LeetCode 2520: Count the Digits That Divide a Number Solution

Master LeetCode problem 2520 (Count the Digits That Divide a Number), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2520. Count the Digits That Divide a Number

Easy

Math

Problem Explanation

Explanation:

To solve this problem, we iterate through each digit of the given number and check if the number is divisible by that digit. If it is, we increment a counter. Finally, we return the counter as the result.

Initialize a counter variable to keep track of the count of divisible digits.
Iterate through each digit of the given number.
Check if the digit is not zero and the number is divisible by that digit.
If the above conditions are met, increment the counter.
Return the counter as the result.

Time Complexity:

The time complexity of this algorithm is O(log n), where n is the given number. This is because we are iterating through each digit of the number.

Space Complexity:

The space complexity of this algorithm is O(1) as we are using only a constant amount of space.

Solution Code

class Solution {
    public int countDividingDigits(int num) {
        int count = 0;
        int n = num;
        while (n > 0) {
            int digit = n % 10;
            if (digit != 0 && num % digit == 0) {
                count++;
            }
            n /= 10;
        }
        return count;
    }
}

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Frequently Asked Questions

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What is the time complexity of LeetCode 2520 (Count the Digits That Divide a Number)?

The time complexity for LeetCode 2520 (Count the Digits That Divide a Number) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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