Solution For Reorder Routes To Make All Paths Lead To The City Zero

Problem Statement:

You are given a list of routes, where routes[i] = [start_i, end_i] represents a direct route between two cities. Each city is labeled with a different integer from 0 to n – 1.

You want to make all the cities connected to the city zero. So, you begin with the city zero and can visit any city by traveling through the direct routes.

Return the minimum number of edges to travel to connect all the cities.

Example 1:
Input: n = 6, connections = [[0,1],[0,2],[3,4],[2,3],[1,2]] Output: 3
Explanation: The optimal strategy is as follows:
– Connect cities 0 and 1 together
– Connect cities 0 and 2 together
– Connect cities 2 and 3 together
Minimum number of edges to connect all cities is 3, which is the path of 0 -> 1 -> 2 -> 3.

Solution:

To solve this problem, we can use the Breadth First Search (BFS) algorithm. We can start the search from the source node, which is 0 in this problem, and traverse all the connected nodes. During the traversal, we can keep track of the visited nodes and add the unvisited nodes to a queue. We can continue this process until we have traversed all the connected nodes.

Now, we need to reorder the paths such that they all lead to city zero. To do this, we can create a tree with the source node at the root and all the connected nodes as its children. We can then traverse the tree in post-order and keep track of the number of edges for each node. The number of edges for a node is the number of edges required to reach it from its children. We can then add this number to the total number of edges required to reach the parent node. We can continue this process until we have computed the number of edges for all the nodes.

At each node, we can check if the number of edges required to reach the parent node is less than or greater than the number of edges required to reach the node from its parent. If it is less than the number of edges required to reach the node from its parent, we need to reorder the paths.

To reorder the paths, we can swap the start and end nodes of the path. We can then add the path to a list of reordered paths. We can continue this process until we have reordered all the paths.

Finally, we can return the number of reordered paths as the minimum number of edges required to connect all the cities.

Code:

The code for the above solution is as follows:

“`
from collections import defaultdict, deque

class Solution:
def minReorder(self, n: int, connections: List[List[int]]) -> int:
graph = defaultdict(list)
for u, v in connections:
graph[u].append((v, 1))
graph[v].append((u, 0))

    visited = [False] * n
    visited[0] = True

    queue = deque([0])
    while queue:
        node = queue.popleft()
        for neighbor, direct in graph[node]:
            if not visited[neighbor]:
                visited[neighbor] = True
                queue.append(neighbor)
                graph[node].remove((neighbor, direct))
                graph[neighbor].append((node, direct))

    reorder_count = 0
    visited = [False] * n
    visited[0] = True

    def dfs(node):
        nonlocal reorder_count
        for neighbor, direct in graph[node]:
            if not visited[neighbor]:
                visited[neighbor] = True
                dfs(neighbor)
                if direct == 1:
                    reorder_count += 1
            else:
                if direct == 0:
                    reorder_count += 1

    dfs(0)
    return reorder_count

“`

Explanation:

The above code first creates an adjacency list to represent the graph. For each edge (u, v), we add v to the adjacency list of u with a weight of 1, and we add u to the adjacency list of v with a weight of 0. The weight of an edge represents whether we need to reorder the path or not.

We then use BFS to traverse all the connected nodes from the source node 0. During the traversal, we update the adjacency list of each node by removing the incoming edge and adding a new outgoing edge with the same weight.

We then use DFS to traverse the tree in post-order and compute the number of edges required to reach each node. We also keep track of the number of reordered paths.

At each node, we check whether the number of edges required to reach the parent node is less than or greater than the number of edges required to reach the node from its parent. If it is less, we increment the number of reordered paths and swap the start and end nodes of the path.

Finally, we return the number of reordered paths as the minimum number of edges required to connect all the cities.

Step by Step Implementation For Reorder Routes To Make All Paths Lead To The City Zero

There are a total of n routes that are all going to the city zero. We need to find a way to reorder them so that all the routes start from the city zero.

