Given a Binary Tree with N nodes valued 0 to N – 1 and N-1 edges and an array arr[] consisting of values of edges, the task is to find the maximum cost of splitting the tree into two halves.
The cost of splitting a tree is equal to the product of sum of node values of the splitted subtrees.
Examples:
Input: N = 6, arr[] = {13, 8, 7, 4, 5, 9}, Edges[][] = {{0, 1}, {1, 2}, {1, 4}, {3, 4}, {4, 5}}
Output: 504
Explanation:
Below is the given tree and resultant tree after removing the edge:
Remove the edge between 1st and 4th, then
t1 = valueat[0] + valueat[1] + valueat[2] = 13 + 8 + 7
t1 = valueat[3] + valueat[4] + valueat[5] = 4 + 5 + 9
t1*t2 = (13 + 8 + 7) * (4 + 5 + 9) = 504Input: N = 7, arr[]= {13, 8, 7, 4, 5, 9, 100}, Edges[][] = { {0, 1}, {1, 2}, {1, 4}, {3, 4}, {4, 5}, {2, 6}}
Output: 4600
Explanation:
Below is the given tree and resultant tree after removing the edge:
Remove the edge between 2nd and 6th, then
t1 = valueat[0] + valueat[1] + valueat[2] + valueat[3] + valueat[4] + valueat[5]= 13 + 8 + 7 + 4 + 5 + 9
t2 = valueat[6] = 100
t1*t2 = (13 + 8 + 7 + 5 + 4 + 9) * (100) = 4600
Approach: The idea is to traverse the given tree and try to break the tree at every possible edge and then find the maximum cost of splitting at that edges. After all the above steps print the maximum cost among all the splitting. Below are the steps:
- All the edges are stored using the adjacency list edges and values at each node is stored in the given array arr[].
- For the current-node, find the sum of values in its descendants including itself.
- Suppose if the edge between the current node and its parent is removed then two trees can be formed.
- Now, calculate values of t1, t2, and check the product of t1 and t2 is maximum or not.
- Repeat this process recursively to all the child-nodes of current-node.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// To store the results and sum of// all nodes in the arrayint ans = 0, allsum = 0;// To create adjacency listvector<int> edges[100001];// Function to add edges into the// adjacency listvoid addedge(int a, int b){ edges[a].push_back(b); edges[b].push_back(a);}// Recursive function that calculate// the value of the cost of splitting// the tree recursivelyvoid findCost(int r, int p, int arr[]){ int i, cur; for (i = 0; i < edges[r].size(); i++) { // Fetch the child of node-r cur = edges[r].at(i); // Neglect if cur node is parent if (cur == p) continue; findCost(cur, r, arr); // Add all values of nodes // which are descendants of r arr[r] += arr[cur]; } // The two trees formed are rooted // at 'r' with its descendants int t1 = arr[r]; int t2 = allsum - t1; // Check and replace if current // product t1*t2 is large if (t1 * t2 > ans) { ans = t1 * t2; }}// Function to find the maximum cost// after splitting the tree in 2 halvesvoid maximumCost(int r, int p, int N, int M, int arr[], int Edges[][2]){ // Find sum of values in all nodes for (int i = 0; i < N; i++) { allsum += arr[i]; } // Traverse edges to create // adjacency list for (int i = 0; i < M; i++) { addedge(Edges[i][0], Edges[i][1]); } // Function Call findCost(r, p, arr);}// Driver Codeint main(){ int a, b, N = 6; // Values in each node int arr[] = { 13, 8, 7, 4, 5, 9 }; int M = 5; // Given Edges int Edges[][2] = { { 0, 1 }, { 1, 2 }, { 1, 4 }, { 3, 4 }, { 4, 5 } }; maximumCost(1, -1, N, M, arr, Edges); cout << ans; return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ // To store the results and sum of// all nodes in the arraystatic int ans = 0, allsum = 0; // To create adjacency liststatic Vector<Integer> []edges = new Vector[100001]; // Function to add edges into the// adjacency liststatic void addedge(int a, int b){ edges[a].add(b); edges[b].add(a);} // Recursive function that calculate// the value of the cost of splitting// the tree recursivelystatic void findCost(int r, int p, int arr[]){ int i, cur; for (i = 0; i < edges[r].size(); i++) { // Fetch the child of node-r cur = edges[r].get(i); // Neglect if cur node is parent if (cur == p) continue; findCost(cur, r, arr); // Add all values of nodes // which are descendants of r arr[r] += arr[cur]; } // The two trees formed are rooted // at 'r' with its descendants int t1 = arr[r]; int t2 = allsum - t1; // Check and replace if current // product t1*t2 is large if (t1 * t2 > ans) { ans = t1 * t2; }} // Function to find the maximum cost// after splitting the tree in 2 halvesstatic void maximumCost(int r, int p, int N, int M, int arr[], int Edges[][]){ // Find sum of values in all nodes for (int i = 0; i < N; i++) { allsum += arr[i]; } // Traverse edges to create // adjacency list for (int i = 0; i < M; i++) { addedge(Edges[i][0], Edges[i][1]); } // Function Call findCost(r, p, arr);} // Driver Codepublic static void main(String[] args){ int a, b, N = 6; // Values in each node int arr[] = {13, 8, 7, 4, 5, 9}; int M = 5; // Given Edges int Edges[][] = {{0, 1}, {1, 2}, {1, 4}, {3, 4}, {4, 5}}; for (int i = 0; i < edges.length; i++) edges[i] = new Vector<Integer>(); maximumCost(1, -1, N, M, arr, Edges); System.out.print(ans);}} // This code is contributed by Amit Katiyar |
