refactor(algorithms, graphs): reorder routes

algorithms/graphs/reorder_routes/README.md

@@ -15,15 +15,15 @@ It's guaranteed that each city can reach city 0 after reorder.
 
 Example 1
 
-![](reorder_routes_example_1.png)
+![](images/examples/reorder_routes_example_1.png)
 
 Input: n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
 Output: 3
 Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).
 
 Example 2
 
-![](reorder_routes_example_2.png)
+![](images/examples/reorder_routes_example_2.png)
 
 Input: n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
 Output: 2
@@ -33,6 +33,57 @@ Example 3
 Input: n = 3, connections = [[1,0],[2,0]]
 Output: 0
 
+## Solution
+
+This algorithm works because representing the network as a graph with directional information allows us to easily 
+identify which roads are misoriented. By performing a DFS from the capital (city 0), we traverse the entire network 
+exactly once. Every time we encounter a road directed away from city 0, we know it needs to be reversed. This approach 
+takes advantage of the network’s tree structure to ensure we only count the minimum number of necessary reversals.
+
+Now, let’s look at the solution steps below:
+
+1. Build an adjacency list using a variable graph to store the roads and their directional flags.
+   - For each connection [ai, bi] in connections:
+     - Append (bi, 1) to graph[ai], where the flag 1 indicates that the road goes from ai to bi and might need to be 
+       reversed.
+     - Append (ai, 0) to graph[bi], where the flag 0 represents the reverse edge, which is correctly oriented for the 
+       DFS.
+2. Create a set, visited, to track which cities have been processed, preventing repeated visits.
+3. Call dfs(0, graph, visited) to start the DFS from the capital (city 0) and store the result in a variable result.
+4. Return result as the final answer is the minimum number of road reorientations needed.
+
+**Define the DFS function**
+
+1. Start the DFS from city 0
+   - Add the current city to the visited set.
+   - Initializes a variable, reversals, to count the number of road reversals needed for the subtree rooted at that city.
+   - For each neighbor and its associated flag (represented by need_reverse) in graph[city]:
+     - If the neighbor hasn’t been visited, add need_reverse to result (As need_reverse equals 1 if the road needs reversal).
+     - Recursively call dfs(neighbor, graph, visited) to continue the traversal.
+   - Returns reversals.
+
+Let’s look at the following illustration to get a better understanding of the solution:
+
+![Solution 1](./images/solutions/reorder_routes_solution_1.png)
+![Solution 2](./images/solutions/reorder_routes_solution_2.png)
+![Solution 3](./images/solutions/reorder_routes_solution_3.png)
+![Solution 4](./images/solutions/reorder_routes_solution_4.png)
+![Solution 5](./images/solutions/reorder_routes_solution_5.png)
+![Solution 6](./images/solutions/reorder_routes_solution_6.png)
+![Solution 7](./images/solutions/reorder_routes_solution_7.png)
+![Solution 8](./images/solutions/reorder_routes_solution_8.png)
+![Solution 9](./images/solutions/reorder_routes_solution_9.png)
+
+### Time Complexity
+
+The time complexity of the solution is O(n), where n is the number of cities because every node and its corresponding 
+edges are visited exactly once during the DFS.
+
+### Space Complexity
+
+The solution’s space complexity is O(n) due to the storage needed for the graph (adjacency list), the visited set, and 
+the recursion call stack.
+
 ## Related Topics
 
