Count Even and Odd in Array

Easy Arrays

Count how many even and odd numbers are in an array.

Sample input

[1, 2, 3, 4, 5, 6]

Sample output

Even: 3, Odd: 3

Solution

arr = [1, 2, 3, 4, 5, 6]
even = sum(1 for x in arr if x % 2 == 0)
odd = len(arr) - even
print(f"Even: {even}, Odd: {odd}")

const arr = [1, 2, 3, 4, 5, 6];
let even = 0, odd = 0;
for (const x of arr) {
  if (x % 2 === 0) even++;
  else odd++;
}
console.log(`Even: ${even}, Odd: ${odd}`);

public class Main {
    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4, 5, 6};
        int even = 0, odd = 0;
        for (int x : arr) {
            if (x % 2 == 0) even++;
            else odd++;
        }
        System.out.println("Even: " + even + ", Odd: " + odd);
    }
}

fun main() {
    val arr = intArrayOf(1, 2, 3, 4, 5, 6)
    val even = arr.count { it % 2 == 0 }
    val odd = arr.size - even
    println("Even: $even, Odd: $odd")
}

let arr = [1, 2, 3, 4, 5, 6]
let even = arr.filter { $0 % 2 == 0 }.count
let odd = arr.count - even
print("Even: \(even), Odd: \(odd)")

void main() {
  List<int> arr = [1, 2, 3, 4, 5, 6];
  int even = 0, odd = 0;
  for (var x in arr) {
    if (x % 2 == 0) {
      even++;
    } else {
      odd++;
    }
  }
  print('Even: $even, Odd: $odd');
}

#include <iostream>
using namespace std;

int main() {
    int arr[] = {1, 2, 3, 4, 5, 6};
    int n = 6, even = 0, odd = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] % 2 == 0) even++;
        else odd++;
    }
    cout << "Even: " << even << ", Odd: " << odd << endl;
    return 0;
}

#include <stdio.h>

int main() {
    int arr[] = {1, 2, 3, 4, 5, 6};
    int n = 6, even = 0, odd = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] % 2 == 0) even++;
        else odd++;
    }
    printf("Even: %d, Odd: %d\n", even, odd);
    return 0;
}