Count of odd numbers in array

Beginner Problems Basic Arrays Easy

Given an array of n elements. The task is to return the count of the number of odd numbers in the array.

Examples:

Input: n=5, array = [1,2,3,4,5]

Output: 3

Explanation: The three odd elements are (1,3,5).

Input: n=6, array = [1,2,1,1,5,1]

Output: 5

Explanation: The five odd elements are one 5 and four 1's.

Input: n=5, array = [1,3,5,7,9]

Constraints

1 <= n <= 10⁵
1 <= arr[i] <= 10⁴

Company Tags

TCS Cognizant Accenture Infosys Capgemini Wipro IBM HCL Tech Mahindra MindTree

Editorial

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Intuition:

A simple way to solve this problem is by traversing the whole array and checking each element if it is odd. The count of the odd numbers found will be the answer.

Approach:

Initialize a counter to zero to keep track of odd numbers. Initialize a for loop to iterate through each element in the array.
For each element, check if it is odd by checking it divisibility by 2 and increment counter if odd.
After iterating through all elements, the counter will contain the total count of odd numbers.

Dry Run:

Solution:

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    
    // Function to count the odd numbers in an array 
    int countOdd(int arr[], int n) {
        // Set counter to 0
        int count = 0;
        // Iterate through the array
        for (int i = 0; i < n; i++) {
        // check for odd values and increment
            if (arr[i] % 2 != 0) {
                count++;
            }
        }
        return count;
    }
};

//Main method
int main() {
    // Creating an instance of Solution class
    Solution sol;
    int arr[] = {1, 2, 3, 4, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    
    // Function call to count the odd numbers in an array 
    int count = sol.countOdd(arr, n);
    cout << "Count of odd numbers: " << count << endl;
    return 0;
}
class Solution {
    
    // Function to count the odd numbers in an array 
    public int countOdd(int[] arr, int n) {
        int count = 0;
        // Iterate through the array
        for (int i = 0; i < n; i++) {
        //  Check for odd values and increment
            if (arr[i] % 2 != 0) {
                count++;
            }
        }
        return count;
    }
    
    //Main method
    public static void main(String[] args) {
        
        // Creating an instance of Solution class
        Solution sol = new Solution();
        int[] arr = {1, 2, 3, 4, 5};
        int n = arr.length;
        
        // Function call to count the odd numbers in an array
        int count = sol.countOdd(arr, n);
        System.out.println("Count of odd numbers: " + count);
    }
}
class Solution:
    # Function to count the odd numbers in an array 
    def countOdd(self, arr, n):
        count = 0
        # Iterate through the array
        for num in arr:
        # Check for odd values and increment
            if num % 2 != 0:
                count += 1
        return count
        
# Main method
if __name__ == "__main__":
    # Creating an instance of Solution class
    sol = Solution()
    arr = [1, 2, 3, 4, 5]
    n = len(arr)
    
    # Function call to count the odd numbers in an array 
    count = sol.countOdd(arr, n)
    print("Count of odd numbers:", count)
class Solution {
    // Function to count the odd numbers in an array 
    countOdd(arr, n) {
        let count = 0;
        // Iterate through the array
        for (let i = 0; i < n; i++) {
       // Check for odd values and increment
            if (arr[i] % 2 !== 0) {
                count++;
            }
        }
        return count;
    }
}

// Creating an instance of Solution class
const sol = new Solution();
const arr = [1, 2, 3, 4, 5];
const n = arr.length;

// Function to count the odd numbers in an array
const count = sol.countOdd(arr, n);
console.log("Count of odd numbers:", count);

Complexity Analysis:

Time Complexity : O(N)
Each element in the array has to be inspected once to determine if it's odd, resulting in a linear time complexity where N is the number of elements in the array.

Space Complexity : O(1)
The space used is constant, as we only use a single counter regardless of the size of the input array.

Code

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