Sync LeetCode submission Runtime - 3 ms (89.27%), Memory - 18.8 MB (77.79%)

Author: jimit105

0207-course-schedule/README.md

@@ -0,0 +1,37 @@
+<p>There are a total of <code>numCourses</code> courses you have to take, labeled from <code>0</code> to <code>numCourses - 1</code>. You are given an array <code>prerequisites</code> where <code>prerequisites[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that you <strong>must</strong> take course <code>b<sub>i</sub></code> first if you want to take course <code>a<sub>i</sub></code>.</p>
+
+<ul>
+	<li>For example, the pair <code>[0, 1]</code>, indicates that to take course <code>0</code> you have to first take course <code>1</code>.</li>
+</ul>
+
+<p>Return <code>true</code> if you can finish all courses. Otherwise, return <code>false</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> numCourses = 2, prerequisites = [[1,0]]
+<strong>Output:</strong> true
+<strong>Explanation:</strong> There are a total of 2 courses to take. 
+To take course 1 you should have finished course 0. So it is possible.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> numCourses = 2, prerequisites = [[1,0],[0,1]]
+<strong>Output:</strong> false
+<strong>Explanation:</strong> There are a total of 2 courses to take. 
+To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= numCourses &lt;= 2000</code></li>
+	<li><code>0 &lt;= prerequisites.length &lt;= 5000</code></li>
+	<li><code>prerequisites[i].length == 2</code></li>
+	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; numCourses</code></li>
+	<li>All the pairs prerequisites[i] are <strong>unique</strong>.</li>
+</ul>

0207-course-schedule/solution.py

@@ -0,0 +1,34 @@
+# Approach 1: Topological Sort Using Kahn's Algorithm
+
+# n = no. of coures, m = len(prerequistes)
+# Time: O(m + n)
+# Space: O(m + n)
+
+from collections import deque
+
+class Solution:
+    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
+        indegree = [0] * numCourses
+        adj = [[] for _ in range(numCourses)]
+
+        for prereq in prerequisites:
+            adj[prereq[1]].append(prereq[0])
+            indegree[prereq[0]] += 1
+
+        queue = deque()
+        for i in range(numCourses):
+            if indegree[i] == 0:
+                queue.append(i)
+
+        nodesVisited = 0
+        while queue:
+            node = queue.popleft()
+            nodesVisited += 1
+
+            for neighbor in adj[node]:
+                indegree[neighbor] -= 1
+                if indegree[neighbor] == 0:
+                    queue.append(neighbor)
+
+        return nodesVisited == numCourses
+