chore: clean up content phase organization

apps/docs/content/(01-linear-structures)/stack/20-valid-parentheses.mdx

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+---
+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'
+---
+
+<Badge variant="accent">String</Badge>
+
+### 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
+
+<YouTube id="WTzjTskDFMg" title="Valid Parentheses - Stack - Leetcode 20 - Python" />