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77 lines
1.6 KiB
Plaintext
77 lines
1.6 KiB
Plaintext
---
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title: '20. Valid Parentheses'
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description: Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid
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sidebar:
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label: 'Valid Parentheses'
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badge: 'Easy'
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---
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<Badge variant="accent">String</Badge>
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### Example 1:
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- Input: `s = "()"`
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- Output: `true`
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### Example 2:
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- Input: `s = "()[]{}"`
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- Output: `true`
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### Example 3:
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- Input: `s = "(]"`
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- Output: `false`
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### Example 4:
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- Input: `s = "([])"`
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- Output: `true`
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### Example 5:
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- Input: `s = "([)]"`
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- Output: `false`
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### Constraints:
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- `1 <= s.length <= 10^4`
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- `s` consists of parentheses only '()[]{}'.
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## Approach
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```mermaid
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flowchart TD
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S(["isValid(s)"]) --> I["stack = [], pairs = closer to opener"]
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I --> L{"more char c?"}
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L -- no --> Z{"stack empty?"}
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Z -- yes --> T(["return True"])
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Z -- no --> U(["return False — unclosed openers left"])
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L -- yes --> Q{"c is a closer?"}
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Q -- no --> A["stack.append(c) — it is an opener"]
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A --> L
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Q -- yes --> M{"stack non-empty and pairs[c] == stack[-1]?"}
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M -- yes --> P["stack.pop() — matched"]
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P --> L
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M -- no --> F(["return False — mismatch"])
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```
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## Solution
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```py
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class Solution:
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def isValid(self, s: str) -> bool:
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stack = []
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pairs = {")": "(", "]": "[", "}": "{"}
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for c in s:
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if c in pairs:
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if stack and pairs[c] == stack[-1]:
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stack.pop()
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else:
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return False
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else:
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stack.append(c)
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return True if not stack else False
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```
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## Explanation
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<YouTube id="WTzjTskDFMg" title="Valid Parentheses - Stack - Leetcode 20 - Python" />
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