 - Depth First Search

algorithms/graphs/reorder_routes/__init__.py

@@ -2,65 +2,85 @@
 from collections import defaultdict
 
 
-class Solution:
+def min_reorder(n: int, connections: List[List[int]]) -> int:
+    """Reorders the edges of the directed graph represented as an adjacency list in connections such that each vertex
+    is connected and has a path to the first vertex marked with 0. n represents the number of vertices.
+
+    Complexity;
+    n is the number of vertices/nodes
+
+    Time Complexity: O(n)
+        O(n) to initialize the adjacency list
+        The dfs function visits each node once, which takes O(n) time in total. Because we have undirected edges,
+        each edge can only be iterated twice (by nodes at the end), resulting in O(e) operations total while
+        visiting all nodes, where e is the number of edges. Because the given graph is a tree, there are n−1
+        undirected edges, so O(n+e)=O(n).
+
+    Space Complexity: O(n)
+        Building the adjacency list takes O(n) space.
+        The recursion call stack used by dfs can have no more than n elements in the worst-case scenario.
+         It would take up O(n) space in that case.
+
+    Args:
+        n(int): number of vertices or in this case, number of cities
+        connections(list): adjacency matrix for a directed graph or in this case, representation of cities
+    Returns:
+        int: minimum number of edges to re-arrange to ensure that each vertex is directly or indirectly connected to
+        the initial vertex
+    """
+
+    # Adjacency list that contains list of pairs of nodes such that adj[node] contains all the neighbours of node in the
+    # form of [neighbour, sign] where neighbour is the neighbouring node and sign is the direction of the edge. If the
+    # sign is 0, it's an 'artificial' edge, meaning it was added by the algorithm in order to get to this vertex, and 1
+    # denotes that it's an 'original' edge, meaning that it's the original edge and no need to re-order that connection
+    adj: Dict[int, List[List[int]]] = defaultdict(lambda: [])
+
     # keep track of number of reorders made
     reorder_count = 0
 
-    def min_reorder(self, n: int, connections: List[List[int]]) -> int:
-        """Reorders the edges of the directed graph represented as an adjacency list in connections such that each vertex
-        is connected and has a path to the first vertex marked with 0. n represents the number of vertices.
-
-        Complexity;
-        n is the number of vertices/nodes
-
-        Time Complexity: O(n)
-            O(n) to initialize the adjacency list
-            The dfs function visits each node once, which takes O(n) time in total. Because we have undirected edges,
-            each edge can only be iterated twice (by nodes at the end), resulting in O(e) operations total while
-            visiting all nodes, where e is the number of edges. Because the given graph is a tree, there are n−1
-            undirected edges, so O(n+e)=O(n).
-
-        Space Complexity: O(n)
-            Building the adjacency list takes O(n) space.
-            The recursion call stack used by dfs can have no more than n elements in the worst-case scenario.
-             It would take up O(n) space in that case.
-
-        Args:
-            n(int): number of vertices or in this case, number of cities
-            connections(list): adjacency matrix for a directed graph or in this case, representation of cities
-        Returns:
-            int: minimum number of edges to re-arrange to ensure that each vertex is directly or indirectly connected to
-            the initial vertex
-        """
-
-        # Adjacency list that contains list of pairs of nodes such that adj[node] contains all the neighbours of node in the
-        # form of [neighbour, sign] where neighbour is the neighbouring node and sign is the direction of the edge. If the
-        # sign is 0, it's an 'artificial' edge, meaning it was added by the algorithm in order to get to this vertex, and 1
-        # denotes that it's an 'original' edge, meaning that it's the original edge and no need to re-order that connection
-        adj: Dict[int, List[List[int]]] = defaultdict(lambda: [])
-
-        def dfs(node: int, parent: int, adjacency: Dict[int, List[List[int]]]):
-            if node not in adjacency:
-                return
-
-            # iterate over all children of node(nodes that share an edge)
-            # for every child, sign, check if child is equal to parent. if child is equal to parent, we will not visit it
-            # again
-            # if child is not equal to parent, we perform count+=sign and recursively call the dfs with node = child and
-            # parent = node
-            for adjacent_node in adjacency.get(node):
-                child = adjacent_node[0]
-                sign = adjacent_node[1]
-
-                if child != parent:
-                    self.reorder_count += sign
-                    dfs(child, node, adjacency)
-
-        for connection in connections:
-            adj[connection[0]].append([connection[1], 1])
-            adj[connection[1]].append([connection[0], 0])
-
-        # we start with node, parent and the adjacency list as 0, and -1 and adj
-        dfs(0, -1, adj)
-
-        return self.reorder_count
+    def dfs(node: int, parent: int, adjacency: Dict[int, List[List[int]]]):
+        nonlocal reorder_count
+        if node not in adjacency:
+            return
+
+        # iterate over all children of node(nodes that share an edge)
+        # for every child, sign, check if child is equal to parent. if child is equal to parent, we will not visit it
+        # again
+        # if child is not equal to parent, we perform count+=sign and recursively call the dfs with node = child and
+        # parent = node
+        for adjacent_node in adjacency.get(node):
+            child = adjacent_node[0]
+            sign = adjacent_node[1]
+
+            if child != parent:
+                reorder_count += sign
+                dfs(child, node, adjacency)
+
+    for connection in connections:
+        adj[connection[0]].append([connection[1], 1])
+        adj[connection[1]].append([connection[0], 0])
+
+    # we start with node, parent and the adjacency list as 0, and -1 and adj
+    dfs(0, -1, adj)
+
+    return reorder_count
+
+
+def min_reorder_2(n: int, connections: List[List[int]]) -> int:
+    def dfs(city, graph, visited):
+        visited.add(city)
+        reversals = 0
+        for neighbor, need_reverse in graph[city]:
+            if neighbor not in visited:
+                reversals += need_reverse
+                reversals += dfs(neighbor, graph, visited)
+        return reversals
+
+    graph = defaultdict(list)
+    for ai, bi in connections:
+        graph[ai].append((bi, 1))
+        graph[bi].append((ai, 0))
+
+    visited = set()
+    result = dfs(0, graph, visited)
+    return result

