Sum of array elements

Beginner Problems Basic Arrays Easy Go Ad-Free - ₹20/mo

Given an array arr of size n, the task is to find the sum of all the elements in the array.

Examples:

Input: n=5, arr = [1,2,3,4,5]

Output: 15

Explanation: Sum of all the elements is 1+2+3+4+5 = 15

Input: n=6, arr = [1,2,1,1,5,1]

Output: 11

Explanation: Sum of all the elements is 1+2+1+1+5+1 = 11

Input: n=3, arr = [2,1,1]

Constraints

1 <= n <= 10⁵
1 <= arr[i] <= 10⁴

Company Tags

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Editorial

Intuition:

To sum an array, begin by initializing the sum to zero, representing the starting point with no accumulated value. Then, sequentially add each element of the array to this sum. This approach ensures that, by the end of the process, the sum reflects the total value of all elements in the array.

Approach:

Initialize a variable sum to 0. This variable will hold the total sum of the array elements.
Iterate through each element in the array using a loop. For each element, add it to the sum variable.
After the loop ends, the sum variable will contain the total sum of all the elements in the array.

Dry Run

Solution

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    int sum(int arr[], int n) {
        // Initialize sum to 0
        int sum = 0;
        // Iterate through each element in the array
        for (int i = 0; i < n; i++) {
            // Add each element to the sum
            sum += arr[i];
        }
        return sum;
    }
};

// The main function
int main() {
    Solution sol;
    int arr[] = {1, 2, 3, 4, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Sum of array elements: " << sol.sum(arr, n) << endl;
    return 0;
}
class Solution {
    public int sum(int[] arr, int n) {
        // Initialize sum to 0
        int sum = 0;
        // Iterate through each element in the array
        for (int i = 0; i < n; i++) {
            // Add each element to the sum
            sum += arr[i];
        }
        return sum;
    }
   // The main method
    public static void main(String[] args) {
        Solution sol = new Solution();
        int[] arr = {1, 2, 3, 4, 5};
        int n = arr.length;
        System.out.println("Sum of array elements: " + sol.sum(arr, n));
    }
}
class Solution:
    def sum(self, arr, n):
        # Initialize sum to 0
        total_sum = 0
        # Iterate through each element in the array
        for num in arr:
            # Add each element to the sum
            total_sum += num
        return total_sum

#The main method
if __name__ == "__main__":
    sol = Solution()
    arr = [1, 2, 3, 4, 5]
    n = len(arr)
    print("Sum of array elements:", sol.sum(arr, n))
class Solution {
    sum(arr, n) {
        // Initialize sum to 0
        let total_sum = 0;
        // Iterate through each element in the array
        for (let i = 0; i < n; i++) {
            // Add each element to the sum
            total_sum += arr[i];
        }
        return total_sum;
    }
}

// Main method
const sol = new Solution();
const arr = [1, 2, 3, 4, 5];
const n = arr.length;
console.log("Sum of array elements:", sol.sum(arr, n));

Time Complexity: O(N) – Array with N elements is traversed, once

Space Complexity: O(1) – A single variable is used to store the sum, regardless of the array size.

Notes

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