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Problem solution · Python

Network Recovery Pathways

Network Recovery Pathways: a Python solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Binary search
Source
Kamyu LeetCode Solutions
Length
65 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

Binary search

For Network Recovery Pathways, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 65 lines of Python from the credited upstream file network-recovery-pathways.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

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 codeNetwork Recovery Pathways · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O((n + e) * logr), r = max(e[2] for e in edges)# Space: O(n + e) # binary search, topological sort, dpclass Solution(object):    def findMaxPathScore(self, edges, online, k):        """        :type edges: List[List[int]]        :type online: List[bool]        :type k: int        :rtype: int        """        INF = float("inf")        def binary_search_right(left, right, check):            while left <= right:                mid = left+(right-left)//2                if not check(mid):                    right = mid-1                else:                    left = mid+1            return right         def topological_sort():            in_degree = [0]*len(adj)            for u in xrange(len(adj)):                for v, _ in adj[u]:                    in_degree[v] += 1            result = []            q = [u for u in xrange(len(adj)) if not in_degree[u]]            while q:                new_q = []                for u in q:                    result.append(u)                    for v, _ in adj[u]:                        in_degree[v] -= 1                        if in_degree[v]:                            continue                        new_q.append(v)                q = new_q            return result         def check(x):            dist = [INF]*len(adj)            dist[0] = 0            for u in order:                if dist[u] == INF:                    continue                for v, c in adj[u]:                    if not (c >= x and online[v]):                        continue                    dist[v] = min(dist[v], dist[u]+c)            return dist[-1] <= k         adj = [[] for _ in xrange(len(online))]        for u, v, c in edges:            adj[u].append((v, c))        order = topological_sort()        left, right = INF, 0        for u in xrange(len(adj)):            for _, c in adj[u]:                left = min(left, c)                right = max(right, c)        result = binary_search_right(left, right, check)        return result if result >= left else -1

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