feat: refactored docs structure

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

@@ -0,0 +1,50 @@
+---
+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]
+```