feat(work): add Minimum Path Sum & Longest Common Subsequence solutions; refactor Unique Paths DP implementation

work/1/Medium/Array/64.minimum-path-sum.py

@@ -0,0 +1,61 @@
+"""
+64. Minimum Path Sum
+Difficulty: Medium
+https://leetcode.com/problems/minimum-path-sum/
+
+──────────────────────────────────────────────────
+
+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.
+
+Note: You can only move either down or right at any point in time.
+
+
+
+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
+"""
+
+
+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]

work/1/Medium/Math/62.unique-paths.py

@@ -43,11 +43,16 @@ Constraints:
 
 class Solution:
     def uniquePaths(self, m: int, n: int) -> int:
-        row = [1] * n
+        dp = [[0 for _ in range(n)] for _ in range(m)]
 
-        for i in range(m - 1):
-            newRow = [1] * n
-            for j in range(n - 2, -1, -1):
-                newRow[j] = newRow[j + 1] + row[j]
-            row = newRow
-        return row[0]
+        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]

work/1/Medium/String/1143.longest-common-subsequence.py

@@ -0,0 +1,63 @@
+"""
+1143. Longest Common Subsequence
+Difficulty: Medium
+https://leetcode.com/problems/longest-common-subsequence/
+
+──────────────────────────────────────────────────
+
+Given two strings text1 and text2, return the length of their longest
+common subsequence. If there is no common subsequence, return 0.
+
+A subsequence of a string is a new string generated from the original
+string with some characters (can be none) deleted without changing the
+relative order of the remaining characters.
+
+        • For example, "ace" is a subsequence of "abcde".
+
+A common subsequence of two strings is a subsequence that is common
+to both strings.
+
+
+
+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.
+"""
+
+
+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]