HackerRank Largest Permutation problem solution

In this HackerRank Largest Permutation problem solution, You are given an unordered array of unique integers incrementing from 1. 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.

Problem solution in Python.

#!/usr/bin/env python

import sys


if __name__ == '__main__':
    N, K = list(map(int, sys.stdin.readline().split()))
    A = list(map(int, sys.stdin.readline().split()))
    
    if K >= N - 1:
        print(*sorted(A, reverse = True), sep = ' ')
    
    else:
        X = sorted(A)
        i, swaps = 0, 0
        
        while swaps < K:
            x = X.pop()
            
            if A[i] != x:
                A[A.index(x)] = A[i]
                A[i] = x
                swaps += 1
            
            i += 1
        
        print(*A, sep = ' ')

Problem solution in Java.

import java.io.*;
import java.util.*;

public class Solution {

    static void swap(int[]a ,int ind1, int ind2){
        int tmp = a[ind1];
        a[ind1]=a[ind2];
        a[ind2]=tmp;
    }
    
    static int find(int[]a,int ind1,int ind2){
        int max=0;
        int maxV = Integer.MIN_VALUE;
        for(int i=ind1;i<=ind2;i++){
            if(a[i]>maxV){
                maxV=a[i];
                max=i;
            }
        }
        
        return max;
    }

    
    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner in = new Scanner(System.in);
        int N=in.nextInt();
        int K=in.nextInt();
        int[]a=new int[N];
        for(int i=0;i<N;i++)
            a[i]=in.nextInt();
        
        int ind=0;
        int max;
        int k=K;
        while(ind<a.length && k>0){
            max=find(a,ind,a.length-1);
            if(a[max]>a[ind]){
                swap(a,ind,max);
                k--;
            }
            ind++;

        }
        
        for(int i=0;i<N;i++)
            System.out.print(a[i]+" ");

    }
}

Problem solution in C++.

#include<iostream>
#include<sstream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include<cmath>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<numeric>
#include<functional>
#include<complex>

using namespace std;
#define BET(a,b,c) ((a)<=(b)&&(b)<(c))
#define FOR(i,n) for(int i=0,i##_end=(int(n));i<i##_end;i++)
#define SZ(x) (int)(x.size())
#define ALL(x) (x).begin(),(x).end()
#define MP make_pair
#define CLS(x,val) memset((x) , val , sizeof(x)) 
#define FOR_EACH(it,v) for(__typeof(v.begin()) it=v.begin(),it_end=v.end() ; it != it_end ; it++)
typedef long long ll_t;
typedef long double ld_t;
typedef vector<int> VI;
typedef vector<VI> VVI;

typedef complex<int> xy;

int P[100000+10];
int IndexFromValue[100000+10];
int main()
{
    int N,K;
    cin>>N>>K;
    FOR(i,N) scanf("%d",&P[i]);
    FOR(i,N) IndexFromValue[P[i]] = i;
    for(int i=N;i>=1 && K > 0;i--){
        int bestPos = N - i;
        if(bestPos == IndexFromValue[i]) continue;
        int p = IndexFromValue[i];
        swap(IndexFromValue[i], IndexFromValue[P[bestPos]]);
        swap(P[bestPos], P[p]);
        
        K--;
    }
    FOR(i,N) printf("%s%d", i?" ":"", P[i]);
    return 0;
}

Problem solution in C.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
        long long int n,k,i,a[100000],temp,j,l;
        scanf("%lld%lld",&n,&k);
       for(i=0;i<n;i++)
        {
        scanf("%lld",&a[i]);
    }
    l=n;
    for(i=0;i<k;i++)
        {
        if(a[i]==l)
            {
        k++;
        }
            else
            {
            for(j=i;j<n;j++)
                {
                if(a[j]==l)
                    {
                    temp=a[j];
                a[j]=a[i];
                a[i]=temp;                

                }
            }
        }
    l--;
    }
    for(i=0;i<n;i++)
        printf("%lld ",a[i]);
        return 0;
}