We can use a greedy algorithm to solve this problem. We can keep track of the starting city for each route and sort the routes based on the starting city. Then, we can iterate through the routes and swap the starting city with the city zero if the route does not start with the city zero.

def minReorder(n, connections): 
    
    # Create a set to keep track of visited nodes 
    visited = set() 
    
    # Create a dictionary to keep track of the number of 
    # reordered connections 
    reordered = {} 
    
    # Create a queue for BFS and add the starting node 
    queue = collections.deque() 
    queue.append(0) 
    
    # Perform BFS 
    while queue: 
        node = queue.popleft() 
        
        # If the node has not been visited yet, 
        # mark it as visited 
        if node not in visited: 
            visited.add(node) 
            
            # Iterate over all the neighbors 
            for neighbor in connections[node]: 
                
                # If the neighbor has not been visited yet, 
                # add it to the queue 
                if neighbor[0] not in visited: 
                    queue.append(neighbor[0]) 
                    
                    # If the connection was not reordered, 
                    # increment the counter 
                    if neighbor[0] not in reordered: 
                        reordered[neighbor[0]] = 1
                        
                    # Otherwise, increment the counter by 2 
                    else: 
                        reordered[neighbor[0]] += 2
                        
    # Return the sum of all the reordered connections 
    return sum(reordered.values())

There are a total of n routes that are labelled from 0 to n - 1. You start at city 0 with label 0 and you can visit every other city in any order you want as long as you visit city 0 again to return to your starting point.

Let's say the distance between city i and city j is dis[i][j]. The total distance of your trip is the sum of the distances of all the routes you take, for example if your trip consists of routes [1,3,2] then your total distance will be dis[1][3] + dis[3][2].

Return the minimum total distance of your trip.

/**
 * @param {number[][]} dis
 * @return {number}
 */
var minReorder = function(dis) {
    
};

There are a total of n routes that are labelled from 0 to n - 1. You start at city 0 with distance 0 and can move to any one of the n cities. The cost of moving from city i to any other city j is given by the matrix W[i][j].

Find the minimum cost to reach city 0 from all the other cities.

This problem is similar to the classic Floyd Warshall algorithm for finding the shortest paths between all pairs of nodes in a graph. However, in this problem, we also need to keep track of the order in which the nodes are visited.

We can use a similar approach as the Floyd Warshall algorithm, but instead of using a 2D array for the distances, we will use a 3D array. The first two dimensions of the array will be the same as the Floyd Warshall algorithm, but the third dimension will be used to store the order in which the nodes are visited.

We can initialize the 3D array with the following values:

The cost of going from city i to city j with k stops will be infinity if k < 0.
The cost of going from city i to city j with 0 stops will be the cost of going from city i to city j without any stops, which is given by the matrix W.
The cost of going from city i to city j with 1 stop will be the cost of going from city i to city k with 0 stops, plus the cost of going from city k to city j with 0 stops.

We can then use a similar algorithm as the Floyd Warshall algorithm to find the minimum cost of going from city i to city j with k stops.

The time complexity of this algorithm will be O(n^3), where n is the number of cities.

//Solution for https://leetcode.com/problems/reorder-routes-to-make-all-paths-lead-to-the-city-zero/

public class Solution {
    public int minReorder(int n, int[][] connections) {
        //We will use a set to keep track of the cities we have visited
        //We will also use a queue to keep track of the cities we need to visit
        HashSet visited = new HashSet();
        Queue queue = new Queue();
        int result = 0;
        
        //We start from city 0
        queue.Enqueue(0);
        visited.Add(0);
        
        //We do a breadth first search
        while(queue.Count > 0){
            int city = queue.Dequeue();
            
            //For each connection, we check if the other city has been visited
            foreach(int[] connection in connections){
                //If the other city has not been visited, we add it to the queue and mark it as visited
                if(connection[0] == city && !visited.Contains(connection[1])){
                    queue.Enqueue(connection[1]);
                    visited.Add(connection[1]);
                }
                //If the other city has been visited, we check if the connection goes from the current city to the other city
                //If it does, we don't need to reverse the connection, so we don't increment the result
                else if(connection[1] == city && !visited.Contains(connection[0])){
                    queue.Enqueue(connection[0]);
                    visited.Add(connection[0]);
                    //We increment the result because we need to reverse the connection
                    result++;
                }
            }
        }
        
        return result;
    }
}