LeetCode #415 — EASY

Add Strings

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

The Problem

Problem Statement

Given two non-negative integers, num1 and num2 represented as string, return the sum of num1 and num2 as a string.

You must solve the problem without using any built-in library for handling large integers (such as BigInteger). You must also not convert the inputs to integers directly.

Example 1:

Input: num1 = "11", num2 = "123"
Output: "134"

Example 2:

Input: num1 = "456", num2 = "77"
Output: "533"

Example 3:

Input: num1 = "0", num2 = "0"
Output: "0"

Constraints:

1 <= num1.length, num2.length <= 10⁴
num1 and num2 consist of only digits.
num1 and num2 don't have any leading zeros except for the zero itself.

Step 01

Brute Force Baseline

Problem summary: Given two non-negative integers, num1 and num2 represented as string, return the sum of num1 and num2 as a string. You must solve the problem without using any built-in library for handling large integers (such as BigInteger). You must also not convert the inputs to integers directly.

Baseline thinking

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

Pattern signal: Math

Example 1

"11"
"123"

Example 2

"456"
"77"

Example 3

"0"
"0"

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 #415: Add Strings
class Solution {
    public String addStrings(String num1, String num2) {
        int i = num1.length() - 1, j = num2.length() - 1;
        StringBuilder ans = new StringBuilder();
        for (int c = 0; i >= 0 || j >= 0 || c > 0; --i, --j) {
            int a = i < 0 ? 0 : num1.charAt(i) - '0';
            int b = j < 0 ? 0 : num2.charAt(j) - '0';
            c += a + b;
            ans.append(c % 10);
            c /= 10;
        }
        return ans.reverse().toString();
    }

    public String subStrings(String num1, String num2) {
        int m = num1.length(), n = num2.length();
        boolean neg = m < n || (m == n && num1.compareTo(num2) < 0);
        if (neg) {
            String t = num1;
            num1 = num2;
            num2 = t;
        }
        int i = num1.length() - 1, j = num2.length() - 1;
        StringBuilder ans = new StringBuilder();
        for (int c = 0; i >= 0; --i, --j) {
            c = (num1.charAt(i) - '0') - c - (j < 0 ? 0 : num2.charAt(j) - '0');
            ans.append((c + 10) % 10);
            c = c < 0 ? 1 : 0;
        }
        while (ans.length() > 1 && ans.charAt(ans.length() - 1) == '0') {
            ans.deleteCharAt(ans.length() - 1);
        }
        if (neg) {
            ans.append('-');
        }
        return ans.reverse().toString();
    }
}

// Accepted solution for LeetCode #415: Add Strings
func addStrings(num1 string, num2 string) string {
	i, j := len(num1)-1, len(num2)-1
	ans := []byte{}
	for c := 0; i >= 0 || j >= 0 || c > 0; i, j = i-1, j-1 {
		if i >= 0 {
			c += int(num1[i] - '0')
		}
		if j >= 0 {
			c += int(num2[j] - '0')
		}
		ans = append(ans, byte(c%10+'0'))
		c /= 10
	}
	for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
		ans[i], ans[j] = ans[j], ans[i]
	}
	return string(ans)
}

func subStrings(num1 string, num2 string) string {
	m, n := len(num1), len(num2)
	neg := m < n || (m == n && num1 < num2)
	if neg {
		num1, num2 = num2, num1
	}
	i, j := len(num1)-1, len(num2)-1
	ans := []byte{}
	for c := 0; i >= 0; i, j = i-1, j-1 {
		c = int(num1[i]-'0') - c
		if j >= 0 {
			c -= int(num2[j] - '0')
		}
		ans = append(ans, byte((c+10)%10+'0'))
		if c < 0 {
			c = 1
		} else {
			c = 0
		}
	}
	for len(ans) > 1 && ans[len(ans)-1] == '0' {
		ans = ans[:len(ans)-1]
	}
	if neg {
		ans = append(ans, '-')
	}
	for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
		ans[i], ans[j] = ans[j], ans[i]
	}
	return string(ans)
}

# Accepted solution for LeetCode #415: Add Strings
class Solution:
    def addStrings(self, num1: str, num2: str) -> str:
        i, j = len(num1) - 1, len(num2) - 1
        ans = []
        c = 0
        while i >= 0 or j >= 0 or c:
            a = 0 if i < 0 else int(num1[i])
            b = 0 if j < 0 else int(num2[j])
            c, v = divmod(a + b + c, 10)
            ans.append(str(v))
            i, j = i - 1, j - 1
        return "".join(ans[::-1])

    def subStrings(self, num1: str, num2: str) -> str:
        m, n = len(num1), len(num2)
        neg = m < n or (m == n and num1 < num2)
        if neg:
            num1, num2 = num2, num1
        i, j = len(num1) - 1, len(num2) - 1
        ans = []
        c = 0
        while i >= 0:
            c = int(num1[i]) - c - (0 if j < 0 else int(num2[j]))
            ans.append(str((c + 10) % 10))
            c = 1 if c < 0 else 0
            i, j = i - 1, j - 1
        while len(ans) > 1 and ans[-1] == '0':
            ans.pop()
        if neg:
            ans.append('-')
        return ''.join(ans[::-1])

// Accepted solution for LeetCode #415: Add Strings
impl Solution {
    pub fn add_strings(num1: String, num2: String) -> String {
        let mut res = vec![];
        let s1 = num1.as_bytes();
        let s2 = num2.as_bytes();
        let (mut i, mut j) = (s1.len(), s2.len());
        let mut is_over = false;
        while i != 0 || j != 0 || is_over {
            let mut sum = if is_over { 1 } else { 0 };
            if i != 0 {
                sum += (s1[i - 1] - b'0') as i32;
                i -= 1;
            }
            if j != 0 {
                sum += (s2[j - 1] - b'0') as i32;
                j -= 1;
            }
            is_over = sum >= 10;
            res.push((sum % 10).to_string());
        }
        res.into_iter().rev().collect()
    }
}

// Accepted solution for LeetCode #415: Add Strings
function addStrings(num1: string, num2: string): string {
    let i = num1.length - 1;
    let j = num2.length - 1;
    const ans: number[] = [];
    for (let c = 0; i >= 0 || j >= 0 || c; --i, --j) {
        c += i < 0 ? 0 : +num1[i];
        c += j < 0 ? 0 : +num2[j];
        ans.push(c % 10);
        c = Math.floor(c / 10);
    }
    return ans.reverse().join('');
}

function subStrings(num1: string, num2: string): string {
    const m = num1.length;
    const n = num2.length;
    const neg = m < n || (m == n && num1 < num2);
    if (neg) {
        const t = num1;
        num1 = num2;
        num2 = t;
    }
    let i = num1.length - 1;
    let j = num2.length - 1;
    const ans: number[] = [];
    for (let c = 0; i >= 0; --i, --j) {
        c = +num1[i] - c;
        if (j >= 0) {
            c -= +num2[j];
        }
        ans.push((c + 10) % 10);
        c = c < 0 ? 1 : 0;
    }
    while (ans.length > 1 && ans.at(-1) === 0) {
        ans.pop();
    }
    return (neg ? '-' : '') + ans.reverse().join('');
}

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.