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Problem solution · Python

Reorder Routes to Make All Paths Lead to the City Zero

Reorder Routes to Make All Paths Lead to the City Zero: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
57 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Reorder Routes to Make All Paths Lead to the City Zero, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 57 lines of Python from the credited upstream file reorder-routes-to-make-all-paths-lead-to-the-city-zero.py.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeReorder Routes to Make All Paths Lead to the City Zero · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# Space: O(n) import collections  class Solution(object):    def minReorder(self, n, connections):        """        :type n: int        :type connections: List[List[int]]        :rtype: int        """        lookup, graph = set(), collections.defaultdict(list)        for u, v in connections:            lookup.add(u*n+v)            graph[v].append(u)            graph[u].append(v)         result = 0        stk = [(-1, 0)]        while stk:            parent, u = stk.pop()            result += (parent*n+u in lookup)            for v in reversed(graph[u]):                if v == parent:                    continue                stk.append((u, v))        return result  # Time:  O(n)# Space: O(n)import collections  class Solution2(object):    def minReorder(self, n, connections):        """        :type n: int        :type connections: List[List[int]]        :rtype: int        """        def dfs(n, lookup, graph, parent, u):            result = (parent*n+u in lookup)            for v in graph[u]:                if v == parent:                    continue                result += dfs(n, lookup, graph, u, v)              return result         lookup, graph = set(), collections.defaultdict(list)        for u, v in connections:            lookup.add(u*n+v)            graph[v].append(u)            graph[u].append(v)         return dfs(n, lookup, graph, -1, 0)

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