Count number of even and odd elements in an array
For the given array of integers, count even and odd elements.
Examples:
Input:
int arr[5] = {2, 3, 4, 5, 6}
Output:
Number of even elements = 3
Number of odd elements = 2
Input:
int arr[5] = {22, 32, 42, 52, 62}
Output:
Number of even elements = 5
Number of odd elements = 0
Solution:
We can also check if a number is odd or even
- By doing AND of 1 and that digit, if the result comes out to be 1 then the number is odd otherwise even.
- By its divisibility by 2. A number is said to be odd if it is not divisible by 2, otherwise its even.
Here, we will check if a number is odd, then we will increment the odd counter otherwise we will increment the even counter.
Below is the implementation of the above approach:
C++
// CPP program to count number of even// and odd elements in an array#include <iostream>using namespace std;void CountingEvenOdd(int arr[], int arr_size){ int even_count = 0; int odd_count = 0; // loop to read all the values in the array for (int i = 0; i < arr_size; i++) { // checking if a number is completely // divisible by 2 if (arr[i] & 1 == 1) odd_count++; else even_count++; } cout << "Number of even elements = " << even_count << "\nNumber of odd elements = " << odd_count;}// Driver Codeint main(){ int arr[] = { 2, 3, 4, 5, 6 }; int n = sizeof(arr) / sizeof(arr[0]); // Function Call CountingEvenOdd(arr, n);} |
Java
// JAVA program to count number of even// and odd elements in an arrayimport java.io.*;class GFG { static void CountingEvenOdd(int arr[], int arr_size) { int even_count = 0; int odd_count = 0; // loop to read all the values in // the array for (int i = 0; i < arr_size; i++) { // checking if a number is // completely divisible by 2 if ((arr[i] & 1) == 1) odd_count++; else even_count++; } System.out.println("Number of even" + " elements = " + even_count + " Number of odd elements = " + odd_count); } // Driver Code public static void main(String[] args) { int arr[] = { 2, 3, 4, 5, 6 }; int n = arr.length; // Function Call CountingEvenOdd(arr, n); }}// This code is Contributed by anuj_67. |
Python3
# Python3 program to count number of# even and odd elements in an arraydef CountingEvenOdd(arr, arr_size): even_count = 0 odd_count = 0 # loop to read all the values # in the array for i in range(arr_size): # checking if a number is # completely divisible by 2 if (arr[i] & 1 == 1): odd_count += 1 else: even_count += 1 print("Number of even elements = ", even_count) print("Number of odd elements = ", odd_count)# Driver Codearr = [2, 3, 4, 5, 6]n = len(arr)# Function CallCountingEvenOdd(arr, n)# This code is contributed by sahishelangia |
C#
// C# program to count number of even// and odd elements in an arrayusing System;class GFG { static void CountingEvenOdd(int[] arr, int arr_size) { int even_count = 0; int odd_count = 0; // loop to read all the values in // the array for (int i = 0; i < arr_size; i++) { // checking if a number is // completely divisible by 2 if ((arr[i] & 1) == 1) odd_count++; else even_count++; } Console.WriteLine("Number of even" + " elements = " + even_count + " Number of odd elements = " + odd_count); } // Driver Code public static void Main() { int[] arr = { 2, 3, 4, 5, 6 }; int n = arr.Length; // Function Call CountingEvenOdd(arr, n); }}// This code is Contributed by anuj_67. |
PHP
<?php// PHP program to count number of even// and odd elements in an arrayfunction CountingEvenOdd( $arr, $arr_size){ $even_count = 0; $odd_count = 0; // loop to read all the values in // the array for( $i = 0 ; $i < $arr_size ; $i++) { // checking if a number is // completely divisible by 2 if ($arr[$i] & 1 == 1) $odd_count ++ ; else $even_count ++ ; } echo "Number of even elements = " , $even_count," Number of odd " , "elements = " ,$odd_count ; }// Driver Code $arr = array(2, 3, 4, 5, 6); $n = count($arr); // Function Call CountingEvenOdd($arr, $n);// This code is Contributed by anuj_67.?> |
Output
Number of even elements = 3 Number of odd elements = 2
Time Complexity: O(n)
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