C program to count numbers of even and odd elements in an array
Count the number of even and odd elements in C
Here, in this page we will discuss the program to count the number of even and odd elements in C programming language. We are given with an array and need to print the count of even elements and odd elements.
Method Discussed :
- Method 1 : Using Modulo Operator
- Method 2 : Using Bit-wise operator
Method 1 :
- Declare two variable say even_count=0 and odd_count = 0.
- Start Iterating over the array.
- Check if(arr[i]%2==0) then increment even_count by 1
- Else odd_count by 1
Time and Space Complexity :
- Time Complexity : O(n)
- Space Complexity : O(1)
Modulo Operator
The modulo operator, denoted by %, is an arithmetic operator. The modulo division operator produces the remainder of an integer division. Syntax: If x and y are integers, then the expression: x % y. produces the remainder when x is divided by y.So, if a number is even then it return 0 on modulo it by 2, otherwise it return 1
Method 1 : Code in C
Run
#include<stdio.h> int main(){ int arr[] = {1, 7, 8, 4, 5, 16, 8}; int n = sizeof(arr)/sizeof(arr[0]); int even_count=0, odd_count=0; for(int i=0; i<n; i++){ if(arr[i]%2==0) even_count++; else odd_count++; } printf("Even Elements count : %d \nOdd Elements count : %d", even_count, odd_count); }
Output :
Even Elements count : 4 Odd Elements count : 3
Method 2 :
In this method we will use bit-wise AND operator. By doing AND of 1 with array element, if the result comes out to be 1 then the number is odd otherwise even.
Method 2 : Code in C
Run
#include<stdio.h> int main(){ int arr[] = {1, 7, 8, 4, 5, 16, 8}; int n = sizeof(arr)/sizeof(arr[0]); int even_count=0, odd_count=0; for(int i=0; i<n; i++){ if((arr[i]&1)==0) even_count++; else odd_count++; } printf("Even Elements count : %d \nOdd Elements count : %d", even_count, odd_count); }
Output :
Even Elements count : 4 Odd Elements count : 3
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