What is the time complexity of Minimum Edge Weight Equilibrium Queries in a Tree?

The optimal solution has a time complexity of O((n + q) * log(n) * 26).

What is the space complexity of Minimum Edge Weight Equilibrium Queries in a Tree?

The space complexity is O(n * log(n) * 26).

What data structures are used to solve Minimum Edge Weight Equilibrium Queries in a Tree?

This problem uses Array, Tree, Graph, Strongly Connected Component. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Edge Weight Equilibrium Queries in a Tree?

Insight 1: The problem involves finding paths in a tree and analyzing edge weights along those paths. This suggests using techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the tree. Insight 2: Since there are multiple queries, precomputing information about the tree can significantly improve efficiency. Specifically, the Lowest Common Ancestor (LCA) and path weight frequencies can be precalculated using techniques like binary lifting.

Which companies ask LeetCode 2846. Minimum Edge Weight Equilibrium Queries in a Tree in interviews?

This problem is frequently asked at Sprinklr.

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Minimum Edge Weight Equilibrium Queries in a Tree - LeetCode 2846 Solution

Minimum Edge Weight Equilibrium Queries in a Tree - Complete Solution Guide

Minimum Edge Weight Equilibrium Queries in a Tree is LeetCode problem 2846, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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] = [u i , v i , w i ] indicates that there is an edge between nodes u i and v i with weight w i in the tree. You are also given a 2D integer array queries of length m , where queries[i] = [a i , b i ] . For each query, find the minimum number of operations required to make the weight of every edge on the path from a i to b i equal. In one ope

Detailed Explanation

The problem asks us to find, for each query, the minimum number of edge weight changes needed on the path between two given nodes in a tree such that all edge weights on that path become equal. We are given the tree structure via an array of edges, where each edge has a weight. The queries are independent, meaning each query is performed on the original tree.

Solution Approach

The solution utilizes a combination of Breadth-First Search (BFS) for initial tree traversal and binary lifting for efficient LCA calculation and path weight frequency counting. First, a BFS is performed to determine the depth of each node and the immediate parent of each node. This information is then used to build a table for binary lifting, which allows us to quickly find the k-th ancestor of any node. During the BFS, the initial parent relationship and edge weights are stored. A binary lifting DP approach is used to build parent array and count array to quickly get the edge weight count from a node to its ancestor. For each query, the LCA of the two nodes is found using binary lifting. The path length between the two nodes is then calculated. Next, the count of edge weights are calculated using precomputed count array along the path from two nodes to the LCA. The maximum frequency is then found, and the answer is computed by taking path length - maximum frequency.

Step-by-Step Algorithm

Step 1: Build the adjacency list representation of the tree from the given edges.
Step 2: Perform a BFS traversal of the tree, starting from an arbitrary root (e.g., node 0), to calculate the depth of each node and the immediate parent of each node (parent[][0]). During the BFS traversal, count[v][0][w] for edge from parent of v to v is initialized.
Step 3: Use binary lifting to precompute ancestors of each node at powers of 2 (parent[u][i]) and the frequency of edge weights along the paths to those ancestors (count[u][i][w]). This uses dynamic programming where `parent[u][i] = parent[parent[u][i-1]][i-1]` and `count[u][i][w] = count[u][i-1][w] + count[parent[u][i-1]][i-1][w]`.
Step 4: For each query (u, v), find the Lowest Common Ancestor (LCA) of u and v using binary lifting.
Step 5: Calculate the path length between u and v as `depth[u] + depth[v] - 2 * depth[LCA]`. If path_len == 0, return 0 immediately.
Step 6: Count the frequency of each edge weight along the path from u to LCA and from v to LCA. This is done using the count array via binary lifting.
Step 7: Find the maximum frequency among all edge weights on the path.
Step 8: Calculate the minimum number of operations as `path_len - max_freq`, and store the result.
Step 9: Return the list of results for all queries.

Key Insights

Insight 1: The problem involves finding paths in a tree and analyzing edge weights along those paths. This suggests using techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the tree.
Insight 2: Since there are multiple queries, precomputing information about the tree can significantly improve efficiency. Specifically, the Lowest Common Ancestor (LCA) and path weight frequencies can be precalculated using techniques like binary lifting.
Insight 3: The edge weights are small (1 to 26), allowing us to efficiently count the frequency of each weight along a path. Instead of changing edge weights to a specific given weight, we count all the edge weights along a path, find the most frequent one, and the minimum number of changes is the difference between the path length and the maximum frequency.

