Given a Generic Tree consisting of N nodes valued from 0 to (N – 1) where P[i]th in the array P[] denotes ith nodes parent(1-based indexing). Each ith node has a weight attached to it, given in the array W[]. The task is to find a pair of nodes (u, v), such that u is an ancestor of v, and Wu – Wv is maximized.
Note: In the array P[], -1 denotes the root node. If there’s a single node, then print -1.
Examples:
Input: N = 4, W[] = {5, 10, 6, 12}, P[] = {2, -1, 4, 2}
Output: 6
Explanation: The tree with weight will be:
Here, 4th node having weight 12 and 3rd node having weight 6, the difference will be (12 – 6) = 6.
Input: N = 1, W = { 30 }, P = { -1 }
Output: -1
Approach: The given problem can be solved by using the Breadth-First Search on the N-ary Tree to mark the ancestor number for the given tree P[], and then using DFS Traversal and find the maximum difference maxDiff, by considering every node as an ancestor with its corresponding nodes having less number ancestor value. Follow the steps below to solve the problem:
- Define a function dfs(int src, int val, vector<int> & W) and perform the following tasks:
- Set the value of visited[src] as true.
- Iterate over the range [0, size) where size is the size of the row tree[cur] using the variable neighbor and performs the following tasks:
- If visited[neighbor] is false and ancestor[neighbor] is greater than ancestor[src] then set the value of maxDiff as the maximum of maxDiff or val – W[neighbor-1].
- Call the function dfs(neighbor, val, W).
- Define a function bfs(int src, int N) and perform the following tasks:
- Assign the vector visited[N + 1] with values false.
- Initialize a queue q[].
- Set the value of ancestorNum[src] as 0 and visited[src] as true.
- Enqueue the value src into the queue q[].
- Traverse in a while loop till queue q[] is not empty and perform the following tasks:
- Initialize the variable cur as the front element of the queue q[] and dequeue from the queue q[].
- Iterate over the range [0, size) where size is the size of the row tree[cur] using the variable neighbor and if visited[neighbor] is false then set it as true and enqueue it into the queue q[] and set the value of ancestor[neighbor] as (ancestor[cur] + 1).
- Initialize the vectors tree[][], visited[] and ancestorNum[].
- Initialize the variable maxDiff as INT_MIN to store the answer.
- Resize the vector tree[][] to size (N + 1).
- Assign vectors visited[N + 1] with value false and ancestorNum[N+1] with value 0.
- Initialize the variable src.
- Iterate over the range [0, N) using the variable i and if P[I] is -1, then set the value of src as i. Otherwise, push the value P[I] in the row i+1 and value i + 1 in the row P[I] into the vector tree[][].
- Call the function bfs(src, N+1) to perform breadth-first search.
- Assign vector visited[N+1] with value false.
- Call the function dfs(src, W[src], W) to perform depth-first search.
- Iterate over the range [0, N) using the variable i and perform the following steps:
- If i equals src, then continue. Otherwise, assign vector visited[N+1] with value false.
- Call the function dfs(i+1, W[i], W).
