In this HackerRank Simple Array Sum problem solution, Given an array of integers, find the sum of its elements.
For example, if the array ar = [1,2,3],1+2+3 = 6, so return 6.
Function Description
Complete the simpleArraySum function in the editor below. It must return the sum of the array elements as an integer.
simpleArraySum has the following parameter(s):
ar: an array of integers
Input Format
The first line contains an integer, n, denoting the size of the array.
The second line contains n space-separated integers representing the array’s elements.
Constraints
0 < n, ar[i] <= 1000
Output Format
Print the sum of the array’s elements as a single integer.

Problem solution in Python programming.
#!/bin/python3
import os
import sys
#
# Complete the simpleArraySum function below.
#
def simpleArraySum(ar):
x=0
for i in range(0,ar_count):
x = x + ar[i]
return x
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
ar_count = int(input())
ar = list(map(int, input().rstrip().split()))
result = simpleArraySum(ar)
fptr.write(str(result) + 'n')
fptr.close()
Problem solution in Java Programming.
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) throws Exception
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
br.readLine();
int output=0;
String[] input = br.readLine().split(" ");
for(String value:input)
{
output += Integer.parseInt(value);
}
System.out.println(output);
}
}
Problem solution in C++ programming.
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int numLines;
int currNumber, total = 0;
cin >> numLines;
for (int i=0; i<numLines;i++) {
cin >> currNumber;
total += currNumber;
}
cout << total;
return 0;
}
Problem solution in C programming.
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int N;
int map[1000];
int main() {
long int sum =0;
scanf("%d",&N);
for(int i = 0; i<N; i++)
{
scanf("%d",&map[i]);
}
for(int i=0; i<N;i++)
{
sum += map[i];
}
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
printf("%ld",sum);
return 0;
}
Problem solution in JavaScript programming.
function processData(input) {
console.log(input.split(/n/)[1].split(/s/).map(Number).reduce(function(a, b) {
return a + b;
}))
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});