docs: rename category folders to remove parentheses and update heap meta title

apps/docs/content/matrix-index-math/48-rotate-image.mdx

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+---
+title: '48. Rotate Image'
+description: You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise)
+sidebar:
+  label: 'Rotate Image'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Array</Badge>
+
+### Example 1:
+- Input: `matrix = [[1,2,3],[4,5,6],[7,8,9]]`
+- Output: `[[7,4,1],[8,5,2],[9,6,3]]`
+
+### Example 2:
+- Input: `matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]`
+- Output: `[[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]`
+
+### Constraints:
+
+- `n == matrix.length == matrix[i].length`
+- `1 <= n <= 20`
+- `-1000 <= matrix[i][j] <= 1000`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["rotate(matrix)"]) --> N["n = len(matrix)"]
+  N --> F{"more r in range(n)?"}
+  F -- yes --> G{"more c in range(r + 1, n)?"}
+  G -- yes --> T["swap matrix[r][c] and matrix[c][r] — transpose across the diagonal"]
+  T --> G
+  G -- no --> F
+  F -- no --> H{"more r in range(n)?"}
+  H -- yes --> R["matrix[r].reverse() — transpose then mirror is a clockwise turn"]
+  R --> H
+  H -- no --> E(["return None — rotated in place"])
+```
+
+## Solution
+
+```py
+class Solution:
+    def rotate(self, matrix: List[List[int]]) -> None:
+        n = len(matrix)
+        for r in range(n):
+            for c in range(r + 1, n):
+                matrix[r][c], matrix[c][r] = matrix[c][r], matrix[r][c]
+
+        for r in range(n):
+            matrix[r].reverse()
+```
+
+## Explanation
+
+<YouTube id="fMSJSS7eO1w" title="Rotate Image - Matrix - Leetcode 48" />

apps/docs/content/matrix-index-math/54-spiral-matrix.mdx

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+---
+title: '54. Spiral Matrix'
+description: Given an m x n matrix, return all elements of the matrix in spiral order
+sidebar:
+  label: 'Spiral Matrix'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Array</Badge>
+
+### Example 1:
+- Input: `matrix = [[1,2,3],[4,5,6],[7,8,9]]`
+- Output: `[1,2,3,6,9,8,7,4,5]`
+
+### Example 2:
+- Input: `matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]`
+- Output: `[1,2,3,4,8,12,11,10,9,5,6,7]`
+
+### Constraints:
+
+- `m == matrix.length`
+- `n == matrix[i].length`
+- `1 <= m, n <= 10`
+- `-100 <= matrix[i][j] <= 100`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["spiralOrder(matrix)"]) --> I["res = [], top = 0, bottom = len(matrix) - 1, left = 0, right = len(matrix[0]) - 1"]
+  I --> W{"top <= bottom and left <= right?"}
+  W -- no --> E(["return res"])
+  W -- yes --> T["append matrix[top][c] for c in left..right, then top += 1 — top row, shrink from above"]
+  T --> R["append matrix[r][right] for r in top..bottom, then right -= 1 — right column"]
+  R --> B{"top <= bottom?"}
+  B -- yes --> C["append matrix[bottom][c] for c in right down to left, then bottom -= 1 — bottom row"]
+  C --> L{"left <= right?"}
+  B -- no --> L
+  L -- yes --> D["append matrix[r][left] for r in bottom down to top, then left += 1 — left column"]
+  D --> W
+  L -- no --> W
+```
+
+## Solution
+
+```py
+class Solution:
+    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
+        res = []
+        top, bottom = 0, len(matrix) - 1
+        left, right = 0, len(matrix[0]) - 1
+
+        while top <= bottom and left <= right:
+            # 1. Top Row
+            for c in range(left, right + 1):
+                res.append(matrix[top][c])
+            top += 1
+
+            # 2. Right Column
+            for r in range(top, bottom + 1):
+                res.append(matrix[r][right])
+            right -= 1
+
+            # 3. Bottom Row
+            if top <= bottom:
+                for c in range(right, left - 1, -1):
+                    res.append(matrix[bottom][c])
+                bottom -= 1
+
+            # 4. Left Column
+            if left <= right:
+                for r in range(bottom, top - 1, -1):
+                    res.append(matrix[r][left])
+                left += 1
+
+        return res
+```
+
+## Explanation
+
+<YouTube id="BJnMZNwUk1M" title="Spiral Matrix - Microsoft Interview Question - Leetcode 54" />