algorithms/graphs/reorder_routes/test_reorder_routes.py

@@ -1,30 +1,25 @@
 import unittest
-from . import Solution
+from typing import List
+from parameterized import parameterized
+from algorithms.graphs.reorder_routes import min_reorder
 
+REORDER_ROUTES_TEST_CASES = [
+    (6, [[0, 1], [1, 3], [2, 3], [4, 0], [4, 5]], 3),
+    (5, [[1, 0], [1, 2], [3, 2], [3, 4]], 2),
+    (3, [[1, 0], [2, 0]], 0),
+    (4, [[0, 1], [1, 2], [2, 3]], 3),
+    (6, [[0, 1], [2, 0], [3, 2], [4, 3], [5, 4]], 1),
+    (5, [[1, 0], [2, 1], [3, 2], [4, 3]], 0),
+    (7, [[0, 1], [2, 0], [3, 2], [4, 3], [5, 3], [6, 5]], 1),
+]
 
-class ReorderRoutesTestCase(unittest.TestCase):
-    def test_1(self):
-        """should return 3 from input n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]"""
-        connections = [[0, 1], [1, 3], [2, 3], [4, 0], [4, 5]]
-        n = 6
-        expected = 3
-        actual = Solution().min_reorder(n, connections)
-        self.assertEqual(expected, actual)
 
-    def test_2(self):
-        """should return 2 from input n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]"""
-        connections = [[1, 0], [1, 2], [3, 2], [3, 4]]
-        n = 5
-        expected = 2
-        actual = Solution().min_reorder(n, connections)
-        self.assertEqual(expected, actual)
-
-    def test_3(self):
-        """should return 0 from input n = 3, connections = [[1,0],[2,0]]"""
-        connections = [[1, 0], [2, 0]]
-        n = 3
-        expected = 0
-        actual = Solution().min_reorder(n, connections)
+class ReorderRoutesTestCase(unittest.TestCase):
+    @parameterized.expand(REORDER_ROUTES_TEST_CASES)
+    def test_min_reorder_routes(
+        self, n: int, connections: List[List[int]], expected: int
+    ):
+        actual = min_reorder(n, connections)
         self.assertEqual(expected, actual)