Count number of odd digits in a number

Beginner Problems Basic Maths Easy

You are given an integer n. You need to return the number of odd digits present in the number.

The number will have no leading zeroes, except when the number is 0 itself.

Examples:

Input: n = 5

Output: 1

Explanation: 5 is an odd digit.

Input: n = 25

Output: 1

Explanation: The only odd digit in 25 is 5.

Input: n = 15

Constraints

0 <= n <= 5000
n will contain no leading zeroes except when it is 0 itself.

Company Tags

TCS Cognizant Accenture Infosys Capgemini Wipro IBM HCL Tech Mahindra MindTree

Editorial

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Intuition:

Given a number, all the digits in the number can be extracted one by one from right to left which can be checked for even and odd.

Approach:

The last digit of the given number can be found by using the modulus operator (used to find the remainder for any division) with the number 10.
Iterate on the original number till there are digits left, extract the last (rightmost) digit, and check whether the digit is odd or not. In every iteration, divide the original number by 10 so that the remaining digits can be extracted in the next iterations.
Keep a counter to count the number of odd digits found in the number and every time an odd digit is encountered, increment the counter.

Dry Run:

Solution:

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

class Solution {
public: 
    /* Function to count number
    of odd digits in N */
    int countOddDigit(int n) {
        /* Counter to store the 
        number of odd digits */
        int oddDigits = 0;

        // Iterate till there are digits left
        while (n > 0) {
            // Extract last digit
            int lastDigit = n % 10;
            
            // Check if digit is odd
            if (lastDigit % 2 != 0) {
                // Increment counter
                oddDigits = oddDigits + 1;
            }
            n = n / 10;
        }

        return oddDigits;
    }
};

int main() {
    int n = 6678;
    
    /* Creating an instance of 
    Solution class */
    Solution sol; 
    
    // Function call to get count of odd digits in n
    int ans = sol.countOddDigit(n);
    cout << "The count of odd digits in the given number is: " << ans;
    
    return 0;
}
class Solution {
    /* Function to count number
    of odd digits in N */
    public int countOddDigit(int n) {
        /* Counter to store the 
        number of odd digits */
        int oddDigits = 0;

        // Iterate till there are digits left
        while (n > 0) {
            // Extract last digit
            int lastDigit = n % 10;
            
            // Check if digit is odd
            if (lastDigit % 2 != 0) {
                // Increment counter
                oddDigits = oddDigits + 1;
            }
            n = n / 10;
        }

        return oddDigits;
    }

    public static void main(String[] args) {
        int n = 6678;
        
        /* Creating an instance of 
        Solution class */
        Solution sol = new Solution(); 
        
        // Function call to get count of odd digits in n
        int ans = sol.countOddDigit(n);
        System.out.println("The count of odd digits in the given number is: " + ans);
    }
}
class Solution:
    # Function to count number
    # of odd digits in N
    def countOddDigit(self, n):
        # Counter to store the 
        # number of odd digits
        oddDigits = 0

        # Iterate till there are digits left
        while n > 0:
            # Extract last digit
            lastDigit = n % 10
            
            # Check if digit is odd
            if lastDigit % 2 != 0:
                # Increment counter
                oddDigits = oddDigits + 1
            n = n // 10

        return oddDigits

# Input number
n = 6678

# Creating an instance of 
# Solution class
sol = Solution()

# Function call to get count of odd digits in n
ans = sol.countOddDigit(n)
print("The count of odd digits in the given number is:", ans)
class Solution {
    /* Function to count number
    of odd digits in N */
    countOddDigit(n) {
        /* Counter to store the 
        number of odd digits */
        let oddDigits = 0;

        // Iterate till there are digits left
        while (n > 0) {
            // Extract last digit
            let lastDigit = n % 10;
            
            // Check if digit is odd
            if (lastDigit % 2 !== 0) {
                // Increment counter
                oddDigits = oddDigits + 1;
            }
            n = Math.floor(n / 10);
        }

        return oddDigits;
    }
}

// Input number
let n = 6678;

/* Creating an instance of 
Solution class */
let sol = new Solution();

// Function call to get count of odd digits in n
let ans = sol.countOddDigit(n);
console.log("The count of odd digits in the given number is: " + ans);

Complexity Analysis:

Time Complexity: O(log₁₀(N)) – In every iteration we are dividing N by 10 (equivalent to the number of digits in N).

Space Complexity: O(1) – Using only couple of variables i.e., constant space.

Code

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