#2846

Hard

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Minimum Edge Weight Equilibrium Queries in a Tree

Find the minimum number of operations to equalize edge weights in a tree between given pairs of nodes.

array tree graph strongly connected component

Problem Statement

You are given an undirected tree with n nodes labeled from 0 to n - 1, and a 2D integer array edges representing the tree. Each edge is described by a triple [ui, vi, wi], where ui and vi are the nodes connected by the edge, and wi is its weight. Your task is to process multiple queries where each query is a pair of nodes [ai, bi]. For each query, find the minimum number of operations to make the weight of every edge along the path between nodes ai and bi equal.

In each operation, you can choose any edge of the tree and change its weight to any value. Your goal is to minimize the number of such operations. The tree is guaranteed to be valid, and you need to return the result for each query. The problem emphasizes binary-tree traversal and state tracking to efficiently determine the solution.

Examples

Example 1

Input: n = 7, edges = [[0,1,1],[1,2,1],[2,3,1],[3,4,2],[4,5,2],[5,6,2]], queries = [[0,3],[3,6],[2,6],[0,6]]

Output: [0,0,1,3]

In the first query, all the edges in the path from 0 to 3 have a weight of 1. Hence, the answer is 0. In the second query, all the edges in the path from 3 to 6 have a weight of 2. Hence, the answer is 0. In the third query, we change the weight of edge [2,3] to 2. After this operation, all the edges in the path from 2 to 6 have a weight of 2. Hence, the answer is 1. In the fourth query, we change the weights of edges [0,1], [1,2] and [2,3] to 2. After these operations, all the edges in the path from 0 to 6 have a weight of 2. Hence, the answer is 3. For each queries[i], it can be shown that answer[i] is the minimum number of operations needed to equalize all the edge weights in the path from ai to bi.

Example 2

Input: n = 8, edges = [[1,2,6],[1,3,4],[2,4,6],[2,5,3],[3,6,6],[3,0,8],[7,0,2]], queries = [[4,6],[0,4],[6,5],[7,4]]

Output: [1,2,2,3]

In the first query, we change the weight of edge [1,3] to 6. After this operation, all the edges in the path from 4 to 6 have a weight of 6. Hence, the answer is 1. In the second query, we change the weight of edges [0,3] and [3,1] to 6. After these operations, all the edges in the path from 0 to 4 have a weight of 6. Hence, the answer is 2. In the third query, we change the weight of edges [1,3] and [5,2] to 6. After these operations, all the edges in the path from 6 to 5 have a weight of 6. Hence, the answer is 2. In the fourth query, we change the weights of edges [0,7], [0,3] and [1,3] to 6. After these operations, all the edges in the path from 7 to 4 have a weight of 6. Hence, the answer is 3. For each queries[i], it can be shown that answer[i] is the minimum number of operations needed to equalize all the edge weights in the path from ai to bi.

Constraints

1 <= n <= 104
edges.length == n - 1
edges[i].length == 3
0 <= ui, vi < n
1 <= wi <= 26
The input is generated such that edges represents a valid tree.
1 <= queries.length == m <= 2 * 104
queries[i].length == 2
0 <= ai, bi < n

Solution Approach

Binary Tree Traversal

Start by rooting the tree at any node, which transforms the problem into a series of path queries. Use a DFS or BFS traversal to precompute information about the path from the root to every other node. This will help track the edge weights along the path between any two nodes.

Path Compression and Optimization

For each query, compress the path from the source to the destination. Use a path compression technique to minimize recalculations. Track the weight frequencies along the path and identify how many distinct weights need to be changed to achieve uniformity across the path.

Efficient Query Handling

Store precomputed data (like edge weights and their frequencies) during the traversal process. For each query, use this data to quickly calculate the minimum operations required to make the path uniform in terms of edge weight. The approach should be optimized for handling large numbers of queries efficiently.

Complexity Analysis

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

The time complexity mainly depends on the number of nodes and edges in the tree, and the number of queries. Efficient tree traversal ensures that the preprocessing time is O(n), and handling each query can be done in logarithmic time with path compression techniques, resulting in an overall complexity of O(n + m log n), where n is the number of nodes and m is the number of queries.

What Interviewers Usually Probe

The candidate should demonstrate a clear understanding of tree traversal techniques and path compression.
Look for the ability to optimize query handling and precompute useful data during tree traversal.
The candidate should be able to discuss the trade-offs between preprocessing and query handling in large datasets.

Common Pitfalls or Variants

Common pitfalls

Failing to optimize for large numbers of queries by neglecting path compression or efficient data handling.
Overcomplicating the problem by not leveraging tree properties or by using brute force for path queries.
Incorrectly calculating the minimum number of operations needed due to misunderstandings about the path and weight frequency tracking.

Follow-up variants

Consider cases with more complex tree structures, like binary search trees or trees with high depth.
How would the approach change if the edges had weights beyond the simple range of 1-26?
What optimizations can be made if the tree has additional constraints such as a high number of queries relative to nodes?

FAQ

How do I solve the "Minimum Edge Weight Equilibrium Queries in a Tree" problem efficiently?

Use binary-tree traversal, precompute information about paths from the root, and apply path compression to handle each query in an optimized manner.

What is the primary strategy for solving this problem?

The primary strategy involves tree traversal, path compression, and efficiently using precomputed data to minimize the number of operations for each query.

Can I solve the problem without traversing the tree?

No, you need to traverse the tree to precompute useful information, as each query depends on the path between two nodes.

What kind of optimizations can I implement for large queries in this problem?

Focus on minimizing the operations by compressing paths and using efficient data structures to track edge weights and frequencies.

What is the time complexity of the solution for this problem?

The time complexity is O(n + m log n), where n is the number of nodes and m is the number of queries, with efficient preprocessing and query handling.

terminal

Solution

Solution 1: Binary Lifting for LCA

The problem asks for the minimum number of operations to make all edge weights the same on the path between any two points. This is essentially the length of the path between these two points, minus the number of times the most frequently occurring edge appears on the path.

class Solution:
    def minOperationsQueries(
        self, n: int, edges: List[List[int]], queries: List[List[int]]
    ) -> List[int]:
        m = n.bit_length()
        g = [[] for _ in range(n)]
        f = [[0] * m for _ in range(n)]
        p = [0] * n
        cnt = [None] * n
        depth = [0] * n
        for u, v, w in edges:
            g[u].append((v, w - 1))
            g[v].append((u, w - 1))
        cnt[0] = [0] * 26
        q = deque([0])
        while q:
            i = q.popleft()
            f[i][0] = p[i]
            for j in range(1, m):
                f[i][j] = f[f[i][j - 1]][j - 1]
            for j, w in g[i]:
                if j != p[i]:
                    p[j] = i
                    cnt[j] = cnt[i][:]
                    cnt[j][w] += 1
                    depth[j] = depth[i] + 1
                    q.append(j)
        ans = []
        for u, v in queries:
            x, y = u, v
            if depth[x] < depth[y]:
                x, y = y, x
            for j in reversed(range(m)):
                if depth[x] - depth[y] >= (1 << j):
                    x = f[x][j]
            for j in reversed(range(m)):
                if f[x][j] != f[y][j]:
                    x, y = f[x][j], f[y][j]
            if x != y:
                x = p[x]
            mx = max(cnt[u][j] + cnt[v][j] - 2 * cnt[x][j] for j in range(26))
            ans.append(depth[u] + depth[v] - 2 * depth[x] - mx)
        return ans

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