Valid Parentheses
Detailed guide and Python implementation for the 'Valid Parentheses' problem.
1. Concept Overview
The 'Valid Parentheses' problem is a key challenge in the Stack section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Valid Parentheses.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Valid Parentheses carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
An input string is valid if:
1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every close bracket has a corresponding open bracket of the same type.
Write a function isValid(s: str) -> bool.
- •1 <= len(s) <= 10^4
- •s consists of parentheses only: '()[]{}'
Examples
s = "()"
True
A single pair of matching parentheses is valid.
s = "()[]{}"True
Three pairs of matching brackets, each closed in order.
s = "(]"
False
Opening '(' is closed by ']' which is the wrong type.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def is_valid_opt(s):
Map = {")": "(", "]": "[", "}": "{"}
stack = []
for c in s:
if c not in Map:
stack.append(c)
continue
if not stack or stack[-1] != Map[c]:
return False
stack.pop()
return not stackBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def is_valid_brute(s):
while "()" in s or "[]" in s or "{}" in s:
s = s.replace("()", "").replace("[]", "").replace("{}", "")
return s == ""Algorithm Pattern Checklist
When dealing with Stack data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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