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Problem solution · Python

Minimum Cost to Repair Edges to Traverse a Graph

Minimum Cost to Repair Edges to Traverse a Graph: a Python solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Breadth-first search
Source
Kamyu LeetCode Solutions
Length
51 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

Breadth-first search

For Minimum Cost to Repair Edges to Traverse a Graph, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 51 lines of Python from the credited upstream file minimum-cost-to-repair-edges-to-traverse-a-graph.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

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 codeMinimum Cost to Repair Edges to Traverse a Graph · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O((n + m) * logr)# Space: O(n + m) # binary search, bfsclass Solution(object):    def minCost(self, n, edges, k):        """        :type n: int        :type edges: List[List[int]]        :type k: int        :rtype: int        """        def binary_search(left, right, check):            while left <= right:                mid = left+(right-left)//2                if check(mid):                    right = mid-1                else:                    left = mid+1            return left                def bfs(x):            lookup = [False]*len(adj)            lookup[0] = True            q = [0]            d = 0            while q:                if d == k+1:                    break                new_q = []                for u in q:                    if u == n-1:                        return True                    for v, w in adj[u]:                        if w > x or lookup[v]:                            continue                        lookup[v] = True                        new_q.append(v)                q = new_q                d += 1            return False                    adj = [[] for _ in xrange(n)]        for u, v, w in edges:            adj[u].append((v, w))            adj[v].append((u, w))        left = min(w for _, _, w in edges)        right = max(w for _, _, w in edges)        result = binary_search(left, right, bfs)        return result if result != right+1 else -1

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