Count Odd Even

Problem

Given an array A[] of N elements. The task is to return the count of the number of odd and even elements in the array.

Examples:

Example 1:

Input:
N = 5
A[] = 1 2 3 4 5
Output:
3 2
Explanation:
There are 3 odd elements (1, 3, 5) and 2 even elements (2 and 4).

Your task:

Your task is to complete the function countOddEven() which should return the count of odd and even elements of the array.

• Expected Time Complexity: $O(N)$
• Expected Auxiliary Space: $O(1)$

Constraints:

• $1<=N<=10^6$
• $1<=A_j<=10^6$

Solution

Python

def countOddEven(self, arr, n):
    odd, even = 0,0
    for i in range(0, n):
        if arr[i]%2==0:
            even+=1
        else:
            odd+=1
    return odd, even

Java

public int[] countOddEven(int[] arr, int n) {
    int odd = 0, even = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] % 2 == 0) {
            even++;
        } else {
            odd++;
        }
    }
    int[] result = { odd, even };
    return result;
}

C++

vector<int> countOddEven(int arr[], int sizeof_array) {
    int odd = 0, even = 0;
    for (int i = 0; i < sizeof_array; i++) {
        if (arr[i] % 2 == 0) {
            even++;
        } else {
            odd++;
        }
    }
    vector<int> result = { odd, even };
    return result;
}

C

void countOddEven(int arr[], int sizeof_array, int *odd, int *even) {
    *odd = 0;
    *even = 0;
    for (int i = 0; i < sizeof_array; i++) {
        if (arr[i] % 2 == 0) {
            (*even)++;
        } else {
            (*odd)++;
        }
    }
}

• Time Complexity: $O(N)$
• Auxiliary Space: $O(1)$