feat: refactored docs structure

apps/docs/content/(05-decision-space)/multi-d-grid-dp/(problems)/1143_longest-common-subsequence.mdx

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+---
+title: '1143. Longest Common Subsequence'
+description: Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0
+icon: dot
+topics:
+  - { name: "String", slug: "string" }
+  - { name: "Dynamic Programming", slug: "dynamic-programming" }
+  - { name: "Longest Common Subsequence", slug: "longest-common-subsequence" }
+issue: "https://github.com/prdlk/leetcode/issues/168"
+sidebar:
+  label: 'Longest Common Subsequence'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `text1 = "abcde", text2 = "ace"`
+- Output: `3`
+- Explanation: The longest common subsequence is "ace" and its length is `3`.
+
+### Example 2:
+- Input: `text1 = "abc", text2 = "abc"`
+- Output: `3`
+- Explanation: The longest common subsequence is "abc" and its length is `3`.
+
+### Example 3:
+- Input: `text1 = "abc", text2 = "def"`
+- Output: `0`
+- Explanation: There is no such common subsequence, so the result is `0`.
+
+### Constraints:
+
+- `1 <= text1.length, text2.length <= 1000`
+- `text1` and `text2` consist of only lowercase English characters.
+
+## Solution
+
+```py
+class Solution:
+    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
+        dp = [[0 for j in range(len(text2) + 1)] for i in range(len(text1) + 1)]
+
+        for i in range(len(text1) - 1, -1, -1):
+            for j in range(len(text2) - 1, -1, -1):
+                if text1[i] == text2[j]:
+                    dp[i][j] = 1 + dp[i + 1][j + 1]
+                else:
+                    dp[i][j] = max(dp[i][j + 1], dp[i + 1][j])
+
+        return dp[0][0]
+```

apps/docs/content/(05-decision-space)/multi-d-grid-dp/(problems)/62_unique-paths.mdx

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+---
+title: '62. Unique Paths'
+description: There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time
+icon: dot
+topics:
+  - { name: "Math", slug: "math" }
+  - { name: "Dynamic Programming", slug: "dynamic-programming" }
+  - { name: "Combinatorics", slug: "combinatorics" }
+issue: "https://github.com/prdlk/leetcode/issues/166"
+sidebar:
+  label: 'Unique Paths'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `m = 3, n = 7`
+- Output: `28`
+
+### Example 2:
+- Input: `m = 3, n = 2`
+- Output: `3`
+- Explanation: From the top-left corner, there are a total of `3` ways to reach the bottom-right corner: `1`. Right -> Down -> Down `2`. Down -> Down -> Right `3`. Down -> Right -> Down
+
+### Constraints:
+
+- `1 <= m, n <= 100`
+
+## Solution
+
+```py
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        dp = [[0 for _ in range(n)] for _ in range(m)]
+
+        for c in range(n):
+            dp[0][c] = 1
+
+        for r in range(m):
+            dp[r][0] = 1
+
+        for r in range(1, m):
+            for c in range(1, n):
+                dp[r][c] = dp[r - 1][c] + dp[r][c - 1]
+
+        return dp[-1][-1]
+```

apps/docs/content/(05-decision-space)/multi-d-grid-dp/(problems)/64_minimum-path-sum.mdx

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+---
+title: '64. Minimum Path Sum'
+description: Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Dynamic Programming", slug: "dynamic-programming" }
+  - { name: "Matrix", slug: "matrix" }
+issue: "https://github.com/prdlk/leetcode/issues/167"
+sidebar:
+  label: 'Minimum Path Sum'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `grid = [[1,3,1],[1,5,1],[4,2,1]]`
+- Output: `7`
+- Explanation: Because the path `1` → `3` → `1` → `1` → `1` minimizes the sum.
+
+### Example 2:
+- Input: `grid = [[1,2,3],[4,5,6]]`
+- Output: `12`
+
+### Constraints:
+
+- `m == grid.length`
+- `n == grid[i].length`
+- `1 <= m, n <= 200`
+- `0 <= grid[i][j] <= 200`
+
+## Solution
+
+```py
+class Solution:
+    def minPathSum(self, grid: list[list[int]]) -> int:
+        if not grid or not grid[0]:
+            return 0
+
+        m = len(grid)
+        n = len(grid[0])
+        dp = [[0 for _ in range(n)] for _ in range(m)]
+
+        for c in range(n):
+            dp[0][c] = dp[0][c - 1] + grid[0][c] if c else grid[0][c]
+
+        for r in range(m):
+            dp[r][0] = dp[r - 1][0] + grid[r][0] if r else grid[r][0]
+
+        for r in range(1, m):
+            for c in range(1, n):
+                dp[r][c] = min(dp[r - 1][c], dp[r][c - 1]) + grid[r][c]
+
+        return dp[-1][-1]
+```

apps/docs/content/(05-decision-space)/multi-d-grid-dp/index.mdx

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+---
+title: Multi-D / Grid DP
+description: 'The result for each state is stored in a table with two or more dimensions, and each cell is computed from its neighbors.'
+hidden: true
+---

apps/docs/content/(05-decision-space)/multi-d-grid-dp/meta.ts

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+import { defineMeta } from "blume";
+
+export default defineMeta({ title: "Multi-D / Grid DP", order: 22, icon: "table", display: "page" });