**Largest Permutation**

### Problem Statement :

You are given an unordered array of unique integers incrementing from . You can swap any two elements a limited number of times. Determine the largest lexicographical value array that can be created by executing no more than the limited number of swaps. Example The following arrays can be formed by swapping the with the other elements: [2,1,3,4] [3,2,1,4] [4,2,3,1] The highest value of the four (including the original) is . If , we can swap to the highest possible value: . Function Description Complete the largestPermutation function in the editor below. It must return an array that represents the highest value permutation that can be formed. largestPermutation has the following parameter(s): int k: the maximum number of swaps int arr[n]: an array of integers Input Format The first line contains two space-separated integers and , the length of and the maximum swaps that can be performed. The second line contains distinct space-separated integers from to as where .

### Solution :

` ````
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int n,l,k,i,j=0;
scanf("%d %d",&n,&k);
int arr[n];
for(i=0;i<n;i++)
{
scanf("%d",&arr[i]);
}
i=0;
while(j<k && i<n)
{
if(arr[i]!=n-i)
{
for(l=i;l<n;l++)
{if(arr[l]==n-i) //swap
{
int tmp=arr[l];
arr[l]=arr[i];
arr[i]=tmp;
break;
}
}
j++;
}
i++;
}
for(i=0;i<n;i++)
{
printf("%d ",arr[i]);
}
return 0;
}
```

` ````
Solution in C++ :
In C++ :
#include<iostream>
using namespace std;
int main()
{
int t,n,k;
{
cin>>n>>k;
int a[n+1],pos[n+1];
for(int i=1;i<=n;i++)
{
cin>>a[i];
pos[a[i]]=i; //position of a[i] will be i
}
for(int i=1;i<=n;i++)
{
if(!k) //no more swapping can be done
break;
else
{
if(a[i]!=n-i+1)
{
k--;
int temp = a[i]; //swapping a[i] with n-i+1
a[i] = n-i+1;
a[pos[n-i+1]] = temp;
pos[temp] = pos[n-i+1]; //swapping a[i]'s and a[n-i+1]'s positions
pos[n-i+1] = i;
}
}
}
for(int i=1;i<=n;i++) //output array after swapping is done
cout<<a[i]<<" ";
}
}
```

` ````
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner scc = new Scanner(System.in);
int n = scc.nextInt();
int k = scc.nextInt();
int arr[] = new int[n];
for(int i=0; i<n; i++){
arr[i] = scc.nextInt();
}
for(int i=0; i<k && i<n; i++){
int j;
for(j=i; j<n; j++){
if(arr[j]==n-i){
break;
}
}
if(j!=i){
int temp = arr[j];
arr[j] = arr[i];
arr[i] = temp;
}
else{
k++;
}
}
for(int i=0; i<n; i++)
System.out.print(arr[i]+" ");
}
}
```

` ````
Solution in Python :
In Python3 :
n, k = map(int, input().split())
p = list(map(lambda s: int(s) - 1, input().split()))
where = [-1] * n
for i, x in enumerate(p):
where[x] = i
for i in range(n):
if k == 0:
break
if p[i] != n - 1 - i:
bad_pos = where[n - 1 - i]
bad_num = p[i]
p[i], p[bad_pos] = p[bad_pos], p[i]
where[bad_num], where[n - 1 - i] = where[n - 1 - i], where[bad_num]
k -= 1
print(' '.join(map(lambda x: str(x + 1), p)))
```

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