Leetcode #415: Add Strings
In this guide, we solve Leetcode #415 Add Strings in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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).
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Math, String, Simulation
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
Input: num1 = "11", num2 = "123"
Output: "134"
Python Solution
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])
Complexity
The time complexity is , where and are the lengths of the two strings respectively. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.