LeetCode #2520 — EASY

Count the Digits That Divide a Number

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

Problem Statement

Given an integer num, return the number of digits in num that divide num.

An integer val divides nums if nums % val == 0.

Example 1:

Input: num = 7
Output: 1
Explanation: 7 divides itself, hence the answer is 1.

Example 2:

Input: num = 121
Output: 2
Explanation: 121 is divisible by 1, but not 2. Since 1 occurs twice as a digit, we return 2.

Example 3:

Input: num = 1248
Output: 4
Explanation: 1248 is divisible by all of its digits, hence the answer is 4.

Constraints:

1 <= num <= 10⁹
num does not contain 0 as one of its digits.

Step 01

Brute Force Baseline

Problem summary: Given an integer num, return the number of digits in num that divide num. An integer val divides nums if nums % val == 0.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

Example 2

Example 3

Core Insight

What unlocks the optimal approach

Use mod by 10 to retrieve the least significant digit of the number
Divide the number by 10, then round it down so that the second least significant digit becomes the least significant digit of the number
Use your language’s mod operator to see if a number is a divisor of another.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2520: Count the Digits That Divide a Number
class Solution {
    public int countDigits(int num) {
        int ans = 0;
        for (int x = num; x > 0; x /= 10) {
            if (num % (x % 10) == 0) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2520: Count the Digits That Divide a Number
func countDigits(num int) (ans int) {
	for x := num; x > 0; x /= 10 {
		if num%(x%10) == 0 {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #2520: Count the Digits That Divide a Number
class Solution:
    def countDigits(self, num: int) -> int:
        ans, x = 0, num
        while x:
            x, val = divmod(x, 10)
            ans += num % val == 0
        return ans

// Accepted solution for LeetCode #2520: Count the Digits That Divide a Number
impl Solution {
    pub fn count_digits(num: i32) -> i32 {
        let mut ans = 0;
        let mut cur = num;
        while cur != 0 {
            if num % (cur % 10) == 0 {
                ans += 1;
            }
            cur /= 10;
        }
        ans
    }
}

// Accepted solution for LeetCode #2520: Count the Digits That Divide a Number
function countDigits(num: number): number {
    let ans = 0;
    for (let x = num; x; x = (x / 10) | 0) {
        if (num % (x % 10) === 0) {
            ++ans;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log num)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.