Algorithm
Problem Name: Algorithms -
In this HackerRank Functions in Algorithms - Java programming 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
nput Format
The first line contains an integer, n , denoting the size of the array.
The second line contains n pace-separated integers representing the array's elements.
Constraints
0 <= ni ,ar [i] <= 1000
Output Format
Print the sum of the array's elements as a single integer.
Sample Input
6
1 2 3 4 10 11
Sample Output
31
Explanation
1 + 2 + 3 + 4 + 10 + 11 = 31.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <assert.h>
#include <limits.h>
#include <math.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* readline();
char** split_string(char*);
/*
* Complete the simpleArraySum function below.
*/
int simpleArraySum(int ar_count, int* ar) {
/*
* Write your code here.
*/
int sum = 0;
for(int i = 0; i < ar_count; i++){
sum += ar[i];
}
return sum;
}
int main()
{
FILE* fptr = fopen(getenv("OUTPUT_PATH"), "w");
char* ar_count_endptr;
char* ar_count_str = readline();
int ar_count = strtol(ar_count_str, &ar_count_endptr, 10);
if (ar_count_endptr == ar_count_str || *ar_count_endptr != '\0') { exit(EXIT_FAILURE); }
char** ar_temp = split_string(readline());
int ar[ar_count];
for (int ar_itr = 0; ar_itr < ar_count; ar_itr++) {
char* ar_item_endptr;
char* ar_item_str = ar_temp[ar_itr];
int ar_item = strtol(ar_item_str, &ar_item_endptr, 10);
if (ar_item_endptr == ar_item_str || *ar_item_endptr != '\0') { exit(EXIT_FAILURE); }
ar[ar_itr] = ar_item;
}
int result = simpleArraySum(ar_count, ar);
fprintf(fptr, "%d\n", result);
fclose(fptr);
return 0;
}
char* readline() {
size_t alloc_length = 1024;
size_t data_length = 0;
char* data = malloc(alloc_length);
while (true) {
char* cursor = data + data_length;
char* line = fgets(cursor, alloc_length - data_length, stdin);
if (!line) { break; }
data_length += strlen(cursor);
if (data_length < alloc_length - 1 || data[data_length - 1] == '\n') { break; }
size_t new_length = alloc_length << 1;
data = realloc(data, new_length);
if (!data) { break; }
alloc_length = new_length;
}
if (data[data_length - 1] == '\n') {
data[data_length - 1] = '\0';
}
data = realloc(data, data_length);
return data;
}
char** split_string(char* str) {
char** splits = NULL;
char* token = strtok(str, " ");
int spaces = 0;
while (token) {
splits = realloc(splits, sizeof(char*) * ++spaces);
if (!splits) {
return splits;
}
splits[spaces - 1] = token;
token = strtok(NULL, " ");
}
return splits;
}
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
#include <numeric>
using namespace std;
vector < string> split_string(string);
/*
* Complete the simpleArraySum function below.
*/
int simpleArraySum(vector<int> ar) {
/*
* Write your code here.
*/
return accumulate(ar.begin(), ar.end(), 0);
}
int main()
{
ofstream fout(getenv("OUTPUT_PATH"));
int ar_count;
cin >> ar_count;
cin.ignore(numeric_limits < streamsize>::max(), '\n');
string ar_temp_temp;
getline(cin, ar_temp_temp);
vector < string> ar_temp = split_string(ar_temp_temp);
vector<int> ar(ar_count);
for (int ar_itr = 0; ar_itr < ar_count; ar_itr++) {
int ar_item = stoi(ar_temp[ar_itr]);
ar[ar_itr] = ar_item;
}
int result = simpleArraySum(ar);
fout << result << "\n";
fout.close();
return 0;
}
vector < string> split_string(string input_string) {
string::iterator new_end = unique(input_string.begin(), input_string.end(), [] (const char &x, const char &y) {
return x == y and x == ' ';
});
input_string.erase(new_end, input_string.end());
while (input_string[input_string.length() - 1] == ' ') {
input_string.pop_back();
}
vector < string> splits;
char delimiter = ' ';
size_t i = 0;
size_t pos = input_string.find(delimiter);
while (pos != string::npos) {
splits.push_back(input_string.substr(i, pos - i));
i = pos + 1;
pos = input_string.find(delimiter, i);
}
splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1));
return splits;
}
#3 Code Example with Java Programming
Code -
Java Programming
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) {
int sum = 0;
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int arr[] = new int[n];
for(int arr_i=0; arr_i < n; arr_i++){
arr[arr_i] = in.nextInt();
sum += arr[arr_i];
}
System.out.println(sum);
}
}
#4 Code Example with Javascript Programming
Code -
Javascript Programming
'use strict';
const fs = require('fs');
process.stdin.resume();
process.stdin.setEncoding('utf-8');
let inputString = '';
let currentLine = 0;
process.stdin.on('data', inputStdin => {
inputString += inputStdin;
});
process.stdin.on('end', _ => {
inputString = inputString.trim().split('\n').map(str => str.trim());
main();
});
function readLine() {
return inputString[currentLine++];
}
/*
* Complete the simpleArraySum function below.
*/
function simpleArraySum(ar) {
let sum = 0;
for (let i = 0; i < ar.length; i++){
sum += ar[i]
}
return sum;
}
function main() {
const ws = fs.createWriteStream(process.env.OUTPUT_PATH);
const arCount = parseInt(readLine(), 10);
const ar = readLine().split(' ').map(arTemp => parseInt(arTemp, 10));
let result = simpleArraySum(ar);
ws.write(result + "\n");
ws.end();
}
#5 Code Example with C# Programming
Code -
C# Programming
using System;
using System.Linq;
class Solution
{
static void Main(String[] args)
{
//no need of the element count as I use LINQ to create the sum instead of iterating the array explicitly in my code.
Console.ReadLine();
var ar_temp = Console.ReadLine().Split(' ');
var ar = Array.ConvertAll(ar_temp, Int32.Parse);
Console.WriteLine(ar.Sum()); //LINQ's sum method has O(n) time complexity.
}
}