Merge pull request #150 from bhyun-kim/main

[bhyun-kim, 김병현] Week 9 Solutions

Author: bhyun-kim

clone-graph/bhyun-kim.py

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+"""
+133. Clone Graph
+https://leetcode.com/problems/clone-graph/
+
+Solution:
+    To clone a graph, we can use a depth-first search (DFS) to explore all nodes and their neighbors.
+    We can create a helper function that takes a node and returns its clone.
+    
+    - We can use a dictionary to map old nodes to new nodes.
+    - We can create a helper function to clone a node and its neighbors.
+        - If the node has already been cloned, we return the clone.
+        - Otherwise, we create a new clone and add it to the dictionary.
+        - We clone all neighbors of the node recursively.
+        - We return the clone.
+    - We start the DFS from the given node and return the clone.
+
+Time complexity: O(n+m)
+    - n is the number of nodes in the graph.
+    - m is the number of edges in the graph.
+    - We explore all nodes and edges once.
+
+Space complexity: O(n)
+    - We use a dictionary to keep track of the mapping between old nodes and new nodes.
+    - The maximum depth of the recursive call stack is the number of nodes in the graph.
+"""
+
+
+# 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"]:
+        if not node:
+            return None
+
+        old_to_new = {}
+
+        def dfs(node):
+            if node in old_to_new:
+                return old_to_new[node]
+
+            clone = Node(node.val)
+            old_to_new[node] = clone
+
+            for neighbor in node.neighbors:
+                clone.neighbors.append(dfs(neighbor))
+
+            return clone
+
+        return dfs(node)

course-schedule/bhyun-kim.py

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+"""
+207. Course Schedule
+https://leetcode.com/problems/course-schedule/
+
+Solution: 
+    If there is a cycle in the graph, it is impossible to finish all courses.
+    We can detect a cycle by using DFS and keeping track of the state of each node.
+    
+    - We can create an adjacency list to represent the graph.
+    - We can use a state array to keep track of the state of each node.
+        - 0: unvisited, 1: visiting, 2: visited
+    - We can create a helper function to check if there is a cycle starting from a node.
+        - If the node is being visited, we have a cycle.
+        - If the node has been visited, there is no cycle.
+        - If not, we mark the node as visiting and explore its neighbors.
+        - After exploring all neighbors, we mark the node as visited.
+    - We can iterate through all nodes and check for cycles.
+    - If we find a cycle, we return False.
+    - If no cycle is found, we return True.
+
+Time complexity: O(m + n)
+    - m is the number of prerequisites.
+    - n is the number of courses.
+    - We explore all prerequisites and courses once.
+
+Space complexity: O(m + n)
+    - We use an adjacency list to represent the graph.
+    - We use a state array to keep track of the state of each node.
+"""
+
+from collections import defaultdict
+from typing import List
+
+
+class Solution:
+    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
+        adj_list = defaultdict(list)
+        for dest, src in prerequisites:
+            adj_list[src].append(dest)
+
+        # State: 0 = unvisited, 1 = visiting, 2 = visited
+        state = [0] * numCourses
+
+        def has_cycle(v):
+            if state[v] == 1:  # Node is being visited, so we have a cycle
+                return True
+            if state[v] == 2:  # Node has been visited, no cycle here
+                return False
+
+            state[v] = 1
+            for neighbor in adj_list[v]:
+                if has_cycle(neighbor):
+                    return True
+
+            state[v] = 2
+            return False
+
+        for course in range(numCourses):
+            if state[course] == 0:
+                if has_cycle(course):
+                    return False
+
+        return True

