HackerRank Sherlock and Array Problem Solution

In this post, we will solve HackerRank Sherlock and Array Problem Solution.

Watson gives Sherlock an array of integers. His challenge is to find an element of the array such that the sum of all elements to the left is equal to the sum of all elements to the right.
Example
arr = [5, 6, 8, 11]
8 is between two subarrays that sum to 11.
arr = [1]
The answer is [1] since left and right sum to 0.
You will be given arrays of integers and must determine whether there is an element that meets the criterion. If there is, return YES. Otherwise, return NO.

Function Description

Complete the balancedSums function in the editor below.

balancedSums has the following parameter(s):

int arr[n]: an array of integers

Returns

string: either YES or NO

Input Format
The first line contains T, the number of test cases.
The next T pairs of lines each represent a test case.
-The first line contains n, the number of elements in the array arr.
-The second line contains n space-separated integers arr[i] where 0 < i < n.

Sample Input 0

Sample Output 0

NO
YES

Explanation 0
For the first test case, no such index exists.
For the second test case, arr[0] + arr[1] = arr[3], therefore index 2 satisfies the given
conditions.

Sample Input 1

Sample Output 1

YES
YES
YES

Explanation 1
In the first test case, arr[2] = 4 is between two subarrays summing to 2. In the second case, arr[0] = 2 is between two subarrays summing to 0. In the third case, arr[2] = 2 is between two subarrays summing to 0.

HackerRank Sherlock and Array Problem Solution

Sherlock and Array C Solution

#include <stdio.h>

#define MAXN 100005

int main ()
{
    unsigned long a[MAXN], t, n, i;
    unsigned long long sum, leftSum[MAXN] = { 0 };
    scanf("%ld", &t);
    while ( t-- ) {
        scanf("%lu", &n);
        sum = 0;
        for ( i = 0; i < n; i++ ) {
            scanf("%lu", a+i);
            leftSum[i] = sum;
            sum += a[i];
        }
        for ( i = 0; i < n; i++ )
            if ( leftSum[i] == sum - a[i] - leftSum[i] ) {
                printf("YES\n");
                break;
            }
        if ( i == n ) 
            printf("NO\n");
    }
    return 0;
}

Sherlock and Array C++ Solution

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    int t;
    cin>>t;
    for(int i=0;i<t;i++){
        int n;
        cin>>n;
        vector<int>A;
        vector<long long> left;
        long long sum = 0 ; 
        for(int j=0; j<n; j++){
            int ai ;
            cin>>ai;
            A.push_back(ai);
            left.push_back(sum);
            //cerr<<"j= "<<j<<" , left sum="<<sum<<endl;
            sum += ai;
        }
        sum=0;
        bool found = false;
        for(int j=n-1; j>=0; j--){
            if(sum==left[j]){
                found = true;
                cout<<"YES"<<endl;
                break;
            }
            //cerr<<"j= "<<j<<" , right sum="<<sum<<endl;
            sum += A[j];
        }
       
        if(!found) cout<<"NO"<<endl;
    }
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    return 0;
}

Sherlock and Array C Sharp Solution

using System;

class Solution
{
    private static bool Exist( long [] arrNums )
    {
        // initialize cursors
        int iLeft = 0, iRight = arrNums.Length - 1;

        long diff = arrNums[ iLeft ] - arrNums[ iRight ];
        while ( iLeft < iRight )
        {
            // compare sum difference and move cursors
            if ( diff < 0 )
            {
                diff += arrNums[ ++iLeft ];
            }
            else if ( diff > 0 )
            {
                diff -= arrNums[ --iRight ];
            }
            else
            {
                diff += arrNums[ ++iLeft ] - arrNums[ --iRight ];
            }
        }

        return ( diff == 0 );
    }

    public static void Main( string [] args )
    {
        // input T
        int nCases = Convert.ToInt32( Console.ReadLine() );

        while ( nCases-- > 0 )
        {
            // input N
            int nNums = Convert.ToInt32( Console.ReadLine() );

            // input numbers
            var strNums = Console.ReadLine().Split();
            var arrNums = new long[ nNums ];
            for ( int i = 0; i < nNums; ++i )
            {
                arrNums[ i ] = Convert.ToInt32( strNums[ i ] );
            }

            // output results
            Console.WriteLine( Exist( arrNums ) ? "YES" : "NO" );
        }

        return;
    }
}

Sherlock and Array Java Solution

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

    /*
     * Complete the 'balancedSums' function below.
     *
     * The function is expected to return a STRING.
     * The function accepts INTEGER_ARRAY arr as parameter.
     */

    public static String balancedSums(List<Integer> arr) {
    // Write your code here
        int totalsum = 0,leftsum = 0 , rightsum = 0;
        
        for(int i = 0; i<arr.size(); i++){
            totalsum = totalsum + arr.get(i);
        }
        
        for(int i = 0; i<arr.size(); i++){
            rightsum = totalsum - arr.get(i) - leftsum;
            if(leftsum == rightsum){
                return "YES";
            }
            else{
                leftsum = leftsum +arr.get(i);
            }
        }
        return "NO";
    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int T = Integer.parseInt(bufferedReader.readLine().trim());

        IntStream.range(0, T).forEach(TItr -> {
            try {
                int n = Integer.parseInt(bufferedReader.readLine().trim());

                List<Integer> arr = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
                    .map(Integer::parseInt)
                    .collect(toList());

                String result = Result.balancedSums(arr);

                bufferedWriter.write(result);
                bufferedWriter.newLine();
            } catch (IOException ex) {
                throw new RuntimeException(ex);
            }
        });

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Sherlock and Array JavaScript Solution

function check(arr) {
  var sum = arr.reduce(function(el, s){ return s + el; });
  var p1 = 0, p2;
  var answer = 'NO';
  
  for (var i = 1; i < arr.length; i++) {
    p1 += arr[i-1];
    p2 = sum - p1 - arr[i];
    if (p1 === p2) {
      answer = 'YES';
      break;
    }
    if (p1 > p2) {
      break;
    }
  }
  
  if (arr.length === 1) {
    answer = 'YES';
  }
  
  console.log(answer);
} 

function processData(input) {
    var parse_fun = function (s) { return parseInt(s, 10); };

    var lines = input.split('\n');
    var T = parse_fun(lines.shift());
    
    for (var i=0;i<T;++i) {
        lines.shift();
        var data = lines.shift().split(' ').map(parse_fun);
        check(data);
    }
}

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

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

Sherlock and Array Python Solution

if __name__ == "__main__":
    t = int(input())
    for asdfs in range(t):
        n = int(input())
        tab = list(map(int, input().split()))
        left = 0
        right = 0
        index = 0
        i = 0
        j = n-1
        while True:
            if index == n-1:
                if left == right:
                    print("YES")
                else:
                    print("NO")
                break
            if left > right:
                right = right + tab[j]
                j = j - 1
            else:
                left = left + tab[i]
                i = i + 1
            index = index + 1

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