Graph

Table of Contents

• 200. Number of Islands (Medium)
• 994. Rotting Oranges (Medium)
• 207. Course Schedule (Medium)
• 208. Implement Trie (Prefix Tree) (Medium)

200. Number of Islands

• LeetCode | LeetCode CH (Medium)

• Tags: array, depth first search, breadth first search, union find, matrix

• Count the number of islands in a 2D grid.
• Method 1: DFS

• Method 2: BFS (use a queue to traverse the grid)

• How to keep track of visited cells?

  1. Mark the visited cell as 0 (or any other value) to avoid revisiting it.
  2. Use a set to store the visited cells.

• Steps:

  1. Init: variables
  2. DFS/BFS: starting from the cell with 1, turn all the connected 1s to 0.
  3. Traverse the grid, and if the cell is 1, increment the count and call DFS/BFS.

200. Number of Islands - Python Solution

from collections import deque
from copy import deepcopy
from typing import List


# DFS
def numIslandsDFS(grid: List[List[str]]) -> int:
    if not grid:
        return 0

    m, n = len(grid), len(grid[0])
    res = 0

    def dfs(r, c):
        if r < 0 or r >= m or c < 0 or c >= n or grid[r][c] != "1":
            return

        grid[r][c] = "2"

        dfs(r + 1, c)
        dfs(r - 1, c)
        dfs(r, c + 1)
        dfs(r, c - 1)

    for r in range(m):
        for c in range(n):
            if grid[r][c] == "1":
                dfs(r, c)
                res += 1

    return res


# BFS + Set
def numIslandsBFS1(grid: List[List[str]]) -> int:
    if not grid:
        return 0

    m, n = len(grid), len(grid[0])
    dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)]
    visited = set()
    res = 0

    def bfs(r, c):
        q = deque([(r, c)])

        while q:
            row, col = q.popleft()

            for dr, dc in dirs:
                nr, nc = row + dr, col + dc
                if (
                    nr < 0
                    or nr >= m
                    or nc < 0
                    or nc >= n
                    or grid[nr][nc] == "0"
                    or (nr, nc) in visited
                ):
                    continue

                visited.add((nr, nc))
                q.append((nr, nc))

    for r in range(m):
        for c in range(n):
            if grid[r][c] == "1" and (r, c) not in visited:
                visited.add((r, c))
                bfs(r, c)
                res += 1

    return res


# BFS + Grid
def numIslandsBFS2(grid: List[List[str]]) -> int:
    if not grid:
        return 0

    m, n = len(grid), len(grid[0])
    dirs = [[0, 1], [0, -1], [1, 0], [-1, 0]]
    res = 0

    def bfs(r, c):
        q = deque([(r, c)])

        while q:
            row, col = q.popleft()

            for dr, dc in dirs:
                nr, nc = dr + row, dc + col
                if (
                    nr < 0
                    or nr >= m
                    or nc < 0
                    or nc >= n
                    or grid[nr][nc] != "1"
                ):
                    continue
                grid[nr][nc] = "2"
                q.append((nr, nc))

    for i in range(m):
        for j in range(n):
            if grid[i][j] == "1":
                grid[i][j] = "2"
                bfs(i, j)
                res += 1

    return res


grid = [
    ["1", "1", "1", "1", "0"],
    ["1", "1", "0", "1", "0"],
    ["1", "1", "0", "0", "0"],
    ["0", "0", "0", "0", "0"],
]

print(numIslandsDFS(deepcopy(grid)))  # 1
print(numIslandsBFS1(deepcopy(grid)))  # 1
print(numIslandsBFS2(deepcopy(grid)))  # 1

200. Number of Islands - C++ Solution

#include <vector>
#include <iostream>
using namespace std;

class Solution
{
private:
    void dfs(vector<vector<char>> &grid, int r, int c)
    {
        int row = grid.size();
        int col = grid[0].size();

        if (r < 0 || r >= row || c < 0 || c >= col || grid[r][c] != '1')
        {
            return;
        }
        grid[r][c] = '0';

        dfs(grid, r - 1, c);
        dfs(grid, r + 1, c);
        dfs(grid, r, c - 1);
        dfs(grid, r, c + 1);
    }

public:
    int numIslands(vector<vector<char>> &grid)
    {
        int m = grid.size(), n = grid[0].size();
        int res = 0;
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (grid[i][j] == '1')
                {
                    res++;
                    dfs(grid, i, j);
                }
            }
        }
        return res;
    }
};

int main()
{
    Solution s;
    vector<vector<char>> grid = {
        {'1', '1', '0', '0', '0'},
        {'1', '1', '0', '0', '0'},
        {'0', '0', '1', '0', '0'},
        {'0', '0', '0', '1', '1'}};
    cout << s.numIslands(grid) << endl;
    return 0;
}

994. Rotting Oranges

• LeetCode | LeetCode CH (Medium)

• Tags: array, breadth first search, matrix

• Return the minimum number of minutes that must elapse until no cell has a fresh orange.
• Hint: Multi-source BFS to count the level.

