fix: Add more test cases to improve code coverage

Author: sairamsharan

src/main/java/com/thealgorithms/graph/HierholzerAlgorithm.java

@@ -1,90 +1,118 @@
 package com.thealgorithms.graph;
 
-import java.util.*;
+import java.util.Collections;
+import java.util.HashMap;
+import java.util.HashSet;
+import java.util.LinkedList;
+import java.util.List;
+import java.util.Map;
+import java.util.Set;
+import java.util.Stack;
 
 /**
  * An implementation of Hierholzer's Algorithm to find an Eulerian Path or Circuit in an undirected graph.
- * This algorithm finds a trail in a graph that visits every edge exactly once.
- * Think of it like solving a puzzle where you trace every line without lifting your pen.
+ * An Eulerian path is a trail in a graph that visits every edge exactly once.
+ * An Eulerian circuit is an Eulerian path that starts and ends at the same vertex.
  *
  * Wikipedia: https://en.wikipedia.org/wiki/Eulerian_path#Hierholzer's_algorithm
  */
 public class HierholzerAlgorithm {
 
+    private final int numVertices;
     private final Map<Integer, LinkedList<Integer>> graph;
 
     /**
-     * Sets up the algorithm with the graph we want to solve.
+     * Constructor for the algorithm.
      * @param graph The graph represented as an adjacency list.
+     * Using a LinkedList for neighbors is efficient for edge removal.
      */
     public HierholzerAlgorithm(Map<Integer, LinkedList<Integer>> graph) {
-        this.graph = (graph == null) ? new HashMap<>() : graph;
+        if (graph == null) {
+            this.graph = new HashMap<>();
+            this.numVertices = 0;
+            return;
+        }
+        this.graph = graph;
+        this.numVertices = graph.size();
     }
 
     /**
-     * Before starting, we have to ask: can this puzzle even be solved?
-     * This method checks the two essential rules for an undirected graph.
-     * @return true if a circuit is possible, false otherwise.
+     * Checks if an Eulerian circuit exists in the undirected graph.
+     * Condition: All vertices with a non-zero degree must be in a single connected component,
+     * and all vertices must have an even degree.
+     * @return true if a circuit exists, false otherwise.
      */
     public boolean hasEulerianCircuit() {
         if (graph.isEmpty()) {
-            return true; // An empty puzzle is trivially solved.
+            return true; // An empty graph has an empty circuit.
         }
 
-        // Rule 1: Every point must have an even number of lines connected to it.
-        // This ensures for every way in, there's a way out.
+        // Check 1: All vertices must have an even degree.
         for (int vertex : graph.keySet()) {
             if (graph.get(vertex).size() % 2 != 0) {
-                return false; // Found a point with an odd number of lines.
+                return false; // Found a vertex with an odd degree.
             }
         }
 
-        // Rule 2: The drawing must be one single, connected piece.
-        // You can't have a separate, floating part of the puzzle.
-        return isCoherentlyConnected();
+        // Check 2: All vertices with edges must be connected.
+        if (!isCoherentlyConnected()) {
+            return false;
+        }
+
+        return true;
     }
 
     /**
-     * This is the main event—finding the actual path.
-     * @return A list of points (vertices) that make up the complete circuit.
+     * Finds the Eulerian circuit.
+     * @return A list of vertices representing the circuit, or an empty list if none exists.
      */
     public List<Integer> findEulerianCircuit() {
         if (!hasEulerianCircuit()) {
-            // If the puzzle can't be solved, return an empty path.
             return Collections.emptyList();
         }
 
-        // We'll work on a copy of the graph so we don't destroy the original.
+        // Create a copy of the graph to avoid modifying the original during traversal.
         Map<Integer, LinkedList<Integer>> tempGraph = new HashMap<>();
         for (Map.Entry<Integer, LinkedList<Integer>> entry : graph.entrySet()) {
             tempGraph.put(entry.getKey(), new LinkedList<>(entry.getValue()));
         }
 
-        // 'currentPath' is our breadcrumb trail as we explore.
+        // Data structures for the algorithm.
         Stack<Integer> currentPath = new Stack<>();
-        // 'circuit' is where we'll lay out the final, complete path.
         List<Integer> circuit = new LinkedList<>();
 
-        // Find any point to start from.
-        int startVertex = graph.keySet().stream().findFirst().orElse(-1);
-        if (startVertex == -1) return Collections.emptyList();
+        // Find a starting vertex (any vertex with edges).
+        int startVertex = -1;
+        for (int vertex : tempGraph.keySet()) {
+            if (!tempGraph.get(vertex).isEmpty()) {
+                startVertex = vertex;
+                break;
+            }
+        }
+
+        if (startVertex == -1) {
+            if (graph.isEmpty()) {
+                return Collections.emptyList();
+            }
+            return Collections.singletonList(graph.keySet().iterator().next()); // Graph with one isolated vertex.
+        }
 
         currentPath.push(startVertex);
 
         while (!currentPath.isEmpty()) {
             int currentVertex = currentPath.peek();
 
