Average numbers in array in C Programming
There are n number of elements stored in an array and this program calculates the average of those numbers. Using different methods.
Input - 1 2 3 4 5 6 7
Output - 4
Explanation - Sum of elements of array 1+2+3+4+5+6+7=28
No of element in array=7
Average=28/7=4
There are two methods
Method 1 −Iterative
In this method we will find sum and divide the sum by the total number of elements.
Given array arr[] and size of array n
Input - 1 2 3 4 5 6 7
Output - 4
Explanation - Sum of elements of array 1+2+3+4+5+6+7=28
No of element in array=7
Average=28/7=4
Example
#include<iostream>
using namespace std;
int main() {
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n=7;
int sum = 0;
for (int i=0; i<n; i++) {
sum += arr[i];
}
float average = sum/n;
cout << average;
return 0;
}Method 2 − Recursive
The idea is to pass index of element as an additional parameter and recursively compute sum. After computing sum, divide the sum by n.
Given array arr[], size of array n and initial index i
Input - 1 2 3 4 5
Output - 3
Explanation - Sum of elements of array 1+2+3+4+5=15
No of element in array=5
Average=15/5=3
Example
#include <iostream>
using namespace std;
int avg(int arr[], int i, int n) {
if (i == n-1) {
return arr[i];
}
if (i == 0) {
return ((arr[i] + avg(arr, i+1, n))/n);
}
return (arr[i] + avg(arr, i+1, n));
}
int main() {
int arr[] = {1, 2, 3, 4, 5};
int n = 5;
cout << avg(arr,0, n) << endl;
return 0;
}
Published on 09-Aug-2019 13:25:59