apps/docs/content/matrix-index-math/73-set-matrix-zeroes.mdx

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+---
+title: '73. Set Matrix Zeroes'
+description: Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0's
+sidebar:
+  label: 'Set Matrix Zeroes'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Array</Badge>
+
+::::warning
+You must do it in place.
+::::
+
+### Example 1:
+- Input: `matrix = [[1,1,1],[1,0,1],[1,1,1]]`
+- Output: `[[1,0,1],[0,0,0],[1,0,1]]`
+
+### Example 2:
+- Input: `matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]`
+- Output: `[[0,0,0,0],[0,4,5,0],[0,3,1,0]]`
+
+### Constraints:
+
+- `m == matrix.length`
+- `n == matrix[0].length`
+- `1 <= m, n <= 200`
+- `-2^31 <= matrix[i][j] <= 2^31 - 1`
+- A straightforward solution using O(mn) space is probably a bad idea.
+- A simple improvement uses O(m + n) space, but still not the best solution.
+- Could you devise a constant space solution?
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["setZeroes(matrix)"]) --> I["m = len(matrix), n = len(matrix[0])"]
+  I --> Z["row0_zero = any(matrix[0][c] == 0), col0_zero = any(matrix[r][0] == 0) — remember before they become markers"]
+  Z --> F{"more r in range(1, m)?"}
+  F -- yes --> G{"more c in range(1, n)?"}
+  G -- yes --> Q{"matrix[r][c] == 0?"}
+  Q -- yes --> M["matrix[r][0] = 0, matrix[0][c] = 0 — row 0 and column 0 are the marker lists"]
+  M --> G
+  Q -- no --> G
+  G -- no --> F
+  F -- no --> H{"more r in range(1, m)?"}
+  H -- yes --> J{"more c in range(1, n)?"}
+  J -- yes --> K{"matrix[r][0] == 0 or matrix[0][c] == 0?"}
+  K -- yes --> W["matrix[r][c] = 0"]
+  W --> J
+  K -- no --> J
+  J -- no --> H
+  H -- no --> A{"row0_zero?"}
+  A -- yes --> B["matrix[0][c] = 0 for c in range(n)"]
+  B --> C{"col0_zero?"}
+  A -- no --> C
+  C -- yes --> D["matrix[r][0] = 0 for r in range(m)"]
+  D --> E(["return None — zeroed in place"])
+  C -- no --> E
+```
+
+## Solution
+
+```py
+class Solution:
+    def setZeroes(self, matrix: List[List[int]]) -> None:
+        """
+        Do not return anything, modify matrix in-place instead.
+        """
+        m, n = len(matrix), len(matrix[0])
+
+        # 1. remember if row 0 or column 0 have their own zeros
+        row0_zero = any(matrix[0][c] == 0 for c in range(n))
+        col0_zero = any(matrix[r][0] == 0 for r in range(m))
+
+        # 2. use row 0 and column 0 as marker lists
+        for r in range(1, m):
+            for c in range(1, n):
+                if matrix[r][c] == 0:
+                    matrix[r][0] = 0
+                    matrix[0][c] = 0
+
+        # 3. zero the inner cells from the markers
+        for r in range(1, m):
+            for c in range(1, n):
+                if matrix[r][0] == 0 or matrix[0][c] == 0:
+                    matrix[r][c] = 0
+
+        # 4. handle row 0 and column 0 last
+        if row0_zero:
+            for c in range(n):
+                matrix[0][c] = 0
+        if col0_zero:
+            for r in range(m):
+                matrix[r][0] = 0
+```
+
+## Explanation
+
+<YouTube id="T41rL0L3Pnw" title="Set Matrix Zeroes - In-place - Leetcode 73" />

apps/docs/content/matrix-index-math/meta.ts

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+import { defineMeta } from "blume";
+
+export default defineMeta({ title: "Matrix Index Math", order: 10 });