- After performing the above steps, print the value of maxDiff as the answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;vector<vector<int> > tree;vector<bool> visited;vector<int> ancestorNum;// Stores the maximum differenceint maxDiff = INT_MIN;// DFS traversal for source node as srcvoid dfs(int src, int val, vector<int>& W){ // Mark src node as visited visited[src] = true; // Traverse the tree for (auto neighbour : tree[src]) { // Check neighbour node is not // visited and ancestorNum should // be greater than the src node if (!visited[neighbour] && (ancestorNum[neighbour] > ancestorNum[src])) { // Update the maxDiff maxDiff = max( val - W[neighbour - 1], maxDiff); // Recurrence call for dfs dfs(neighbour, val, W); } }}// BFS traversal for source node as srcvoid bfs(int src, int N){ // Initially mark all node as // not visited visited.assign(N, false); // Stores the nodes queue<int> q; // Initially for src node mark // ancestorNum as 0 ancestorNum[src] = 0; // Mark src as visited visited[src] = true; // Push src node into the q q.push(src); // Traverse the queue q while (!q.empty()) { // Pop front element of the q int cur = q.front(); q.pop(); // Traverse the tree for (auto neighbour : tree[cur]) { // Check neighbour node is // already not visited if (!visited[neighbour]) { // Mark the neighbour // node as visited visited[neighbour] = true; // Push the neighbour // node into the q q.push(neighbour); // Update the neighbour // node ancestorNum ancestorNum[neighbour] = ancestorNum[cur] + 1; } } }}// Function to find the maximized// difference between two pair of nodes// in rooted tree such that one node// is ancestor of another nodevoid maximumDiff(vector<int> W, vector<int> P, int N){ if (N == 1) { cout << "-1\n"; return; } // Resize the tree tree.resize(N + 1); // Mark all the nodes as not visited visited.assign(N + 1, false); // Assign all the node values // for ancestorNum to 0 ancestorNum.assign(N + 1, 0); // Stores the source node to traverse int src; for (int i = 0; i < N; i++) { // Check P[i] is -1 if (P[i] == -1) // Update the source node src src = i; else { // Store the tree values tree[i + 1].push_back(P[i]); tree[P[i]].push_back(i + 1); } } // BFS from the source node src bfs(src, N + 1); // Mark all the nodes as not visited visited.assign(N + 1, false); // DFS Call for source node src dfs(src, W[src], W); // For every node call dfs function for (int i = 0; i < N; i++) { // Check i is root node if (i == src) continue; // Mark all the nodes as // not visited visited.assign(N + 1, false); // DFS Call for source // node as i+1 dfs(i + 1, W[i], W); } // Print the maxDiff cout << maxDiff << endl;}// Driver Codeint main(){ vector<int> W = { 5, 10, 6, 12 }; vector<int> P = { 2, -1, 4, 2 }; int N = P.size(); maximumDiff(W, P, N); return 0;} |
Java
// Java program for the above approachimport java.util.ArrayList;import java.util.Arrays;import java.util.LinkedList;import java.util.Queue;public class Main { private static ArrayList<ArrayList<Integer> > tree; private static boolean[] visited; private static int[] ancestorNum; private static int maxDiff = Integer.MIN_VALUE; private static void dfs(int src, int val, int[] W) { visited[src] = true; // Check neighbour node is not // visited and ancestorNum should // be greater than the src node for (int neighbour : tree.get(src)) { if (!visited[neighbour] && ancestorNum[neighbour] > ancestorNum[src]) { maxDiff = Math.max(val - W[neighbour - 1], maxDiff); dfs(neighbour, val, W); } } } private static void bfs(int src, int N) { visited = new boolean[N]; Arrays.fill(visited, false); Queue<Integer> q = new LinkedList<>(); ancestorNum[src] = 0; visited[src] = true; q.offer(src); while (!q.isEmpty()) { int cur = q.poll(); for (int neighbour : tree.get(cur)) { if (!visited[neighbour]) { visited[neighbour] = true; q.offer(neighbour); ancestorNum[neighbour] = ancestorNum[cur] + 1; } } } } // Function to find the maximized // difference between two pair of nodes // in rooted tree such that one node // is ancestor of another node private static void maximumDiff(int[] W, int[] P, int N) { if (N == 