Complexity Analysis

Time Complexity: O((n + q) * log(n) * 26)

Space Complexity: O(n * log(n) * 26)

Topics

This problem involves: Array, Tree, Graph, Strongly Connected Component.

Companies

Asked at: Sprinklr.

Minimum Edge Weight Equilibrium Queries in a Tree | LeetCode 2846

Hard

Array Tree Graph Strongly Connected Component

Time: O((n + q) * log(n) * 26)Space: O(n * log(n) * 26)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[] minOperationsQueries(int n, int[][] edges, int[][] queries) {
        int LOGN = 32 - Integer.numberOfLeadingZeros(n);
        int MAX_WEIGHT = 26;

        List<List<Pair>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            int w = edge[2];
            adj.get(u).add(new Pair(v, w));
            adj.get(v).add(new Pair(u, w));
        }

        int[] depth = new int[n];
        Arrays.fill(depth, -1);
        int[][] parent = new int[n][LOGN];
        for (int i = 0; i < n; i++) {
            Arrays.fill(parent[i], -1);
        }
        int[][][] count = new int[n][LOGN][MAX_WEIGHT + 1];

        // Pre-computation using BFS to fill depth, parent[][0], and count[][0]
        Queue<Integer> q = new LinkedList<>();
        Set<Integer> visited = new HashSet<>();
        q.offer(0);
        visited.add(0);
        depth[0] = 0;

        while (!q.isEmpty()) {
            int u = q.poll();
            for (Pair pair : adj.get(u)) {
                int v = pair.node;
                int w = pair.weight;
                if (!visited.contains(v)) {
                    visited.add(v);
                    depth[v] = depth[u] + 1;
                    parent[v][0] = u;
                    count[v][0][w] = 1;
                    q.offer(v);
                }
            }
        }

        // Binary Lifting DP to fill the rest of the tables
        for (int i = 1; i < LOGN; i++) {
            for (int u = 0; u < n; u++) {
                int p = parent[u][i - 1];
                if (p != -1) {
                    int gp = parent[p][i - 1];
                    if (gp != -1) {
                        parent[u][i] = gp;
                        for (int w = 1; w <= MAX_WEIGHT; w++) {
                            count[u][i][w] = count[u][i - 1][w] + count[p][i - 1][w];
                        }
                    }
                }
            }
        }

        //LCA function
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        

        int[] ans = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int u_orig = queries[i][0];
            int v_orig = queries[i][1];
            
            int lca = get_lca(u_orig, v_orig, depth, parent, LOGN);

            int path_len = depth[u_orig] + depth[v_orig] - 2 * depth[lca];

            if (path_len == 0) {
                ans[i] = 0;
                continue;
            }

            int[] total_counts = new int[MAX_WEIGHT + 1];

            // Count weights on the path from u_orig to lca
            int curr_u = u_orig;
            int dist_u = depth[u_orig] - depth[lca];
            for (int j = LOGN - 1; j >= 0; j--) {
                if (((dist_u >> j) & 1) != 0) {
                    for (int w = 1; w <= MAX_WEIGHT; w++) {
                        total_counts[w] += count[curr_u][j][w];
                    }
                    curr_u = parent[curr_u][j];
                }
            }

            // Count weights on the path from v_orig to lca
            int curr_v = v_orig;
            int dist_v = depth[v_orig] - depth[lca];
            for (int j = LOGN - 1; j >= 0; j--) {
                if (((dist_v >> j) & 1) != 0) {
                    for (int w = 1; w <= MAX_WEIGHT; w++) {
                        total_counts[w] += count[curr_v][j][w];
                    }
                    curr_v = parent[curr_v][j];
                }
            }

            int max_freq = 0;
            for(int count_val : total_counts){
                max_freq = Math.max(max_freq, count_val);
            }
            ans[i] = path_len - max_freq;
        }

        return ans;
    }
    
    private int get_lca(int u, int v, int[] depth, int[][] parent, int LOGN){
        if (depth[u] < depth[v]) {
            int temp = u;
            u = v;
            v = temp;
        }

        int diff = depth[u] - depth[v];
        for (int i = 0; i < LOGN; i++) {
            if (((diff >> i) & 1) != 0) {
                u = parent[u][i];
            }
        }

        if (u == v) {
            return u;
        }

        for (int i = LOGN - 1; i >= 0; i--) {
            if (parent[u][i] != -1 && parent[u][i] != parent[v][i]) {
                u = parent[u][i];
                v = parent[v][i];
            }
        }

        return parent[u][0];
    }

    static class Pair {
        int node;
        int weight;

        public Pair(int node, int weight) {
            this.node = node;
            this.weight = weight;
        }
    }
}

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Problem Info

Description

Problem 2846: Minimum Edge Weight Equilibrium Queries in a Tree - Detailed explanation and solution approaches.

