Given an array arr of size N, the task is to calculate the average of all the elements of the given array.
Example
Input: arr[] = {1, 2, 3, 4, 5} Output: 3
Approach
1. We can simply find the sum of all elements of a given array.
2. Divide this sum by the total number of elements present in the array.
Sum of the elements is = 1+2+3+4+5 = 15
The total number of elements = 5.
Average = 15/5 = 3
Pseudocode
procedure average(arr) Declare sum as integer FOR EACH value in arr DO sum ← sum + arr[n] END FOR avg ← sum / size_of_array Display avg end procedure
Program to Calculate Average of All Elements of an Array
Below is the C program to implement the above approach:
C
// C program to demonstrate // average of array elements #include <stdio.h> // Function that return average // of given array. double average( int a[], int n)
{ // Find the sum of array element
int sum = 0;
for ( int i = 0; i < n; i++)
sum += a[i];
return ( double )sum / n;
} // Driver code int main()
{ // Input array
int arr[] = { 1, 2, 3, 4, 5 };
// Size of array
int n = sizeof (arr) / sizeof (arr[0]);
// average(arr, n) function is return the
// average of the array.
int avg = average(arr, n);
// Display average of given array
printf ( "Average = %d " , avg);
return 0;
} |
Output
Average = 3
The complexity of the above method
Time Complexity: O(n)
Space Complexity: O(1)