LeetCode #3099 — EASY

Harshad Number

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

The Problem

Problem Statement

An integer divisible by the sum of its digits is said to be a Harshad number. You are given an integer x. Return the sum of the digits of x if x is a Harshad number, otherwise, return -1.

Example 1:

Input: x = 18

Output: 9

Explanation:

The sum of digits of x is 9. 18 is divisible by 9. So 18 is a Harshad number and the answer is 9.

Example 2:

Input: x = 23

Output: -1

Explanation:

The sum of digits of x is 5. 23 is not divisible by 5. So 23 is not a Harshad number and the answer is -1.

Constraints:

1 <= x <= 100

Step 01

Brute Force Baseline

Problem summary: An integer divisible by the sum of its digits is said to be a Harshad number. You are given an integer x. Return the sum of the digits of x if x is a Harshad number, otherwise, return -1.

Baseline thinking

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

Pattern signal: Math

Example 1

Example 2

Step 02

Core Insight

What unlocks the optimal approach

Use a while loop and divide <code>x</code> by <code>10</code> to find the sum of the digits of <code>x</code>.

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 #3099: Harshad Number
class Solution {
    public int sumOfTheDigitsOfHarshadNumber(int x) {
        int s = 0;
        for (int y = x; y > 0; y /= 10) {
            s += y % 10;
        }
        return x % s == 0 ? s : -1;
    }
}

// Accepted solution for LeetCode #3099: Harshad Number
func sumOfTheDigitsOfHarshadNumber(x int) int {
	s := 0
	for y := x; y > 0; y /= 10 {
		s += y % 10
	}
	if x%s == 0 {
		return s
	}
	return -1
}

# Accepted solution for LeetCode #3099: Harshad Number
class Solution:
    def sumOfTheDigitsOfHarshadNumber(self, x: int) -> int:
        s, y = 0, x
        while y:
            s += y % 10
            y //= 10
        return s if x % s == 0 else -1

// Accepted solution for LeetCode #3099: Harshad Number
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3099: Harshad Number
// class Solution {
//     public int sumOfTheDigitsOfHarshadNumber(int x) {
//         int s = 0;
//         for (int y = x; y > 0; y /= 10) {
//             s += y % 10;
//         }
//         return x % s == 0 ? s : -1;
//     }
// }

// Accepted solution for LeetCode #3099: Harshad Number
function sumOfTheDigitsOfHarshadNumber(x: number): number {
    let s = 0;
    for (let y = x; y; y = Math.floor(y / 10)) {
        s += y % 10;
    }
    return x % s === 0 ? s : -1;
}

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 n)

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.