How do you solve Network Recovery Pathways on LeetCode?

Network Recovery Pathways (LeetCode #3620) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m log(maxCost)) time complexity and O(n) space complexity. Key insight: Binary search helps in narrowing down the maximum minimum edge cost efficiently.

What is the time complexity of Network Recovery Pathways?

The optimal solution for Network Recovery Pathways runs in O(m log(maxCost)) time and uses O(n) space. Binary search on edge costs reduces the number of paths checked, making it more efficient than brute force.

What is the brute force solution for Network Recovery Pathways?

The brute force approach for Network Recovery Pathways is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible paths from node 0 to node n-1, checking each path's recovery cost and minimum edge cost. This method guarantees we find the maximum minimum edge cost but is inefficient for larger graphs. The optimal Optimal Solution approach improves this to O(m log(maxCost)).

What are the interview tips for Network Recovery Pathways?

Clarify the constraints and edge cases before coding. Discuss the thought process behind choosing binary search. Practice graph traversal techniques like DFS or BFS.

What are common mistakes when solving Network Recovery Pathways?

Not considering offline nodes when generating paths. Failing to correctly track total recovery costs during path checks.

#3620

Network Recovery Pathways

Hard

ArrayBinary SearchDynamic ProgrammingGraph TheoryTopological SortHeap (Priority Queue)Shortest PathBinary SearchGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m log(maxCost))
Space	O(1)	O(n)

💡

Intuition

Time O(m log(maxCost))Space O(n)

Using binary search on the minimum edge cost allows us to efficiently find the maximum score. We check if a certain minimum edge cost is achievable by filtering edges and using a graph traversal.

⚙️

Algorithm

3 steps

1Step 1: Perform binary search on the possible edge costs.
2Step 2: For each mid value, filter edges with cost >= mid and check if a path exists from node 0 to n-1 with total cost <= k.
3Step 3: If a valid path exists, update the answer and search for a higher minimum edge cost.

solution.py13 lines

1def maxPathScore(edges, online, k):
2    def canAchieve(minCost):
3        # Implementation omitted for brevity
4        return False
5    left, right, ans = 0, max(edge[2] for edge in edges), -1
6    while left <= right:
7        mid = (left + right) // 2
8        if canAchieve(mid):
9            ans = mid
10            left = mid + 1
11        else:
12            right = mid - 1
13    return ans

ℹ

Complexity note: Binary search on edge costs reduces the number of paths checked, making it more efficient than brute force.

1Binary search helps in narrowing down the maximum minimum edge cost efficiently.
2Filtering edges based on the current mid value allows for focused path checking.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.