-            // If there's an unexplored hallway from our current location...
+            // If the current vertex has unvisited edges
             if (tempGraph.containsKey(currentVertex) && !tempGraph.get(currentVertex).isEmpty()) {
-                // ...let's go down it.
-                int nextVertex = tempGraph.get(currentVertex).pollFirst();
-                // Erase the hallway behind us so we don't use it again.
+                int nextVertex = tempGraph.get(currentVertex).pollFirst(); // Get a neighbor
+
+                // Remove the reverse edge as well (for undirected graph)
                 tempGraph.get(nextVertex).remove(Integer.valueOf(currentVertex));
-                // Add the new location to our breadcrumb trail.
+
+                // Push the neighbor to the stack to continue the tour
                 currentPath.push(nextVertex);
             } else {
-                // If we've hit a dead end, we're done with this part of the tour.
-                // We add our location to the final path and backtrack.
+                // If "stuck" (no more edges), backtrack and add to the final circuit.
                 circuit.add(0, currentPath.pop());
             }
         }
@@ -93,20 +121,37 @@ public List<Integer> findEulerianCircuit() {
     }
 
     /**
-     * A helper to check if the graph is one single piece.
-     * It does a simple walk (DFS) starting from one point and checks if it can reach all other points.
+     * Helper method to check if all vertices with a non-zero degree are connected.
+     * Uses a simple traversal (DFS).
      */
     private boolean isCoherentlyConnected() {
-        if (graph.isEmpty()) return true;
+        if (graph.isEmpty()) {
+            return true;
+        }
+
         Set<Integer> visited = new HashSet<>();
-        int startNode = graph.keySet().stream().findFirst().orElse(-1);
-        if (startNode == -1) return true;
+        int startNode = -1;
+
+        // Find the first vertex with a degree greater than 0
+        for (int vertex : graph.keySet()) {
+            if (!graph.get(vertex).isEmpty()) {
+                startNode = vertex;
+                break;
+            }
+        }
+
+        // If no edges in the graph, it's connected.
+        if (startNode == -1) {
+            return true;
+        }
 
+        // Perform DFS from the start node
         dfs(startNode, visited);
 
+        // Check if all vertices with edges were visited
         for (int vertex : graph.keySet()) {
             if (!graph.get(vertex).isEmpty() && !visited.contains(vertex)) {
-                return false; // Found a part of the puzzle we couldn't reach.
+                return false; // Found a vertex with edges that wasn't visited
             }
         }
         return true;

src/test/java/com/thealgorithms/graph/HierholzerAlgorithmTest.java

@@ -1,9 +1,15 @@
 package com.thealgorithms.graph;
 
 import static org.junit.jupiter.api.Assertions.assertEquals;
+import static org.junit.jupiter.api.Assertions.assertFalse;
 import static org.junit.jupiter.api.Assertions.assertTrue;
 
-import java.util.*;
+import java.util.Arrays;
+import java.util.Collections;
+import java.util.HashMap;
+import java.util.LinkedList;
+import java.util.List;
+import java.util.Map;
 import org.junit.jupiter.api.Test;
 
 public class HierholzerAlgorithmTest {
@@ -31,16 +37,42 @@ public void testFindsEulerianCircuitInSimpleTriangleGraph() {
     }
 
     @Test
-    public void testHandlesGraphWithNoEulerianCircuit() {
-        // Create a graph where a vertex has an odd degree
+    public void testFailsForGraphWithOddDegreeVertices() {
+        // Create a graph where vertices 0 and 1 have an odd degree (1)
         Map<Integer, LinkedList<Integer>> graph = new HashMap<>();
         graph.put(0, new LinkedList<>(Collections.singletonList(1)));
         graph.put(1, new LinkedList<>(Collections.singletonList(0)));
-        graph.put(2, new LinkedList<>(Collections.emptyList())); // Vertex 2 is isolated
 
         HierholzerAlgorithm algorithm = new HierholzerAlgorithm(graph);
 
         // The algorithm should correctly identify that no circuit exists
-        assertEquals(false, algorithm.hasEulerianCircuit());
+        assertFalse(algorithm.hasEulerianCircuit());
+        // The find method should return an empty list
+        assertTrue(algorithm.findEulerianCircuit().isEmpty());
+    }
+
+    @Test
+    public void testFailsForDisconnectedGraph() {
+        // Create a graph with two separate triangles (0-1-2 and 3-4-5)
+        Map<Integer, LinkedList<Integer>> graph = new HashMap<>();
+        graph.put(0, new LinkedList<>(Arrays.asList(1, 2)));
+        graph.put(1, new LinkedList<>(Arrays.asList(0, 2)));
+        graph.put(2, new LinkedList<>(Arrays.asList(0, 1)));
+        graph.put(3, new LinkedList<>(Arrays.asList(4, 5)));
+        graph.put(4, new LinkedList<>(Arrays.asList(3, 5)));
+        graph.put(5, new LinkedList<>(Arrays.asList(3, 4)));
+
+        HierholzerAlgorithm algorithm = new HierholzerAlgorithm(graph);
+
+        // All degrees are even, but the graph is not connected, so no circuit exists
+        assertFalse(algorithm.hasEulerianCircuit());
+    }
+
+    @Test
+    public void testHandlesEmptyGraph() {
+        Map<Integer, LinkedList<Integer>> graph = new HashMap<>();
+        HierholzerAlgorithm algorithm = new HierholzerAlgorithm(graph);
+        assertTrue(algorithm.hasEulerianCircuit());
+        assertTrue(algorithm.findEulerianCircuit().isEmpty());
     }
 }