Leetcode #2846: Minimum Edge Weight Equilibrium Queries in a Tree
In this guide, we solve Leetcode #2846 Minimum Edge Weight Equilibrium Queries in a Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Tree, Graph, Array, Strongly Connected Component
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
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]
Explanation: 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.
Python Solution
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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.