial_index != best_index) {
 
                    h[j] = best_index;
 
                    // Change the index of the element
                    // which was at position 0. Swap
                    // the element basically.
                    int element = arr[best_index];
                    h[element] = initial_index;
                    swap(
                        arr[best_index],
                        arr[initial_index]);
 
                    // decrement number of swaps
                    k--;
                }
            }
        }
    }
}
 
// Driver code
int main()
{
    int arr[] = { 3, 1, 4, 2, 5 };
 
    // K is the number of swaps
    int k = 10;
 
    // n is the size of the array
    int n = sizeof(arr) / sizeof(int);
 
    // Function calling
    bestpermutation(arr, k, n);
 
    cout << "Largest possible permutation after "
         << k << " swaps is ";
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
 
    return 0;
}
// This method is contributed by Kishan Mishra.

Java

// Java Program to find the
// largest permutation after
// at most k swaps using unordered_map.
import java.util.*;
class GFG
{
 
  // Function to find the largest
  // permutation after k swaps
  static void bestpermutation(ArrayList<Integer> arr, int k, int n)
  {
 
    // Storing the elements and
    // their index in map
    HashMap<Integer, Integer> h =  
      new HashMap<Integer, Integer>(); 
    for (int i = 0; i < n; i++)
    {
      h.put(arr.get(i), i);
    }
 
    // If number of swaps allowed
    // are equal to number of elements
    // then the resulting permutation
    // will be descending order of
    // given permutation.
    if (n <= k) {
      Collections.sort(arr, Collections.reverseOrder()); 
    }
    else {
 
      for (int j = n; j >= 1; j--)
      {
        if (k > 0)
        {
 
          int initial_index = h.get(j);
          int best_index = n - j;
 
          // if j is not at it's best index
          if (initial_index != best_index)
          {
            h.put(j, best_index);
 
            // Change the index of the element
            // which was at position 0. Swap
            // the element basically.
            int element = arr.get(best_index);
            h.put(element, initial_index);
            int temp = arr.get(best_index);
            arr.set(best_index, arr.get(initial_index));
            arr.set(initial_index, temp);
 
            // decrement number of swaps
            k--;
          }
        }
      }
    }
  }
 
  // Driver code
  public static void main(String []args)
  {
 
    ArrayList<Integer> arr = new ArrayList<Integer>();
    arr.add(3);
    arr.add(1);
    arr.add(4);
    arr.add(2);
    arr.add(5);
 
    // K is the number of swaps
    int k = 10;
 
    // n is the size of the array
    int n = arr.size();
 
    // Function calling
    bestpermutation(arr, k, n);
 
    System.out.print("Largest possible permutation after " + k + " swaps is ");
 
    for (int i = 0; i < n; i++)
      System.out.print(arr.get(i) + " ");
  }
}
 
// This code is contributed by rutvik_56.

Python3

# Python3 program to find the
# largest permutation after
# at most k swaps using unordered_map.
 
# Function to find the largest
# permutation after k swaps
def bestpermutation(arr, k, n):
     
    # Storing the elements and
    # their index in map
    h = {}
    for i in range(n):
        h[arr[i]] = i
 
    # If number of swaps allowed
    # are equal to number of elements
    # then the resulting permutation
    # will be descending order of
    # given permutation.
    if (n <= k):
        arr.sort()
        arr.reverse()
    else:
         
        for j in range(n, 0, -1):
            if (k > 0):
                initial_index = h[j]
                best_index = n - j
 
                # If j is not at it's best index
                if (initial_index != best_index):
                    h[j] = best_index
 
                    # Change the index of the element
                    # which was at position 0. Swap
                    # the element basically.
                    element = arr[best_index]
                    h[element] = initial_index
                    arr[best_index], arr[initial_index] = (arr[initial_index],
                                                           arr[best_index])
                                                            
