Add Two Integers

The Problem

Problem Statement

Given two integers num1 and num2, return the sum of the two integers.

Example 1:

Input: num1 = 12, num2 = 5
Output: 17
Explanation: num1 is 12, num2 is 5, and their sum is 12 + 5 = 17, so 17 is returned.

Example 2:

Input: num1 = -10, num2 = 4
Output: -6
Explanation: num1 + num2 = -6, so -6 is returned.

Constraints:

-100 <= num1, num2 <= 100

Step 01

Brute Force Baseline

Problem summary: Given two integers num1 and num2, return the sum of the two integers. Example 1: Input: num1 = 12, num2 = 5 Output: 17 Explanation: num1 is 12, num2 is 5, and their sum is 12 + 5 = 17, so 17 is returned. Example 2: Input: num1 = -10, num2 = 4 Output: -6 Explanation: num1 + num2 = -6, so -6 is returned.

Baseline thinking

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

Pattern signal: Math

Example 1

12
5

Example 2

-10
4

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #2235: Add Two Integers
class Solution {
    public int sum(int num1, int num2) {
        return num1 + num2;
    }
}

// Accepted solution for LeetCode #2235: Add Two Integers
func sum(num1 int, num2 int) int {
	return num1 + num2
}

# Accepted solution for LeetCode #2235: Add Two Integers
class Solution:
    def sum(self, num1: int, num2: int) -> int:
        return num1 + num2

// Accepted solution for LeetCode #2235: Add Two Integers
impl Solution {
    pub fn sum(num1: i32, num2: i32) -> i32 {
        num1 + num2
    }
}

// Accepted solution for LeetCode #2235: Add Two Integers
function sum(num1: number, num2: number): number {
    return num1 + num2;
}

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.