1) { System.out.println("-1"); return; } tree = new ArrayList<>(); for (int i = 0; i <= N; i++) tree.add(new ArrayList<>()); int src = 0; for (int i = 0; i < N; i++) { if (P[i] == -1) { src = i; } else { tree.get(i + 1).add(P[i]); tree.get(P[i]).add(i + 1); } } // Assign all the node values // for ancestorNum to 0 ancestorNum = new int[N + 1]; bfs(src, N + 1); Arrays.fill(visited, false); dfs(src, W[src], W); // For every node call dfs function for (int i = 0; i < N; i++) { // Check i is root node if (i == src) continue; Arrays.fill(visited, false); dfs(i + 1, W[i], W); } System.out.println(maxDiff); } public static void main(String[] args) { int[] W = { 5, 10, 6, 12 }; int[] P = { 2, -1, 4, 2 }; int N = P.length; maximumDiff(W, P, N); }} |
Python3
# Python 3 program for the above approachtree = []visited = []ancestorNum = []import sys# Stores the maximum differencemaxDiff = -sys.maxsize - 1# DFS traversal for source node as srcdef dfs(src, val, W): global ancestorNum global visited global tree global maxDiff # Mark src node as visited visited[src] = True # Traverse the tree for neighbour in tree[src]: # Check neighbour node is not # visited and ancestorNum should # be greater than the src node if (visited[neighbour] == False and (ancestorNum[neighbour]> ancestorNum[src])): # Update the maxDiff maxDiff = max(val - W[neighbour - 1],maxDiff) # Recurrence call for dfs dfs(neighbour, val, W)# BFS traversal for source node as srcdef bfs(src,N): global ancestorNum global visited global tree # Initially mark all node as # not visited visited = [False for i in range(N)] # Stores the nodes q = [] # Initially for src node mark # ancestorNum as 0 ancestorNum[src] = 0 # Mark src as visited visited[src] = True # Push src node into the q q.append(src) # Traverse the queue q while (len(q)>0): # Pop front element of the q cur = q[0] q = q[1:] # Traverse the tree for neighbour in tree[cur]: # Check neighbour node is # already not visited if (visited[neighbour]==False): # Mark the neighbour # node as visited visited[neighbour] = True # Push the neighbour # node into the q q.append(neighbour) # Update the neighbour # node ancestorNum ancestorNum[neighbour] = ancestorNum[cur] + 1# Function to find the maximized# difference between two pair of nodes# in rooted tree such that one node# is ancestor of another nodedef maximumDiff(W, P, N): global ancestorNum global visited global tree if (N == 1): print("-1") return # Resize the tree tree = [[] for i in range(N+1)] # Mark all the nodes as not visited visited = [False for i in range(N + 1)] # Assign all the node values # for ancestorNum to 0 ancestorNum = [0 for i in range(N + 1)] # Stores the source node to traverse src = 0 for i in range(N): # Check P[i] is -1 if (P[i] == -1): # Update the source node src src = i else: # Store the tree values tree[i + 1].append(P[i]) tree[P[i]].append(i + 1) # BFS from the source node src bfs(src, N + 1) # Mark all the nodes as not visited visited = [False for i in range(N+1)] # DFS Call for source node src dfs(src, W[src], W) # For every node call dfs function for i in range(N): # Check i is root node if (i == src): continue # Mark all the nodes as # not visited visited = [False for i in range(N+1)] # DFS Call for source # node as i+1 dfs(i + 1, W[i], W) # Print the maxDiff print(maxDiff)# Driver Codeif __name__ == '__main__': W = [5, 10, 6, 12] P = [2, -1, 4, 2] N = len(P) maximumDiff(W, P, N) # This code is contributed by SURENDRA_GANGWAR. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;// Stores the maximum differenceclass Program{ // Stores the maximum difference static int maxDiff = int.MinValue; static List<List<int>> tree = new List<List<int>>(); static bool[] visited; static int[] ancestorNum; // DFS traversal for source node as src static void DFS(int src, int val, int[] W) { // Mark src node as visited visited[src] = true; // Traverse the tree foreach (int neighbour in tree[src]) // Check neighbour node is not // visited and ancestorNum should // be greater than the src node { if (!visited[neighbour] && ancestorNum[neighbour] > ancestorNum[src]) { // Update the maxDiff maxDiff = Math.Max(val - W[neighbour - 1], maxDiff); // Recurrence call for dfs DFS(neighbour, val, W); } } } // BFS traversal for source node as src static void BFS(int src, int N) { // Initially mark all node as // not visited visited = new bool[N]; // Initially for src node mark // ancestorNum as 0 Queue<int> q = new Queue<int>(); ancestorNum = new int[N]; ancestorNum[src] = 0; visited[src] = true; q.Enqueue(src); while (q.Count > 0) { int cur = q.Dequeue(); foreach (int neighbour in tree[cur]) { if (!visited[neighbour]) { visited[neighbour] = true; q.Enqueue(neighbour); ancestorNum[neighbour] = ancestorNum[cur] + 1; } } } } static void MaximumDiff(int[] W, int[] P, int N) { if (N == 1) { Console.WriteLine("-1"); return; } tree = new List<List<int>>(); for (int i = 0; i <= N; i++) { tree.Add(new List<int>()); } int src = 0; for (int i = 0; i < N; i++) { if (P[i] == -1) { src = i; } else { tree[i + 1].Add(P[i]); tree[P[i]].Add(i + 1); } } BFS(src, N + 1); visited = new bool[N + 1]; DFS(src, W[src], W); for (int i = 0; i < N; i++) { if (i == src) continue; visited = new bool[N + 1]; DFS(i + 1, W[i], W); } Console.WriteLine(maxDiff); } // Driver Code static void Main(string[] args) { int[] W = { 5, 10, 6, 12 }; int[] P = { 2, -1, 4, 2 }; int N = P.Length; MaximumDiff(W, P, N); }} |
Javascript
// JavaScript code for the above approach let tree = []; let visited = []; let ancestorNum = []; // Stores the maximum difference let maxDiff = Number.MIN_SAFE_INTEGER; // DFS traversal for source node as src function dfs(src, val, W) { // Mark src node as visited visited[src] = true; // Traverse the tree for (let neighbour of tree[src]) { // Check neighbour node is not // visited and ancestorNum should // be greater than the src node if (!visited[neighbour] && ancestorNum[neighbour] > ancestorNum[src]) { // Update the maxDiff maxDiff = Math.max(val - W[neighbour - 1], maxDiff,6); // Recurrence call for dfs dfs(neighbour, val, W); } } } // BFS traversal for source node as src function bfs(src, N) { // Initially mark all node as // not visited visited = Array(N).fill(false); // Stores the nodes let q = []; // Initially for src node mark // ancestorNum as 0 ancestorNum[src] = 0; // Mark src as visited visited[src] = true; // Push src node into the q q.push(src); // Traverse the queue q while (q.length > 0) { // Pop front element of the q let cur = q.shift(); // Traverse the tree for (let neighbour of tree[cur]) { // Check neighbour node is // already not visited if (!visited[neighbour]) { // Mark the neighbour // node as visited visited[neighbour] = true; // Push the neighbour // node into the q q.push(neighbour); // Update the neighbour // node ancestorNum ancestorNum[neighbour] = ancestorNum[cur] + 1; } } } } // Function to find the maximized // difference between two pair of nodes // in rooted tree such that one node // is ancestor of another node function maximumDiff(W, P, N) { if (N === 1) { console.log("-1"); return; } // Resize the tree tree= new Array(N + 1); tree.fill([]); // Mark all the nodes as not visited visited = Array(N + 1).fill(false); // Assign all the node values // for ancestorNum to 0 ancestorNum.length = N + 1; ancestorNum.fill(0); // Stores the source node to traverse let src = 0; for (let i = 0; i < N; i += 1) { // Check P[i] is -1 if (P[i] === -1) { // Update the source node src src = i; } else { // Store the tree values tree[i + 1].push(P[i]); tree[P[i]].push(i + 1); } } // BFS from the source node src bfs(src, N + 1); // Mark all the nodes as not visited visited = Array(N + 1).fill(false); // DFS Call for source node src dfs(src, W[src], W); // For every node call dfs function for (let i = 0; i < N; i += 1) { // Check i is root node if (i === src) continue; // Mark all the nodes as // not visited visited = Array(N + 1).fill(false); // DFS Call for source // node as i+1 dfs(i + 1, W[i], W); } // Print the maxDiff console.log(maxDiff); } // Test the code maximumDiff([5, 10, 6, 12], [2, -1, 4, 2], 4);// This code is contributed by Potta Lokesh |
Output:
6
Time Complexity: O(N2)
Auxiliary Space: O(N)
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