---
title: '20. Valid Parentheses'
description: Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid
sidebar:
  label: 'Valid Parentheses'
  badge: 'Easy'
---

String

### Example 1:
- Input: `s = "()"`
- Output: `true`

### Example 2:
- Input: `s = "()[]{}"`
- Output: `true`

### Example 3:
- Input: `s = "(]"`
- Output: `false`

### Example 4:
- Input: `s = "([])"`
- Output: `true`

### Example 5:
- Input: `s = "([)]"`
- Output: `false`

### Constraints:

- `1 <= s.length <= 10^4`
- `s` consists of parentheses only '()[]{}'.

## Approach

```mermaid
flowchart TD
  S(["isValid(s)"]) --> I["stack = [], pairs = closer to opener"]
  I --> L{"more char c?"}
  L -- no --> Z{"stack empty?"}
  Z -- yes --> T(["return True"])
  Z -- no --> U(["return False — unclosed openers left"])
  L -- yes --> Q{"c is a closer?"}
  Q -- no --> A["stack.append(c) — it is an opener"]
  A --> L
  Q -- yes --> M{"stack non-empty and pairs[c] == stack[-1]?"}
  M -- yes --> P["stack.pop() — matched"]
  P --> L
  M -- no --> F(["return False — mismatch"])
```

## Solution

```py
class Solution:
    def isValid(self, s: str) -> bool:
        stack = []
        pairs = {")": "(", "]": "[", "}": "{"}

        for c in s:
            if c in pairs:
                if stack and pairs[c] == stack[-1]:
                    stack.pop()
                else:
                    return False
            else:
                stack.append(c)

        return True if not stack else False
```

## Explanation

[Valid Parentheses - Stack - Leetcode 20 - Python](https://www.youtube.com/watch?v=WTzjTskDFMg)