Harshad Number

Harshad Number - Problem

Math Easy

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.

Input & Output

Example 1 — Basic Harshad Number

$ Input: x = 18

› Output: 9

💡 Note: Sum of digits: 1 + 8 = 9. Since 18 % 9 = 0, 18 is divisible by 9, so return 9.

Example 2 — Not a Harshad Number

$ Input: x = 23

› Output: -1

💡 Note: Sum of digits: 2 + 3 = 5. Since 23 % 5 = 3 ≠ 0, 23 is not divisible by 5, so return -1.

Example 3 — Single Digit

$ Input: x = 7

› Output: 7

💡 Note: Sum of digits: 7. Since 7 % 7 = 0, any single digit is always a Harshad number, so return 7.

Constraints

1 ≤ x ≤ 100

Visualization

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Asked in

G Google 15 M Microsoft 12 🍎 Apple 8

The key insight is to calculate the sum of digits and check divisibility. Best approach is string conversion for cleaner code. Time: O(log x), Space: O(log x)

Common Approaches

✓ Mathematical Digit Extraction

⏱️ Time: O(log x) Space: O(1)

Use mathematical operations to extract each digit, calculate their sum, then check if the original number is divisible by this sum.

String-Based Digit Sum

⏱️ Time: O(log x) Space: O(log x)

Convert the integer to a string, then iterate through each character to calculate the digit sum. This avoids mathematical digit extraction.

Mathematical Digit Extraction — Algorithm Steps

Extract digits using modulo 10 and integer division
Sum all extracted digits
Check if original number is divisible by digit sum

Visualization

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Step-by-Step Walkthrough

Extract Digits

Use x % 10 to get last digit, x // 10 to remove it

Sum Digits

Add each extracted digit to running sum

Check Divisibility

Test if original number % digit_sum == 0

Code -

solution.c — C

#include <stdio.h>

int solution(int x) {
    int original = x;
    int digitSum = 0;
    
    // Extract digits and calculate sum
    while (x > 0) {
        digitSum += x % 10;
        x /= 10;
    }
    
    // Check if Harshad number
    if (original % digitSum == 0) {
        return digitSum;
    } else {
        return -1;
    }
}

int main() {
    int x;
    scanf("%d", &x);
    
    int result = solution(x);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log x)

Loop runs for each digit in x, which is log₁₀(x) digits

⚡ Linearithmic

Space Complexity

O(1)

Only uses constant extra variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Mathematical Digit Extraction — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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