**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:3Explanation: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:1.8Explanation: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