Two-Hundred-Twenty-One Commit: Add Tasks Scheduling problem to Topological Sort Section

Author: AAdewunmi

src/Topological_Sort/Problem_2_Tasks_Scheduling.java

@@ -0,0 +1,55 @@
+package Topological_Sort;
+
+// Problem Statement: Tasks Scheduling (medium)
+// LeetCode Question:
+
+import java.util.*;
+
+public class Problem_2_Tasks_Scheduling {
+    public boolean isSchedulingPossible(int tasks, int[][] prerequisites) {
+        List<Integer> sortedOrder = new ArrayList<>();
+        if (tasks <= 0)
+            return false;
+
+        // a. Initialize the graph
+        HashMap<Integer, Integer> inDegree =
+                new HashMap<>(); // count of incoming edges for every vertex
+        HashMap<Integer, List<Integer>> graph = new HashMap<>(); // adjacency list graph
+        for (int i = 0; i < tasks; i++) {
+            inDegree.put(i, 0);
+            graph.put(i, new ArrayList<Integer>());
+        }
+
+        // b. Build the graph
+        for (int i = 0; i < prerequisites.length; i++) {
+            int parent = prerequisites[i][0], child = prerequisites[i][1];
+            graph.get(parent).add(child); // put the child into it's parent's list
+            inDegree.put(child, inDegree.get(child) + 1); // increment child's inDegree
+        }
+
+        // c. Find all sources i.e., all vertices with 0 in-degrees
+        Queue<Integer> sources = new LinkedList<>();
+        for (Map.Entry<Integer, Integer> entry : inDegree.entrySet()) {
+            if (entry.getValue() == 0)
+                sources.add(entry.getKey());
+        }
+
+        // d. For each source, add it to the sortedOrder and subtract one from all of its
+        // children's in-degrees if a child's in-degree becomes zero, add it to sources queue
+        while (!sources.isEmpty()) {
+            int vertex = sources.poll();
+            sortedOrder.add(vertex);
+            // get the node's children to decrement their in-degrees
+            List<Integer> children = graph.get(vertex);
+            for (int child : children) {
+                inDegree.put(child, inDegree.get(child) - 1);
+                if (inDegree.get(child) == 0)
+                    sources.add(child);
+            }
+        }
+
+        // if sortedOrder doesn't contain all tasks, there is a cyclic dependency between
+        // tasks, therefore, we will not be able to schedule all tasks
+        return sortedOrder.size() == tasks;
+    }
+}