994. Rotting Oranges - Python Solution

from collections import deque
from typing import List


# BFS
def orangesRotting(grid: List[List[int]]) -> int:
    m, n = len(grid), len(grid[0])
    fresh = 0
    q = deque()
    dirs = [[1, 0], [0, 1], [0, -1], [-1, 0]]

    for i in range(m):
        for j in range(n):
            if grid[i][j] == 2:
                q.append([i, j])
            elif grid[i][j] == 1:
                fresh += 1
    res = 0

    while q and fresh > 0:
        size = len(q)
        for _ in range(size):
            r, c = q.popleft()
            for dr, dc in dirs:
                nr, nc = dr + r, dc + c
                if 0 <= nr < m and 0 <= nc < n and grid[nr][nc] == 1:
                    q.append([nr, nc])
                    grid[nr][nc] = 2
                    fresh -= 1
        res += 1

    return res if fresh == 0 else -1


grid = [[2, 1, 1], [1, 1, 0], [0, 1, 1]]
assert orangesRotting(grid) == 4

207. Course Schedule

• LeetCode | LeetCode CH (Medium)

• Tags: depth first search, breadth first search, graph, topological sort

• Return true if it is possible to finish all courses, otherwise return false.
• Dependency relationships imply the topological sort algorithm.
• Cycle detection
• Topological Sort
• DAG (Directed Acyclic Graph)
• Time complexity: O(V+E)
• Space complexity: O(V+E)
• Prerequisites: Indegree (Look at the problem 1557. Minimum Number of Vertices to Reach All Nodes)
  • Indegree: Number of incoming edges to a vertex
• Applications: task scheduling, course scheduling, build systems, dependency resolution, compiler optimization, etc.

Course to prerequisites mapping

flowchart LR
    0((0)) --> 1((1))
    0((0)) --> 2((2))
    1((1)) --> 3((3))
    3((3)) --> 4((4))
    1((1)) --> 4((4))

Prerequisites to course mapping

flowchart LR
    1((1)) --> 0((0))
    2((2)) --> 0((0))
    3((3)) --> 1((1))
    4((4)) --> 3((3))
    4((4)) --> 1((1))

course	0	0	1	1	3
prerequisite	1	2	3	4	4

index	0	1	2	3	4
in-degree	0	0	0	0	0

Initialize

• graph

prerequisite	1	2	3	4
course	[0]	[0]	[1]	[1, 3]

• in-degree

	0	1	2	3	4
in-degree	2	2	0	1	0

• queue: [2, 4]
• pop 2 from the queue

flowchart LR
    1((1)) --> 0((0))
    3((3)) --> 1((1))
    4((4)) --> 3((3))
    4((4)) --> 1((1))

	0	1	2	3	4
in-degree	1	2	0	1	0

• queue: [4]
• pop 4 from the queue

flowchart LR
    1((1)) --> 0((0))
    3((3)) --> 1((1))

	0	1	2	3	4
in-degree	1	1	0	0	0

• queue: [3]
• pop 3 from the queue

flowchart LR
    1((1)) --> 0((0))

	0	1	2	3	4
in-degree	1	0	0	0	0

• queue: [1]
• pop 1 from the queue

flowchart LR
    0((0))

	0	1	2	3	4
in-degree	0	0	0	0	0

• queue: [0]
• pop 0 from the queue
• All courses are taken. Return True.

207. Course Schedule - Python Solution

from collections import defaultdict, deque
from typing import List


# BFS (Kahn's Algorithm)
def canFinishBFS(numCourses: int, prerequisites: List[List[int]]) -> bool:
    graph = defaultdict(list)
    indegree = defaultdict(int)

    for crs, pre in prerequisites:
        graph[pre].append(crs)
        indegree[crs] += 1

    q = deque([i for i in range(numCourses) if indegree[i] == 0])
    count = 0

    while q:
        crs = q.popleft()
        count += 1

        for nxt in graph[crs]:
            indegree[nxt] -= 1

            if indegree[nxt] == 0:
                q.append(nxt)

    return count == numCourses


# DFS + Set
def canFinishDFS1(numCourses: int, prerequisites: List[List[int]]) -> bool:
    graph = defaultdict(list)
    for crs, pre in prerequisites:
        graph[crs].append(pre)

    visiting = set()

    def dfs(crs):
        if crs in visiting:  # cycle detected
            return False
        if graph[crs] == []:
            return True

        visiting.add(crs)

        for pre in graph[crs]:
            if not dfs(pre):
                return False

        visiting.remove(crs)
        graph[crs] = []

        return True

    for crs in range(numCourses):
        if not dfs(crs):
            return False
    return True