                    # Decrement number of swaps
                    k -= 1
 
# Driver Code
arr = [ 3, 1, 4, 2, 5 ]
 
# K is the number of swaps
k = 10
 
# n is the size of the array
n = len(arr)
 
# Function calling
bestpermutation(arr, k, n)
 
print("Largest possible permutation after",
      k, "swaps is", end = " ")
       
for i in range(n):
    print(arr[i], end = " ")
 
# This code is contributed by divyesh072019

C#

// C# Program to find the
// largest permutation after
// at most k swaps using unordered_map.
using System;
using System.Collections.Generic;
class GFG {
     
    // Function to find the largest
    // permutation after k swaps
    static void bestpermutation(List<int> arr, int k, int n)
    {
        // Storing the elements and
        // their index in map
        Dictionary<int, int> h =  
                       new Dictionary<int, int>(); 
        for (int i = 0; i < n; i++) {
            h.Add(arr[i], i);
        }
      
        // If number of swaps allowed
        // are equal to number of elements
        // then the resulting permutation
        // will be descending order of
        // given permutation.
        if (n <= k) {
            arr.Sort();
            arr.Reverse();
        }
        else {
      
            for (int j = n; j >= 1; j--) {
                if (k > 0) {
      
                    int initial_index = h[j];
                    int best_index = n - j;
      
                    // if j is not at it's best index
                    if (initial_index != best_index) {
      
                        h[j] = best_index;
      
                        // Change the index of the element
                        // which was at position 0. Swap
                        // the element basically.
                        int element = arr[best_index];
                        h[element] = initial_index;
                        int temp = arr[best_index];
                        arr[best_index] = arr[initial_index];
                        arr[initial_index] = temp;
      
                        // decrement number of swaps
                        k--;
                    }
                }
            }
        }
    }
 
  static void Main() {
    List<int> arr = new List<int>(new int[] {3, 1, 4, 2, 5 });
  
    // K is the number of swaps
    int k = 10;
  
    // n is the size of the array
    int n = arr.Count;
  
    // Function calling
    bestpermutation(arr, k, n);
  
    Console.Write("Largest possible permutation after " + k + " swaps is ");
    for (int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
  }
}
 
// This code is contributed by divyeshrabadiya07

Javascript

<script>
 
// JavaScript Program to find the
// largest permutation after
// at most k swaps using unordered_map.
 
// Function to find the largest
  // permutation after k swaps
function bestpermutation(arr,k,n)
{
    // Storing the elements and
    // their index in map
    let h = 
      new Map();
    for (let i = 0; i < n; i++)
    {
      h.set(arr[i], i);
    }
  
    // If number of swaps allowed
    // are equal to number of elements
    // then the resulting permutation
    // will be descending order of
    // given permutation.
    if (n <= k) {
      arr.sort(function(a,b){return b-a;});
    }
    else {
  
      for (let j = n; j >= 1; j--)
      {
        if (k > 0)
        {
  
          let initial_index = h[j];
          let best_index = n - j;
  
          // if j is not at it's best index
          if (initial_index != best_index)
          {
            h.set(j, best_index);
  
            // Change the index of the element
            // which was at position 0. Swap
            // the element basically.
            let element = arr.get(best_index);
            h.set(element, initial_index);
            let temp = arr[best_index];
            arr.set(best_index, arr[initial_index]);
            arr.set(initial_index, temp);
  
            // decrement number of swaps
            k--;
          }
        }
      }
    }
}
 
// Driver code
let arr=[3,1,4,2,5];
// K is the number of swaps
let k = 10;
 
// n is the size of the array
let n = arr.length;
 
// Function calling
bestpermutation(arr, k, n);
 
document.write(
"Largest possible permutation after " + k + " swaps is "
);
 
for (let i = 0; i < n; i++)
    document.write(arr[i] + " ");
 
 
// This code is contributed by avanitrachhadiya2155
 
</script>

Output

Largest possible permutation after 10 swaps is 5 4 3 2 1

Complexity Analysis:

Time Complexity: O(N).
Only one traversal of the array is required.
Space Complexity: O(n).
To store the new array O(n) space is required.

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

My Personal Notes

Last Updated : 06 Apr, 2023

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