XOR of all the nodes in the sub-tree of the given node
Given an n-ary tree and Q queries where each query consists of an integer u which denotes a node. The task is to print the xor of all the values of nodes in the sub-tree of the given node.
Examples:
Input:
q[] = {0, 1, 4, 5}
Output:
0
3
5
6
Query 1: (1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6 ^ 7) = 0
Query 2: (2 ^ 4 ^ 5) = 3
Query 3: (5) = 5
Query 4: (6) = 6
Naive approach: For each node we have to traverse the tree for every query find the xor of all the nodes of the sub-tree. So the complexity of the code will be O(N * Q).
Efficient approach: If we pre-compute the xor of all the nodes of the sub-tree by traversing the total tree once and first computing the xor of all the nodes of the sub-tree of the child node first and then using the value of the child node for computing the xor of all the nodes of the sub-tree of the parent node. In this way we can compute the xor in O(N) time and store them for queries. When a query is asked we will print the pre-computed value in O(1) time.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Adjacency list of the graph vector<vector<int> > graph; // Value of the node vector<int> values, xor_values; // Function to pre-compute the xor values int pre_compute_xor(int i, int prev) { // xor of the sub-tree int x = values[i]; for (int j = 0; j < graph[i].size(); j++) if (graph[i][j] != prev) { // xor x with xor of the sub-tree // of it child nodes x ^= pre_compute_xor(graph[i][j], i); } xor_values[i] = x; // Return the xor return x; } // Function to return the xor of // the nodes of the sub-tree // rooted at node u int query(int u) { return xor_values[u]; } // Driver code int main() { int n = 7; graph.resize(n); xor_values.resize(n); // Create the graph graph[0].push_back(1); graph[0].push_back(2); graph[1].push_back(3); graph[1].push_back(4); graph[2].push_back(5); graph[2].push_back(6); // Set the values of the nodes values.push_back(1); values.push_back(2); values.push_back(3); values.push_back(4); values.push_back(5); values.push_back(6); values.push_back(7); // Pre-computation pre_compute_xor(0, -1); // Perform queries int queries[] = { 0, 1, 4, 5 }; int q = sizeof(queries) / sizeof(queries[0]); for (int i = 0; i < q; i++) cout << query(queries[i]) << endl; return 0; } |
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Java
// Java implementation of the approach import java.util.*; class GFG { static int n = 7; // Adjacency list of the graph static Vector<Integer> []graph = new Vector[n]; // Value of the node static Vector<Integer> values = new Vector<Integer>(), xor_values = new Vector<Integer>(n); // Function to pre-compute the xor values static int pre_compute_xor(int i, int prev) { // xor of the sub-tree int x = values.get(i); for (int j = 0; j < graph[i].size(); j++) if (graph[i].get(j)!= prev) { // xor x with xor of the sub-tree // of it child nodes x ^= pre_compute_xor(graph[i].get(j), i); } xor_values.remove(i); xor_values.add(i, x); // Return the xor return x; } // Function to return the xor of // the nodes of the sub-tree // rooted at node u static int query(int u) { return xor_values.get(u); } // Driver code public static void main(String[] args) { for(int i = 0; i < n; i++) { graph[i] = new Vector<Integer>(); xor_values.add(0); } // Create the graph graph[0].add(1); graph[0].add(2); graph[1].add(3); graph[1].add(4); graph[2].add(5); graph[2].add(6); // Set the values of the nodes values.add(1); values.add(2); values.add(3); values.add(4); values.add(5); values.add(6); values.add(7); // Pre-computation pre_compute_xor(0, -1); // Perform queries int queries[] = { 0, 1, 4, 5 }; int q = queries.length; for (int i = 0; i < q; i++) System.out.print(query(queries[i]) +"\n"); } } // This code is contributed by Rajput-Ji |
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Python3
# Python3 implementation of the approach # Adjacency list of the graph graph = [] # Value of the node values = [] xor_values = [] # Function to pre-compute the xor values def pre_compute_xor(i, prev): # xor of the sub-tree x = values[i] for j in range(len(graph[i])): if graph[i][j] != prev: # xor x with xor of the sub-tree # of it child nodes x ^= pre_compute_xor(graph[i][j], i) xor_values[i] = x # Return the xor return x # Function to return the xor of # the nodes of the sub-tree # rooted at node u def query(u): return xor_values[u] # Driver code n = 7for i in range(n): graph.append([]) xor_values.append(0) # Create the graph graph[0].append(1) graph[0].append(2) graph[1].append(3) graph[1].append(4) graph[2].append(5) graph[2].append(6) # Set the values of the nodes values.append(1) values.append(2) values.append(3) values.append(4) values.append(5) values.append(6) values.append(7) # Pre-computation pre_compute_xor(0, -1) # Perform queries queries = [ 0, 1, 4, 5 ] q = len(queries) for i in range(q): print(query(queries[i])) # This code is contributed by divyamohan123 |
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C#
// C# implementation of the approach using System; using System.Collections.Generic; class GFG { static int n = 7; // Adjacency list of the graph static List<int> []graph = new List<int>[n]; // Value of the node static List<int> values = new List<int>(), xor_values = new List<int>(n); // Function to pre-compute the xor values static int pre_compute_xor(int i, int prev) { // xor of the sub-tree int x = values[i]; for (int j = 0; j < graph[i].Count; j++) if (graph[i][j] != prev) { // xor x with xor of the sub-tree // of it child nodes x ^= pre_compute_xor(graph[i][j], i); } xor_values.RemoveAt(i); xor_values.Insert(i, x); // Return the xor return x; } // Function to return the xor of // the nodes of the sub-tree // rooted at node u static int query(int u) { return xor_values[u]; } // Driver code public static void Main(String[] args) { for(int i = 0; i < n; i++) { graph[i] = new List<int>(); xor_values.Add(0); } // Create the graph graph[0].Add(1); graph[0].Add(2); graph[1].Add(3); graph[1].Add(4); graph[2].Add(5); graph[2].Add(6); // Set the values of the nodes values.Add(1); values.Add(2); values.Add(3); values.Add(4); values.Add(5); values.Add(6); values.Add(7); // Pre-computation pre_compute_xor(0, -1); // Perform queries int []queries = { 0, 1, 4, 5 }; int q = queries.Length; for (int i = 0; i < q; i++) Console.Write(query(queries[i]) +"\n"); } } // This code is contributed by PrinciRaj1992 |
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Output:
0 3 5 6
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