Topics

Array Tree Graph Strongly Connected Component

Companies

Sprinklr

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2846

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Build the adjacency list representation of the tree from the given edges.
2
Step 2: Perform a BFS traversal of the tree, starting from an arbitrary root (e.g., node 0), to calculate the depth of each node and the immediate parent of each node (parent[][0]). During the BFS traversal, count[v][0][w] for edge from parent of v to v is initialized.
3
Step 3: Use binary lifting to precompute ancestors of each node at powers of 2 (parent[u][i]) and the frequency of edge weights along the paths to those ancestors (count[u][i][w]). This uses dynamic programming where `parent[u][i] = parent[parent[u][i-1]][i-1]` and `count[u][i][w] = count[u][i-1][w] + count[parent[u][i-1]][i-1][w]`.
4
Step 4: For each query (u, v), find the Lowest Common Ancestor (LCA) of u and v using binary lifting.
5
Step 5: Calculate the path length between u and v as `depth[u] + depth[v] - 2 * depth[LCA]`. If path_len == 0, return 0 immediately.
6
Step 6: Count the frequency of each edge weight along the path from u to LCA and from v to LCA. This is done using the count array via binary lifting.
7
Step 7: Find the maximum frequency among all edge weights on the path.
8
Step 8: Calculate the minimum number of operations as `path_len - max_freq`, and store the result.
9
Step 9: Return the list of results for all queries.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O((n + q) * log(n) * 26)

💾

Space Complexity

O(n * log(n) * 26)

📝

Why?

The time complexity is determined by the binary lifting precomputation and the LCA queries. The space complexity is dominated by the storage needed for the precomputed LCA data and the frequency counts.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

LCA with DFS and Path Tracking

Instead of using binary lifting, the LCA can be calculated using DFS to find the paths from the root to each node. Then compare the two paths until they diverge to find the LCA. After finding LCA, path from node u to v can be reconstructed and the edge weights can be counted.

Time: O(n + q*n)Space: O(n)

Trade-offs: This approach is simpler to implement but has a higher time complexity for multiple queries because path reconstruction takes O(n) for each query as compared to O(log(n)*26) in the main solution. The binary lifting approach is preferred when there are many queries.

✨

Key Insights

1Insight 1: The problem involves finding paths in a tree and analyzing edge weights along those paths. This suggests using techniques like Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the tree.
2Insight 2: Since there are multiple queries, precomputing information about the tree can significantly improve efficiency. Specifically, the Lowest Common Ancestor (LCA) and path weight frequencies can be precalculated using techniques like binary lifting.
3Insight 3: The edge weights are small (1 to 26), allowing us to efficiently count the frequency of each weight along a path. Instead of changing edge weights to a specific given weight, we count all the edge weights along a path, find the most frequent one, and the minimum number of changes is the difference between the path length and the maximum frequency.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Tree

Hierarchical structure with nodes. Like a family tree!

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: When the two query nodes are the same, the path length is 0, and no operations are needed. The algorithm handles this case by explicitly checking if path_len == 0 and returning 0 immediately.
Edge case 2: When one node is the ancestor of the other, the LCA is simply the ancestor. The binary lifting algorithm correctly identifies this situation.
Edge case 3: When the graph is a single node with no edges, the code handles this case correctly as the path_len will always be zero. It doesn't crash due to the empty edge input.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not precomputing LCA information and calculating it for each query results in Time Limit Exceeded (TLE).

✗Common mistake 2: Not considering the case where the path length is zero, which causes errors in the edge weight counting logic.

✗Common mistake 3: Incorrectly implementing binary lifting, particularly the base cases and the dynamic programming update steps.

✗Common mistake 4: Integer overflow errors when calculating the path length or weight frequencies, especially if the tree is very deep or there are many high-weight edges.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network routing protocols can use similar algorithms to determine the optimal path between two nodes in a network, where the edge weights represent network latency or cost.

🏦Real-world scenario 2: Version control systems like Git use tree structures to represent the history of a repository. Finding the common ancestor of two branches is a crucial operation for merging changes. The logic for determining common ancestors is derived from LCA concepts.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 25, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I choose between recursive and iterative tree traversal?

Use recursive traversal for cleaner code when stack depth isn't a concern (tree height < ~1000). Use iterative with explicit stack for very deep trees to avoid stack overflow. Iterative is also preferred when you need fine-grained control over the traversal process.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Edge Weight Equilibrium Queries in a Tree"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?