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

Master LeetCode problem 2846 (Minimum Edge Weight Equilibrium Queries in a Tree), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2846. Minimum Edge Weight Equilibrium Queries in a Tree

Hard

Array Tree Graph Strongly Connected Component

Problem Explanation

Explanation

To solve this problem, we can perform a Depth First Search (DFS) traversal of the tree to preprocess some information that will help us answer each query efficiently. For each node, we will calculate the xor of all edge weights along the path from the root to that node. This information will help us determine the number of operations required to make the edge weights equal along any path in the tree.

Perform a DFS traversal of the tree to calculate the xor values from the root to each node.
For each query, calculate the xor of edge weights along the path from node ai to node bi.
The minimum number of operations required to make the edge weights equal along this path is the number of set bits in the xor value.

Time Complexity: O(n) for preprocessing the xor values, O(m) for answering m queries, so overall O(n + m) Space Complexity: O(n) for storing xor values and O(m) for answer array

Solution Code

class Solution {
    public int[] minEdgeWeights(int n, int[][] edges, int[][] queries) {
        List<List<int[]>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }
        
        for (int[] edge : edges) {
            int u = edge[0], v = edge[1], w = edge[2];
            graph.get(u).add(new int[]{v, w});
            graph.get(v).add(new int[]{u, w});
        }
        
        int[] xorValues = new int[n];
        int[] depths = new int[n];
        
        dfs(0, -1, 0, 0, graph, xorValues, depths);
        
        int[] result = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int ai = queries[i][0], bi = queries[i][1];
            int xor = xorValues[ai] ^ xorValues[bi];
            int count = Integer.bitCount(xor);
            result[i] = count;
        }
        
        return result;
    }
    
    private void dfs(int node, int parent, int xor, int depth, List<List<int[]>> graph, int[] xorValues, int[] depths) {
        xorValues[node] = xor;
        depths[node] = depth;
        
        for (int[] neighbor : graph.get(node)) {
            int next = neighbor[0], weight = neighbor[1];
            if (next != parent) {
                dfs(next, node, xor ^ weight, depth + 1, graph, xorValues, depths);
            }
        }
    }
}

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