Simple Array Sum – HackerRank Solution

Simple Array Sum – HackerRank Solution

Problem :

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.

Sample Input :

6
1 2 3 4 10 11

Sample Output :

31

Explanation :

We print the sum of the array’s elements: 1 + 2 + 3 + 4 + 10 + 11 = 31.

Solution :

#include <bits/stdc++.h>

using namespace std;

vector<string> split_string(string);

/*
 * Complete the simpleArraySum function below.
 */
int simpleArraySum(vector<int> ar) {
    int sum = 0;
    for (int i = 0; i < ar.size(); i++) {
        sum += ar[i];
    }
    return sum;
}

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;
}

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