design-add-and-search-words-data-structure/bhyun-kim.py

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+"""
+211. Design Add and Search Words Data Structure
+https://leetcode.com/problems/design-add-and-search-words-data-structure/
+
+Solution:
+    To solve this problem, we can use a trie data structure to store the words.
+    We can create a TrieNode class to represent each node in the trie.
+
+    In addWord, we can iterate through each character in the word and create nodes as needed.
+    We mark the end of the word by setting is_end_of_word to True.
+
+    In search, we can use a depth-first search (DFS) to explore all possible paths.
+    If we encounter a '.', we explore all children of the current node.
+    If we find a mismatch or reach the end of the word, we return False.
+
+
+Time complexity: 
+    - addWord: O(m)
+        - m is the length of the word.
+        - We iterate through each character in the word once.
+    - search: O(n * 26^l)
+        - n is the number of nodes in the trie.
+        - l is the length of the word.
+        - We explore all possible paths in the trie.
+        - The worst-case time complexity is O(n * 26^l) when all nodes have 26 children.
+
+Space complexity: O(n)
+    - n is the number of nodes in the trie.
+    - We use a trie data structure to store the words.
+"""
+
+
+class TrieNode:
+    def __init__(self):
+        self.children = {}
+        self.is_end_of_word = False
+
+
+class WordDictionary:
+    def __init__(self):
+        self.root = TrieNode()
+
+    def addWord(self, word):
+        node = self.root
+        for char in word:
+            if char not in node.children:
+                node.children[char] = TrieNode()
+            node = node.children[char]
+        node.is_end_of_word = True
+
+    def search(self, word):
+        def dfs(j, node):
+            for i in range(j, len(word)):
+                char = word[i]
+                if char == ".":
+                    for child in node.children.values():
+                        if dfs(i + 1, child):
+                            return True
+                    return False
+                else:
+                    if char not in node.children:
+                        return False
+                    node = node.children[char]
+            return node.is_end_of_word
+
+        return dfs(0, self.root)

number-of-islands/bhyun-kim.py

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+"""
+200. Number of Islands
+https://leetcode.com/problems/number-of-islands/
+
+Solution:
+    To solve this problem, we can use a depth-first search (DFS) to explore all connected land cells.
+    We can create a helper function that takes the current cell and marks it as visited.
+    
+    - We can iterate through all cells in the grid.
+    - If the current cell is land, we explore all connected land cells using DFS.
+    - We mark all connected land cells as visited.
+    - We increment the island count by 1.
+
+Time complexity: O(m x n)
+    - m and n are the dimensions of the grid.
+    - We explore all cells in the grid once.
+
+Space complexity: O(m x n)
+    - We use a recursive call stack to explore all connected land cells.
+    - The maximum depth of the call stack is the number of cells in the grid.
+
+"""
+
+from typing import List
+
+
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        if not grid:
+            return 0
+
+        m, n = len(grid), len(grid[0])
+
+        island_count = 0
+
+        def dfs(i, j):
+            if i < 0 or i >= m or j < 0 or j >= n or grid[i][j] != "1":
+                return
+            grid[i][j] = "#"
+
+            dfs(i - 1, j)
+            dfs(i + 1, j)
+            dfs(i, j - 1)
+            dfs(i, j + 1)
+
+        for i in range(m):
+            for j in range(n):
+                if grid[i][j] == "1":
+                    dfs(i, j)
+                    island_count += 1
+
+        return island_count

pacific-atlantic-water-flow/bhyun-kim.py

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+"""
+417. Pacific Atlantic Water Flow
+https://leetcode.com/problems/pacific-atlantic-water-flow/
+
+Solution:
+    To solve this problem, we can use depth-first search (DFS) to explore all possible paths starting from each cell.
+    We can create a helper function that takes the current cell and marks it as reachable.
+
+    - We can create two sets to store the cells that are reachable from the Pacific and Atlantic oceans.
+    - We can start DFS from the cells on the borders of the Pacific and Atlantic oceans.
+    - We can find the intersection of the two sets to get the cells that are reachable from both oceans.
+
+
+Time complexity: O(m x n)
+    - m and n are the dimensions of the grid.
+    - We explore all cells in the grid once.
+
+Space complexity: O(m x n)
+    - We use two sets to store the reachable cells from the Pacific and Atlantic oceans.
+    - The maximum size of the sets is the number of cells in the grid.
+"""
+
+
+from typing import List
+
+
+class Solution:
+    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:
+        if not heights or not heights[0]:
+            return []
+
+        def dfs(matrix, reachable, x, y):
+            directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
+            reachable.add((x, y))
+            for dx, dy in directions:
+                nx, ny = x + dx, y + dy
+                if (
+                    0 <= nx < m
+                    and 0 <= ny < n
+                    and (nx, ny) not in reachable
+                    and matrix[nx][ny] >= matrix[x][y]
+                ):
+                    dfs(matrix, reachable, nx, ny)
+
+        m, n = len(heights), len(heights[0])
+        pacific_reachable = set()
+        atlantic_reachable = set()
+
+        for i in range(m):
+            dfs(heights, pacific_reachable, i, 0)
+            dfs(heights, atlantic_reachable, i, n - 1)
+
+        for j in range(n):
+            dfs(heights, pacific_reachable, 0, j)
+            dfs(heights, atlantic_reachable, m - 1, j)
+
+        return list(pacific_reachable & atlantic_reachable)