# Average of all the elements in the array

Problem Statement: Given an array, we have to find the average of all the elements in the array.

Examples:

```Example 1:
Input: N = 5, array[] = {1,2,3,4,5}
Output: 3
Explanation: Average is the sum of all the elements divided by number of elements.Therefore (1+2+3+4+5)/5 = 3.

Example 2:
Input:  N=6, array[] = {1,2,1,1,5,1}
Output: 1.8
Explanation: Average is the sum of all the elements divided by number of elements.Therefore (1+2+1+1+5+1)/6 = 1.8```

### Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Approach

• Using for loop traverse through the array and while traversing  maintain  a variable for storing sum of the elements in the array.
• After completing the traversal simply divide the sum by no. of elements in the array. Code:

## C++ Code

``````#include<bits/stdc++.h>
using namespace std;
int main() {
int n = 5;
int arr[] = {1, 2, 3, 4, 5};
double sum = 0;
for (int i = 0; i < n; i++) {
sum += (double)arr[i];
}
double average = sum / n;
cout << "The average is "<<average;

}
``````

Output:

The average is 3

Time Complexity: O(n) As we are traversing the array once.

Space Complexity: O(1).

## Java Code

``````public class Main {
public static void main(String args[]) {
int n = 5;
int arr[] = {1,2,3,4,5};
//using double as average can be in decimal.
double sum = 0;
for (int i = 0; i < n; i++) {
sum += (double) arr[i];
}
double average = sum / n;
System.out.println("The average is "+average);
}
}
``````

Output:

The average is 3.0

Time Complexity: O(n) As we are traversing the array once.

Space Complexity: O(1).

Special thanks to Pranav Padawe for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article