{ "data": { "question": { "questionId": "3079", "questionFrontendId": "2846", "boundTopicId": null, "title": "Minimum Edge Weight Equilibrium Queries in a Tree", "titleSlug": "minimum-edge-weight-equilibrium-queries-in-a-tree", "content": "
There is an undirected tree with n
nodes labeled from 0
to n - 1
. You are given the integer n
and a 2D integer array edges
of length n - 1
, where edges[i] = [ui, vi, wi]
indicates that there is an edge between nodes ui
and vi
with weight wi
in the tree.
You are also given a 2D integer array queries
of length m
, where queries[i] = [ai, bi]
. For each query, find the minimum number of operations required to make the weight of every edge on the path from ai
to bi
equal. In one operation, you can choose any edge of the tree and change its weight to any value.
Note that:
\n\nai
to bi
is a sequence of distinct nodes starting with node ai
and ending with node bi
such that every two adjacent nodes in the sequence share an edge in the tree.Return an array answer
of length m
where answer[i]
is the answer to the ith
query.
\n
Example 1:
\n\n\nInput: 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]]\nOutput: [0,0,1,3]\nExplanation: In the first query, all the edges in the path from 0 to 3 have a weight of 1. Hence, the answer is 0.\nIn the second query, all the edges in the path from 3 to 6 have a weight of 2. Hence, the answer is 0.\nIn 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.\nIn 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.\nFor 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.\n\n\n
Example 2:
\n\n\nInput: 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]]\nOutput: [1,2,2,3]\nExplanation: 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.\nIn 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.\nIn 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.\nIn 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.\nFor 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.\n\n\n
\n
Constraints:
\n\n1 <= n <= 104
edges.length == n - 1
edges[i].length == 3
0 <= ui, vi < n
1 <= wi <= 26
edges
represents a valid tree.1 <= queries.length == m <= 2 * 104
queries[i].length == 2
0 <= ai, bi < n
freq[node][weight]
which saves the frequency of each edge weight
on the path from the root to each node
.",
"The frequency of edge weight w
on the path from a
to b
is equal to freq[a][w] + freq[b][w] - freq[lca(a,b)][w] * 2
, where lca(a,b)
is the lowest common ancestor of a
and b
in the tree.",
"lca(a,b)
can be calculated using binary lifting algorithm or Tarjan algorithm."
],
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