feat(work): add solutions for number-of-islands, pacific-atlantic-water-flow, and clone-graph problems

work/1/Medium/Array/200.number-of-islands.py

@@ -47,6 +47,26 @@ Constraints:
         • grid[i][j] is '0' or '1'.
 """
 
+
 class Solution:
     def numIslands(self, grid: List[List[str]]) -> int:
+        rows, cols = len(grid), len(grid[0])
 
+        def dfs(r, c):
+            if r < 0 or r >= rows or c < 0 or c >= cols or grid[r][c] != "1":
+                return
+            else:
+                grid[r][c] = "0"
+                dfs(r, c + 1)  # right
+                dfs(r + 1, c)  # bottom
+                dfs(r, c - 1)  # left
+                dfs(r - 1, c)  # top
+
+        num_islands = 0
+        for r in range(rows):
+            for c in range(cols):
+                if grid[r][c] == "1":
+                    num_islands += 1
+                    dfs(r, c)
+
+        return num_islands

work/1/Medium/Array/417.pacific-atlantic-water-flow.py

@@ -0,0 +1,75 @@
+"""
+417. Pacific Atlantic Water Flow
+Difficulty: Medium
+https://leetcode.com/problems/pacific-atlantic-water-flow/
+
+──────────────────────────────────────────────────
+
+There is an m x n rectangular island that borders both the Pacific
+Ocean and Atlantic Ocean. The Pacific Ocean touches the island's left
+and top edges, and the Atlantic Ocean touches the island's right and
+bottom edges.
+
+The island is partitioned into a grid of square cells. You are given
+an m x n integer matrix heights where heights[r][c] represents the
+height above sea level of the cell at coordinate (r, c).
+
+The island receives a lot of rain, and the rain water can flow to
+neighboring cells directly north, south, east, and west if the
+neighboring cell's height is less than or equal to the current cell's
+height. Water can flow from any cell adjacent to an ocean into the
+ocean.
+
+Return a 2D list of grid coordinates result where result[i] = [ri,
+ci] denotes that rain water can flow from cell (ri, ci) to both the
+Pacific and Atlantic oceans.
+
+ 
+
+Example 1:
+
+Input: heights =
+[[1,2,2,3,5],[3,2,3,4,4],[2,4,5,3,1],[6,7,1,4,5],[5,1,1,2,4]]
+Output: [[0,4],[1,3],[1,4],[2,2],[3,0],[3,1],[4,0]]
+Explanation: The following cells can flow to the Pacific and Atlantic
+oceans, as shown below:
+[0,4]: [0,4] -> Pacific Ocean 
+       [0,4] -> Atlantic Ocean
+[1,3]: [1,3] -> [0,3] -> Pacific Ocean 
+       [1,3] -> [1,4] -> Atlantic Ocean
+[1,4]: [1,4] -> [1,3] -> [0,3] -> Pacific Ocean 
+       [1,4] -> Atlantic Ocean
+[2,2]: [2,2] -> [1,2] -> [0,2] -> Pacific Ocean 
+       [2,2] -> [2,3] -> [2,4] -> Atlantic Ocean
+[3,0]: [3,0] -> Pacific Ocean 
+       [3,0] -> [4,0] -> Atlantic Ocean
+[3,1]: [3,1] -> [3,0] -> Pacific Ocean 
+       [3,1] -> [4,1] -> Atlantic Ocean
+[4,0]: [4,0] -> Pacific Ocean 
+       [4,0] -> Atlantic Ocean
+Note that there are other possible paths for these cells to flow to
+the Pacific and Atlantic oceans.
+
+Example 2:
+
+Input: heights = [[1]]
+Output: [[0,0]]
+Explanation: The water can flow from the only cell to the Pacific and
+Atlantic oceans.
+
+ 
+
+Constraints:
+
+	• m == heights.length
+
+	• n == heights[r].length
+
+	• 1 <= m, n <= 200
+
+	• 0 <= heights[r][c] <= 10^5
+"""
+
+class Solution:
+    def pacificAtlantic(self, heights: list[list[int]]) -> list[list[int]]:
+        

work/1/Medium/Hash Table/133.clone-graph.py

@@ -0,0 +1,108 @@
+"""
+133. Clone Graph
+Difficulty: Medium
+https://leetcode.com/problems/clone-graph/
+
+──────────────────────────────────────────────────
+
+Given a reference of a node in a connected undirected graph.
+
+Return a deep copy (clone) of the graph.
+
+Each node in the graph contains a value (int) and a list (List[Node])
+of its neighbors.
+
+class Node {
+    public int val;
+    public List<Node> neighbors;
+}
+
+
+
+Test case format:
+
+For simplicity, each node's value is the same as the node's index
+(1-indexed). For example, the first node with val == 1, the second
+node with val == 2, and so on. The graph is represented in the test
+case using an adjacency list.
+
+An adjacency list is a collection of unordered lists used to
+represent a finite graph. Each list describes the set of neighbors of
+a node in the graph.
+
+The given node will always be the first node with val = 1. You must
+return the copy of the given node as a reference to the cloned graph.
+
+
+
+Example 1:
+
+Input: adjList = [[2,4],[1,3],[2,4],[1,3]]
+Output: [[2,4],[1,3],[2,4],[1,3]]
+Explanation: There are 4 nodes in the graph.
+1st node (val = 1)'s neighbors are 2nd node (val = 2) and 4th node
+(val = 4).
+2nd node (val = 2)'s neighbors are 1st node (val = 1) and 3rd node
+(val = 3).
+3rd node (val = 3)'s neighbors are 2nd node (val = 2) and 4th node
+(val = 4).
+4th node (val = 4)'s neighbors are 1st node (val = 1) and 3rd node
+(val = 3).
+
+Example 2:
+
+Input: adjList = [[]]
+Output: [[]]
+Explanation: Note that the input contains one empty list. The graph
+consists of only one node with val = 1 and it does not have any
+neighbors.
+
+Example 3:
+
+Input: adjList = []
+Output: []
+Explanation: This an empty graph, it does not have any nodes.
+
+
+
+Constraints:
+
+        • The number of nodes in the graph is in the range [0, 100].
+
+        • 1 <= Node.val <= 100
+
+        • Node.val is unique for each node.
+
+        • There are no repeated edges and no self-loops in the graph.
+
+• The Graph is connected and all nodes can be visited starting from
+the given node.
+"""
+
+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val = 0, neighbors = None):
+        self.val = val
+        self.neighbors = neighbors if neighbors is not None else []
+"""
+
+from typing import Optional
+
+
+class Solution:
+    def cloneGraph(self, node: Optional["Node"]) -> Optional["Node"]:
+        oldToNew = {}
+
+        def dfs(node):
+            if node in oldToNew:
+                return oldToNew[node]
+
+            copy = Node(node.val)
+            oldToNew[node] = copy
+
+            for n in node.neighbors:
+                copy.neighbors.append(dfs(n))
+            return copy
+
+        return dfs(node) if node else None