# Maximum cost of splitting given Binary Tree into two halves

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) = 504

Input: 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:

1. All the edges are stored using the adjacency list edges and values at each node is stored in the given array arr[].
2. For the current-node, find the sum of values in its descendants including itself.
3. Suppose if the edge between the current node and its parent is removed then two trees can be formed.
4. Now, calculate values of t1, t2, and check the product of t1 and t2 is maximum or not.
5. Repeat this process recursively to all the child-nodes of current-node.

Below is the implementation of the above approach:

 // C++ program for the above approach #include using namespace std;   // To store the results and sum of // all nodes in the array int ans = 0, allsum = 0;   // To create adjacency list vector edges[100001];   // Function to add edges into the // adjacency list void 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 recursively 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].at(i);           // Neglect if cur node is parent         if (cur == p)             continue;           findCost(cur, r, arr);           // Add all values of nodes         // which are decendents of r         arr[r] += arr[cur];     }       // The two trees formed are rooted     // at 'r' with its decendents     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 halves void 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 Code int 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 program for the above approach import java.util.*; class GFG{    // To store the results and sum of // all nodes in the array static int ans = 0, allsum = 0;    // To create adjacency list static Vector []edges = new Vector[100001];    // Function to add edges into the // adjacency list static 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 recursively static 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 decendents of r         arr[r] += arr[cur];     }        // The two trees formed are rooted     // at 'r' with its decendents     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 halves static 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 Code public 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();     maximumCost(1, -1, N, M, arr, Edges);     System.out.print(ans); } }    // This code is contributed by Amit Katiyar

 // C# program for the above approach using System; using System.Collections.Generic; class GFG{    // To store the results and sum of // all nodes in the array static int ans = 0, allsum = 0;    // To create adjacency list static List []edges = new List[100001];    // Function to add edges into the // adjacency list static 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 recursively static 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 decendents of r         arr[r] += arr[cur];     }        // The two trees formed are rooted     // at 'r' with its decendents     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 halves static 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 Code public 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();     maximumCost(1, -1, N, M, arr, Edges);     Console.Write(ans); } }    // This code is contributed by Rajput-Ji

Output:
504

Time Complexity: O(N)
Auxiliary Space: O(N)

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Improved By : amit143katiyar, Rajput-Ji

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