Python3
# Python3 program for the above approach# To store the results and sum of# all nodes in the arrayans = 0allsum = 0# To create adjacency listedges = [[] for i in range(100001)]# Function to add edges into the# adjacency listdef addedge(a, b): global edges edges[a].append(b) edges[b].append(a)# Recursive function that calculate# the value of the cost of splitting# the tree recursivelydef findCost(r, p, arr): global edges global ans global allsum i = 0 for i in range(len(edges[r])): # Fetch the child of node-r cur = edges[r][i] # Neglect if cur node is parent if (cur == p): continue findCost(cur, r, arr) # Add all values of nodes # which are descendants of r arr[r] += arr[cur] # The two trees formed are rooted # at 'r' with its descendants t1 = arr[r] t2 = allsum - t1 # Check and replace if current # product t1*t2 is large if (t1 * t2 > ans): ans = t1 * t2# Function to find the maximum cost# after splitting the tree in 2 halvesdef maximumCost(r, p, N, M, arr, Edges): global allsum # Find sum of values in all nodes for i in range(N): allsum += arr[i] # Traverse edges to create # adjacency list for i in range(M): addedge(Edges[i][0], Edges[i][1]) # Function Call findCost(r, p, arr)# Driver Codeif __name__ == '__main__': N = 6 # Values in each node arr = [ 13, 8, 7, 4, 5, 9 ] M = 5 # Given Edges Edges = [ [ 0, 1 ], [ 1, 2 ], [ 1, 4 ], [ 3, 4 ], [ 4, 5 ] ] maximumCost(1, -1, N, M, arr, Edges) print(ans)# This code is contributed by ipg2016107 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // To store the results and sum of// all nodes in the arraystatic int ans = 0, allsum = 0; // To create adjacency liststatic List<int> []edges = new List<int>[100001]; // Function to add edges into the// adjacency liststatic void addedge(int a, int b){ edges[a].Add(b); edges[b].Add(a);} // Recursive function that calculate// the value of the cost of splitting// the tree recursivelystatic void findCost(int r, int p, int []arr){ int i, cur; for (i = 0; i < edges[r].Count; i++) { // Fetch the child of node-r cur = edges[r][i]; // Neglect if cur node is parent if (cur == p) continue; findCost(cur, r, arr); // Add all values of nodes // which are descendants of r arr[r] += arr[cur]; } // The two trees formed are rooted // at 'r' with its descendants int t1 = arr[r]; int t2 = allsum - t1; // Check and replace if current // product t1*t2 is large if (t1 * t2 > ans) { ans = t1 * t2; }} // Function to find the maximum cost// after splitting the tree in 2 halvesstatic void maximumCost(int r, int p, int N, int M, int []arr, int [, ]Edges){ // Find sum of values in all nodes for (int i = 0; i < N; i++) { allsum += arr[i]; } // Traverse edges to create // adjacency list for (int i = 0; i < M; i++) { addedge(Edges[i, 0], Edges[i, 1]); } // Function Call findCost(r, p, arr);} // Driver Codepublic static void Main(String[] args){ int N = 6; // Values in each node int []arr = {13, 8, 7, 4, 5, 9}; int M = 5; // Given Edges int [,]Edges = {{0, 1}, {1, 2}, {1, 4}, {3, 4}, {4, 5}}; for (int i = 0; i < edges.Length; i++) edges[i] = new List<int>(); maximumCost(1, -1, N, M, arr, Edges); Console.Write(ans);}} // This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript program for the above approach // To store the results and sum of // all nodes in the array let ans = 0, allsum = 0; // To create adjacency list let edges = new Array(100001); // Function to add edges into the // adjacency list function addedge(a, b) { edges[a].push(b); edges[b].push(a); } // Recursive function that calculate // the value of the cost of splitting // the tree recursively function findCost(r, p, arr) { let i, cur; for (i = 0; i < edges[r].length; i++) { // Fetch the child of node-r cur = edges[r][i]; // Neglect if cur node is parent if (cur == p) continue; findCost(cur, r, arr); // Add all values of nodes // which are descendants of r arr[r] += arr[cur]; } // The two trees formed are rooted // at 'r' with its descendants let t1 = arr[r]; let t2 = allsum - t1; // Check and replace if current // product t1*t2 is large if (t1 * t2 > ans) { ans = t1 * t2; } } // Function to find the maximum cost // after splitting the tree in 2 halves function maximumCost(r, p, N, M, arr, Edges) { // Find sum of values in all nodes for (let i = 0; i < N; i++) { allsum += arr[i]; } // Traverse edges to create // adjacency list for (let i = 0; i < M; i++) { addedge(Edges[i][0], Edges[i][1]); } // Function Call findCost(r, p, arr); } let N = 6; // Values in each node let arr = [13, 8, 7, 4, 5, 9]; let M = 5; // Given Edges let Edges = [[0, 1], [1, 2], [1, 4], [3, 4], [4, 5]]; for (let i = 0; i < edges.length; i++) edges[i] = []; maximumCost(1, -1, N, M, arr, Edges); document.write(ans); </script> |
Output:
504
Time Complexity: O(N)
Auxiliary Space: O(N)
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