Reorder Routes to Make All Paths Lead to the

Tested Python solution for LeetCode 1466 with 29 pytest cases. Generate a practice environment with lcpy.

LeetCode 1466, Medium. Topics: Depth-First Search, Breadth-First Search, Graph Theory. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 29 parametrized pytest cases, and a playground notebook:

lcpy gen -n 1466   # by problem number
lcpy gen -s reorder_routes_to_make_all_paths_lead_to_the_city_zero   # by problem name

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Problem

There are n cities numbered from 0 to n - 1 and n - 1 roads such that there is only one way to travel between two different cities (this network form a tree). Last year, The ministry of transport decided to orient the roads in one direction because they are too narrow.

Roads are represented by connections where connections[i] = [ai, bi] represents a road from city ai to city bi.

This year, there will be a big event in the capital (city 0), and many people want to travel to this city.

Your task consists of reorienting some roads such that each city can visit the city 0. Return the minimum number of edges changed.

It’s guaranteed that each city can reach city 0 after reorder.

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Examples

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).

Input: n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
Output: 2

Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

Input: n = 3, connections = [[1,0],[2,0]]
Output: 0

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Constraints

• 2 <= n <= 5 * 10^4
• connections.length == n - 1
• connections[i].length == 2
• 0 <= ai, bi <= n - 1
• ai != bi

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

from collections import defaultdict, deque


class Solution:
    # Time: O(n)
    # Space: O(n)
    def min_reorder(self, n: int, connections: list[list[int]]) -> int:
        adjacency: dict[int, list[tuple[int, int]]] = defaultdict(list)
        for src, dst in connections:
            adjacency[src].append((dst, 1))
            adjacency[dst].append((src, 0))

        flips = 0
        visited = [False] * n
        visited[0] = True
        queue: deque[int] = deque([0])
        while queue:
            city = queue.popleft()
            for neighbor, directed_out in adjacency[city]:
                if visited[neighbor]:
                    continue
                flips += directed_out
                visited[neighbor] = True
                queue.append(neighbor)
        return flips

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Complexity

Time	Space
O(n)	O(n)

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Tags

NeetCode All.