#3600

Hard

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Maximize Spanning Tree Stability with Upgrades

Maximizing the stability of a spanning tree with upgrades requires careful optimization of edge strengths using binary search and graph algorithms.

binary search greedy union find graph minimum spanning tree

Problem Statement

You are given n nodes, numbered from 0 to n - 1, and a list of edges where each edge is represented as [ui, vi, si, musti]. The edge connects nodes ui and vi with a strength score si. The value musti indicates whether the edge must be included in the spanning tree (musti == 1) or if it can be optionally included (musti == 0). You are also given an integer k, the maximum number of upgrades allowed. Each upgrade doubles the strength of an edge with musti == 0, and each edge can be upgraded at most once.

The stability of a spanning tree is defined as the minimum edge strength among all edges included in the tree. Your task is to find the maximum possible stability of a spanning tree that can be formed with at most k upgrades, or return -1 if no spanning tree can be formed.

Examples

Example 1

Input: n = 3, edges = [[0,1,2,1],[1,2,3,0]], k = 1

Output: 2

Example 2

Input: n = 3, edges = [[0,1,4,0],[1,2,3,0],[0,2,1,0]], k = 2

Output: 6

Example 3

Input: n = 3, edges = [[0,1,1,1],[1,2,1,1],[2,0,1,1]], k = 0

Output: -1

Constraints

2 <= n <= 105
1 <= edges.length <= 105
edges[i] = [ui, vi, si, musti]
0 <= ui, vi < n
ui != vi
1 <= si <= 105
musti is either 0 or 1.
0 <= k <= n
There are no duplicate edges.

Solution Approach

Binary Search Over Stability

The problem can be approached by performing a binary search over the possible values of the minimum edge strength in the spanning tree. The lower bound of the search is the smallest edge strength, while the upper bound is the largest. For each value of stability, check if it's possible to form a spanning tree with that minimum edge strength using Union Find and Greedy techniques.

Greedy Strategy with Union Find

At each step of the binary search, use a greedy approach to select edges with strength at least equal to the current mid-point. Use Union Find to efficiently check if the selected edges can form a valid spanning tree. The edges with musti == 0 can be upgraded during this process to increase the chances of forming a valid spanning tree.

Edge Sorting

Sort the edges array in descending order of their strength scores. This ensures that we consider stronger edges first, which is crucial for maximizing the minimum strength in the spanning tree. This sorting helps in efficiently applying the Union Find and greedy approach.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution depends on the binary search and the Union Find operations. Binary search will run for log(max edge strength) iterations, and for each iteration, Union Find operations with path compression and union by rank will perform nearly constant time operations. Thus, the time complexity is approximately O(E log S), where E is the number of edges and S is the maximum edge strength. Space complexity is O(n + E) due to the Union Find data structure and the storage of edges.

What Interviewers Usually Probe

Candidate demonstrates a clear understanding of binary search over a range of values.
Candidate can efficiently apply Union Find in combination with greedy strategies.
Candidate understands edge sorting as a preprocessing step to improve efficiency.

Common Pitfalls or Variants

Common pitfalls

Overlooking the requirement to sort edges in descending order of strength before applying Union Find.
Failing to handle the case when no valid spanning tree can be formed and returning an incorrect result.
Not properly considering edge upgrades during the greedy selection of edges.

Follow-up variants

Allowing multiple upgrades per edge instead of just one.
Limiting the number of edges that can be used in the spanning tree.
Varying the structure of the graph, such as introducing cycles or disconnected components.

FAQ

What is the best way to approach maximizing spanning tree stability with upgrades?

The key approach is binary search over the minimum strength of the spanning tree, combined with greedy edge selection and Union Find to efficiently build the tree.

How does the Union Find data structure help in this problem?

Union Find helps efficiently check whether adding an edge will create a cycle and helps in determining if the selected edges form a valid spanning tree.

What does the musti value in the edges represent?

musti indicates whether an edge must be included in the spanning tree (musti == 1) or if it can optionally be included (musti == 0).

How do upgrades work in this problem?

Each upgrade doubles the strength of an edge with musti == 0. Upgrades can only be applied to edges that are optional, and each edge can be upgraded at most once.

Why is edge sorting crucial for this problem?

Sorting the edges in descending order of strength ensures that we consider the strongest edges first, maximizing the minimum strength in the spanning tree.

terminal

Solution

Solution 1: Binary Search + Union-Find

According to the problem description, the stability of a spanning tree is determined by the minimum strength edge in it. If a stability $x$ is feasible, then for any $y < x$, stability $y$ is also feasible. Therefore, we can use binary search to find the maximum stability.

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class UnionFind:
    def __init__(self, n):
        self.p = list(range(n))
        self.size = [1] * n
        self.cnt = n

    def find(self, x):
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]

    def union(self, a, b):
        pa, pb = self.find(a), self.find(b)
        if pa == pb:
            return False
        if self.size[pa] > self.size[pb]:
            self.p[pb] = pa
            self.size[pa] += self.size[pb]
        else:
            self.p[pa] = pb
            self.size[pb] += self.size[pa]
        self.cnt -= 1
        return True


class Solution:
    def maxStability(self, n: int, edges: List[List[int]], k: int) -> int:
        def check(lim: int) -> bool:
            uf = UnionFind(n)
            for u, v, s, _ in edges:
                if s >= lim:
                    uf.union(u, v)
            rem = k
            for u, v, s, _ in edges:
                if s * 2 >= lim and rem > 0:
                    if uf.union(u, v):
                        rem -= 1
            return uf.cnt == 1

        uf = UnionFind(n)
        mn = 10**6
        for u, v, s, must in edges:
            if must:
                mn = min(mn, s)
                if not uf.union(u, v):
                    return -1
        for u, v, _, _ in edges:
            uf.union(u, v)
        if uf.cnt > 1:
            return -1
        l, r = 1, mn
        while l < r:
            mid = (l + r + 1) >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return l

Continue Practicing

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