LeetCode #258 — EASY

Add Digits

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

The Problem

Problem Statement

Given an integer num, repeatedly add all its digits until the result has only one digit, and return it.

Example 1:

Input: num = 38
Output: 2
Explanation: The process is
38 --> 3 + 8 --> 11
11 --> 1 + 1 --> 2 
Since 2 has only one digit, return it.

Example 2:

Input: num = 0
Output: 0

Constraints:

0 <= num <= 2³¹ - 1

Follow up: Could you do it without any loop/recursion in O(1) runtime?

Step 01

Brute Force Baseline

Problem summary: Given an integer num, repeatedly add all its digits until the result has only one digit, and return it.

Baseline thinking

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

Pattern signal: Math

Example 1

Example 2

Core Insight

What unlocks the optimal approach

A naive implementation of the above process is trivial. Could you come up with other methods?
What are all the possible results?
How do they occur, periodically or randomly?
You may find this <a href="https://en.wikipedia.org/wiki/Digital_root" target="_blank">Wikipedia article</a> useful.

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 #258: Add Digits
class Solution {
    public int addDigits(int num) {
        return (num - 1) % 9 + 1;
    }
}

// Accepted solution for LeetCode #258: Add Digits
func addDigits(num int) int {
	if num == 0 {
		return 0
	}
	return (num-1)%9 + 1
}

# Accepted solution for LeetCode #258: Add Digits
class Solution:
    def addDigits(self, num: int) -> int:
        return 0 if num == 0 else (num - 1) % 9 + 1

// Accepted solution for LeetCode #258: Add Digits
impl Solution {
    pub fn add_digits(mut num: i32) -> i32 {
        ((num - 1) % 9) + 1
    }
}

// Accepted solution for LeetCode #258: Add Digits
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #258: Add Digits
// class Solution {
//     public int addDigits(int num) {
//         return (num - 1) % 9 + 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.