Leetcode #772: Basic Calculator III
In this guide, we solve Leetcode #772 Basic Calculator III 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
Implement a basic calculator to evaluate a simple expression string. The expression string contains only non-negative integers, '+', '-', '*', '/' operators, and open '(' and closing parentheses ')'.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Stack, Recursion, Math, String
Intuition
The problem has a natural nested or last-in-first-out structure.
A stack lets us resolve matches in the correct order as we scan.
Approach
Push items as they appear and pop when you can finalize a decision.
The stack captures the unresolved part of the input.
Steps:
- Push elements as you scan.
- Pop when a rule or match is satisfied.
- Use the stack to compute results.
Example
Input: s = "1+1"
Output: 2
Python Solution
class Solution:
def calculate(self, s: str) -> int:
def dfs(q):
num, sign, stk = 0, "+", []
while q:
c = q.popleft()
if c.isdigit():
num = num * 10 + int(c)
if c == "(":
num = dfs(q)
if c in "+-*/)" or not q:
match sign:
case "+":
stk.append(num)
case "-":
stk.append(-num)
case "*":
stk.append(stk.pop() * num)
case "/":
stk.append(int(stk.pop() / num))
num, sign = 0, c
if c == ")":
break
return sum(stk)
return dfs(deque(s))
Complexity
The time complexity is O(n). The space complexity is O(n).
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.