# DFS + List
def canFinishDFS2(numCourses: int, prerequisites: List[List[int]]) -> bool:
    graph = defaultdict(list)
    for pre, crs in prerequisites:
        graph[crs].append(pre)

    # 0: init, 1: visiting, 2: visited
    status = [0] * numCourses

    def dfs(crs):
        if status[crs] == 1:  # cycle detected
            return False
        if status[crs] == 2:
            return True

        status[crs] = 1

        for pre in graph[crs]:
            if not dfs(pre):
                return False

        status[crs] = 2
        return True

    for crs in range(numCourses):
        if not dfs(crs):
            return False
    return True


prerequisites = [[0, 1], [0, 2], [1, 3], [1, 4], [3, 4]]
print(canFinishBFS(5, prerequisites))  # True
print(canFinishDFS1(5, prerequisites))  # True
print(canFinishDFS2(5, prerequisites))  # True

207. Course Schedule - C++ Solution

#include <functional>
#include <iostream>
#include <queue>
#include <vector>
using namespace std;

class Solution {
   public:
    // BFS
    bool canFinishBFS(int numCourses, vector<vector<int>> &prerequisites) {
        vector<vector<int>> graph(numCourses);
        vector<int> indegree(numCourses, 0);
        for (auto &pre : prerequisites) {
            graph[pre[1]].push_back(pre[0]);
            indegree[pre[0]]++;
        }

        queue<int> q;
        for (int i = 0; i < numCourses; i++) {
            if (indegree[i] == 0) {
                q.push(i);
            }
        }

        int cnt = 0;
        while (!q.empty()) {
            int cur = q.front();
            q.pop();
            cnt++;

            for (int nxt : graph[cur]) {
                indegree[nxt]--;
                if (indegree[nxt] == 0) {
                    q.push(nxt);
                }
            }
        }
        return cnt == numCourses;
    }

    // DFS
    bool canFinishDFS(int numCourses, vector<vector<int>> &prerequisites) {
        vector<vector<int>> graph(numCourses);
        for (auto &pre : prerequisites) {
            graph[pre[1]].push_back(pre[0]);
        }
        // 0: not visited, 1: visiting, 2: visited
        vector<int> state(numCourses, 0);

        function<bool(int)> dfs = [&](int pre) -> bool {
            state[pre] = 1;  // visiting
            for (int crs : graph[pre]) {
                if (state[crs] == 1 || (state[crs] == 0 && dfs(crs))) {
                    return true;
                }
            }
            state[pre] = 2;  // visited
            return false;
        };

        for (int i = 0; i < numCourses; i++) {
            if (state[i] == 0 && dfs(i)) {
                return false;
            }
        }
        return true;
    }
};

int main() {
    Solution sol;
    vector<vector<int>> prerequisites = {{1, 0}, {2, 1}, {3, 2}, {4, 3},
                                         {5, 4}, {6, 5}, {7, 6}, {8, 7},
                                         {9, 8}, {10, 9}};
    int numCourses = 11;
    cout << sol.canFinishBFS(numCourses, prerequisites) << endl;
    cout << sol.canFinishDFS(numCourses, prerequisites) << endl;
    return 0;
}

208. Implement Trie (Prefix Tree)

• LeetCode | LeetCode CH (Medium)

• Tags: hash table, string, design, trie

Trie

• A trie is a tree-like data structure whose nodes store the letters of an alphabet.

flowchart TD
Root(( ))
Root --- C1(("C"))
Root --- D((D))
C1 --- A1(("A"))
A1 --- T1(("T"))
A1 --- R1(("R"))
A1 --- N((N))
Root --- B1((B))
B1 --- A2((A))
A2 --- T2((T))
A2 --- R2((R))

208. Implement Trie (Prefix Tree) - Python Solution

class TrieNode:
    def __init__(self):
        self.children = {}
        self.endOfWord = None


class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, word: str) -> None:
        node = self.root

        for c in word:
            if c not in node.children:
                node.children[c] = TrieNode()
            node = node.children[c]
        node.endOfWord = True

    def search(self, word: str) -> bool:
        node = self.root

        for c in word:
            if c not in node.children:
                return False
            node = node.children[c]
        return node.endOfWord

    def startsWith(self, prefix: str) -> bool:
        node = self.root

        for c in prefix:
            if c not in node.children:
                return False
            node = node.children[c]
        return True


# Your Trie object will be instantiated and called as such:
obj = Trie()
obj.insert("apple")
print(obj.search("word"))  # False
print(obj.startsWith("app"))  # True

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