HackerRank Larry’s Array problem solution

In this HackerRank Larry’s Array problem solution, Larry has been given a permutation of a sequence of natural numbers incrementing from 1 as an array. He must determine whether the array can be sorted using the following operation any number of times:

• Choose any 3 consecutive indices and rotate their elements in such a way that 
  ABC -> BCA -> CAB -> ABC.

For example, if A = {1,6,5,2,4,3}:

A		rotate 
[1,6,5,2,4,3]	[6,5,2]
[1,5,2,6,4,3]	[5,2,6]
[1,2,6,5,4,3]	[5,4,3]
[1,2,6,3,5,4]	[6,3,5]
[1,2,3,5,6,4]	[5,6,4]
[1,2,3,4,5,6]

YES

On a new line for each test case, print YES if A can be fully sorted. Otherwise, print NO.

HackerRank Larry’s Array problem solution in Python programming.

#!/bin/python3

import sys

def canSort(l):
    s = sorted(l)
    for i in s[:-2]:
        if l.index(i) % 2:
            l.remove(i)
            l[0],l[1]=l[1],l[0]
        else:
            l.remove(i)
    if l[0]<l[1]:
        return True
    else:
        return False

if __name__ == "__main__":
    t = int(sys.stdin.readline().strip())
    for _ in range(t):
        n = int(sys.stdin.readline().strip())
        A = list(map(int, sys.stdin.readline().split()))
        if canSort(A):
            print("YES")
        else:
            print("NO")

Larry’s Array problem solution in Java Programming.

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

public class Solution {

    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 scan = new Scanner(System.in);
        int c = Integer.parseInt(scan.nextLine());
        for(int i=0;i<c;i++){
            int n = Integer.parseInt(scan.nextLine());
            String[] sp = scan.nextLine().split("\s+");
            LinkedList<Integer> ll = new LinkedList<Integer>();
            for(int j=0;j<n;j++)ll.add(Integer.parseInt(sp[j]));
            for(int j=0;j<n-2;j++){
                if(ll.get(j)!=j+1){
                    int index = findIndex(ll,j+1);
                    if((index-j)%2==0){
                        int val = ll.remove(index);
                        ll.add(j,val);
                    }
                    else{
                        int val =ll.remove(index);
                        ll.add(j,val);
                        int tmp = ll.remove(j+2);
                        ll.add(j+1,tmp);
                        
                    }
                }
            }
            String res = "YES";
            if(ll.get(n-1) != n)res = "NO";
            System.out.println(res);
        }
    }
    public static int findIndex(LinkedList<Integer> ll, int val){
        for(int i=val;i<ll.size();i++)if(ll.get(i)==val)return i;
            return -1;
    }
}

Problem solution in C++ programming.

#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <queue>
#include <map>
#include <set>
#include <list>
#include <utility>
#include <numeric>
#include <fstream>

using namespace std;

#define		ALL(c)	(c).begin(),(c).end()
#define		SZ(c)	int((c).size())
#define		LEN(s)	int((s).length())
#define		FOR(i,n)	for(int i=0;i<(n);++i)
#define		FORD(i,a,b)	for(int i=(a);i<=(b);++i)
#define		FORR(i,a,b)	for(int i=(b);i>=(a);--i)

typedef istringstream iss;
typedef ostringstream oss;
typedef long double ld;
typedef long long i64;
typedef pair<int,int> pii;

typedef vector<i64> vi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<vvvi> vvvvi;

typedef vector<string> vs;

int main()
{
	//freopen("input.txt", "r", stdin);
	ios_base::sync_with_stdio(0);

	int T;
	cin >> T;
	FOR(tt, T)
	{
		int n;
		cin >> n;
		vi a(n);
		FOR(i, n) cin >> a[i];
		int r = 0;
		FOR(i, n) FORD(j, i+1, n-1) if (a[i] > a[j]) r++;
		cout << (r % 2 == 0 ? "YES" : "NO") << endl;
	}

	return 0;	
}

Problem solution in C programming.

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

int main() {
    int t;
    scanf("%d", &t);
    for (int a0 = 0; a0 < t; a0++){
        int n;
        scanf("%d", &n);
        int A[n];
        for (int i = 0; i < n; i++){
            scanf("%d", &A[i]);
        }
        int aux, permutation = 0;
        for (int i = 0; i < n - 1; i++){
            if (A[i] != i + 1){
                for (int j = i + 1; j < n; j++){
                    if (A[j] == i + 1){
                        aux = A[i];
                        A[i] = A[j];
                        A[j] = aux;
                        permutation++;
                    }
                }
            }
        }
        if (permutation % 2 == 0){
            printf("YES");
        }
        else {
            printf("NO");
        }
        printf("n");
    } 
    return 0;
}

Problem solution in JavaScript programming.

function processData(input) {
    var lines = input.split('n');
    for (var c=0; c<parseInt(lines[0]); c++) {
        var len = parseInt(lines[c*2+1]);
        var arr = lines[c*2+2].split(' ').map(function(n){return parseInt(n)});
        
        var inv = 0;
        for (var i=0; i<arr.length; i++) {
            for (var j=i; j<arr.length; j++) {
                if (arr[i]>arr[j]) {
                    inv++
                }
            }
        }
        console.log((inv%2===0)?'YES':'NO');
    }
} 


process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});