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LeetCode #207 (Course Schedule)

1// Topological Sort using Kahn's BFS Algorithm

2class Solution {

3public:

4 bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {

5 vector<int> inDegree(numCourses, 0);

6 vector<vector<int>> adj(numCourses);

7 queue<int> q;

8 Type here to test live autocomplete

CodePilot Suggestions

push(val) Queue Method

pop() Method

prerequisites Parameter

9 return visitedCount == numCourses;

10 }

11};

AI Coach (Level 1 Hint) Step 1 of 4

This problem is about finding cycles in a directed graph. Start by finding courses that have no prerequisites (in-degree is 0), then work forward.

Test Code (Judge0 Runner) Ready

// Click "Run Code" to test this solution
Test 1: [[1,0]] → true
Test 2: [[1,0],[0,1]] → false

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269. Alien Dictionary

Hard 🔒 Unlocked Free

Google (38) Meta (29) Amazon (24) Microsoft (16)

There is a new alien language that uses the English alphabet, but the order of the letters is different.

You are given a list of words from the alien dictionary sorted in their alphabet's order.

Return a string showing the letters sorted in the correct order. If there is no valid order (a cycle exists), return "".

Example 1:

Input: words = ["wrt","wrf","er","ett","rftt"]

Output: "wertf"

Example 2:

Input: words = ["z","x"]

Output: "zx"

Rules & Limits:
• 1 to 100 words in the dictionary
• Each word is 1 to 100 characters long
• All characters are lowercase English letters

🔗 Free Community Mirror: Also available on LintCode Open on LintCode ↗

Starter Code Loaded • Ready to Run Python 3.10

class Solution: def alienOrder(self, words: List[str]) -> str: adj = {c: set() for w in words for c in w} in_degree = {c: 0 for c in adj} # Build directed graph of letter order for first, second in zip(words, words[1:]): for c1, c2 in zip(first, second): if c1 != c2: if c2 not in adj[c1]: adj[c1].add(c2) in_degree[c2] += 1 break else: if len(second) < len(first): return "" # BFS Topological Sort q = deque([c for c in in_degree if in_degree[c] == 0]) res = [] while q: c = q.popleft() res.append(c) for neighbor in adj[c]: in_degree[neighbor] -= 1 if in_degree[neighbor] == 0: q.append(neighbor) return "".join(res) if len(res) == len(in_degree) else ""

Test Results (Judge0 Compiler) ✓ 2/2 Test Cases Passed

Test 1: ["wrt","wrf","er","ett","rftt"] → Output: "wertf"
Test 2: ["z","x"] → Output: "zx"
Memory: 14.1 MB • Speed: 31ms (Faster than 98.4%)

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queue<int> q; q.p|

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