# Find the root of the sub-tree whose weighted sum XOR with X is maximum

Given a tree, and the weights of all the nodes, the task is to find the root of the sub-tree whose weighted sum XOR with given integer X is maximum.

Examples:

Input: X = 15
Output: 4
Weight of sub-tree for parent 1 = ((-1) + (5) + (-2) + (-1) + (3)) XOR 15 = 4 XOR 15 = 11
Weight of sub-tree for parent 2 = ((5) + (-1) + (3)) XOR 15 = 7 XOR 15 = 8
Weight of sub-tree for parent 3 = -1 XOR 15 = -16
Weight of sub-tree for parent 4 = 3 XOR 15 = 12
Weight of sub-tree for parent 5 = -2 XOR 15 = -15
Node 4 gives the maximum sub-tree weighted sum XOR X.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Perform dfs on the tree, and for every node calculate the sub-tree weighted sum rooted at the current node then find the maximum (sum XOR X) value for a node.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `int` `ans = 0, maxi = INT_MIN; ` ` `  `vector<``int``> graph; ` `vector<``int``> weight(100); ` ` `  `// Function to perform dfs and update the tree ` `// such that every node's weight is the sum of ` `// the weights of all the nodes in the sub-tree ` `// of the current node including itself ` `void` `dfs(``int` `node, ``int` `parent) ` `{ ` `    ``for` `(``int` `to : graph[node]) { ` `        ``if` `(to == parent) ` `            ``continue``; ` `        ``dfs(to, node); ` ` `  `        ``// Calculating the weighted ` `        ``// sum of the subtree ` `        ``weight[node] += weight[to]; ` `    ``} ` `} ` ` `  `// Function to find the node ` `// having maximum sub-tree sum XOR x ` `void` `findMaxX(``int` `n, ``int` `x) ` `{ ` ` `  `    ``// For every node ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` ` `  `        ``// If current node's weight XOR x ` `        ``// is maximum so far ` `        ``if` `(maxi < (weight[i] ^ x)) { ` `            ``maxi = (weight[i] ^ x); ` `            ``ans = i; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `x = 15; ` `    ``int` `n = 5; ` ` `  `    ``// Weights of the node ` `    ``weight = -1; ` `    ``weight = 5; ` `    ``weight = -1; ` `    ``weight = 3; ` `    ``weight = -2; ` ` `  `    ``// Edges of the tree ` `    ``graph.push_back(2); ` `    ``graph.push_back(3); ` `    ``graph.push_back(4); ` `    ``graph.push_back(5); ` ` `  `    ``dfs(1, 1); ` `    ``findMaxX(n, x); ` ` `  `    ``cout << ans; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `    ``static` `int` `ans = ``0``, maxi = Integer.MIN_VALUE; ` ` `  `    ``static` `Vector[] graph = ``new` `Vector[``100``]; ` `    ``static` `Integer[] weight = ``new` `Integer[``100``]; ` ` `  `    ``// Function to perform dfs and update the tree ` `    ``// such that every node's weight is the sum of ` `    ``// the weights of all the nodes in the sub-tree ` `    ``// of the current node including itself ` `    ``static` `void` `dfs(``int` `node, ``int` `parent) ` `    ``{ ` `        ``for` `(``int` `to : graph[node]) ` `        ``{ ` `            ``if` `(to == parent) ` `                ``continue``; ` `            ``dfs(to, node); ` ` `  `            ``// Calculating the weighted ` `            ``// sum of the subtree ` `            ``weight[node] += weight[to]; ` `        ``} ` `    ``} ` ` `  `    ``// Function to find the node ` `    ``// having maximum sub-tree sum XOR x ` `    ``static` `void` `findMaxX(``int` `n, ``int` `x) ` `    ``{ ` ` `  `        ``// For every node ` `        ``for` `(``int` `i = ``1``; i <= n; i++) ` `        ``{ ` ` `  `            ``// If current node's weight XOR x ` `            ``// is maximum so far ` `            ``if` `(maxi < (weight[i] ^ x)) ` `            ``{ ` `                ``maxi = (weight[i] ^ x); ` `                ``ans = i; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `x = ``15``; ` `        ``int` `n = ``5``; ` `        ``for` `(``int` `i = ``0``; i < ``100``; i++) ` `            ``graph[i] = ``new` `Vector(); ` `         `  `        ``// Weights of the node ` `        ``weight[``1``] = -``1``; ` `        ``weight[``2``] = ``5``; ` `        ``weight[``3``] = -``1``; ` `        ``weight[``4``] = ``3``; ` `        ``weight[``5``] = -``2``; ` ` `  `        ``// Edges of the tree ` `        ``graph[``1``].add(``2``); ` `        ``graph[``2``].add(``3``); ` `        ``graph[``2``].add(``4``); ` `        ``graph[``1``].add(``5``); ` ` `  `        ``dfs(``1``, ``1``); ` `        ``findMaxX(n, x); ` ` `  `        ``System.out.print(ans); ` `    ``} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python

 `# Python implementation of the approach ` `from` `sys ``import` `maxsize ` ` `  `# Function to perform dfs and update the tree ` `# such that every node's weight is the sum of ` `# the weights of all the nodes in the sub-tree ` `# of the current node including itself ` `def` `dfs(node, parent): ` `    ``global` `maxi, graph, weight, x, ans ` `    ``for` `to ``in` `graph[node]: ` `        ``if` `(to ``=``=` `parent): ` `            ``continue` `        ``dfs(to, node) ` `         `  `        ``# Calculating the weighted ` `        ``# sum of the subtree ` `        ``weight[node] ``+``=` `weight[to] ` `         `  `# Function to find the node ` `# having maximum sub-tree sum XOR x ` `def` `findMaxX(n, x): ` `    ``global` `maxi, graph, weight, ans ` `     `  `    ``# For every node ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``): ` `         `  `        ``# If current node's weight XOR x ` `        ``# is maximum so far ` `        ``if` `(maxi < (weight[i] ^ x)): ` `            ``maxi ``=` `(weight[i] ^ x) ` `            ``ans ``=` `i ` ` `  `# Driver code ` `ans ``=` `0` `maxi ``=` `-``maxsize ` ` `  `graph ``=` `[[] ``for` `i ``in` `range``(``100``)] ` `weight ``=` `[``0``]``*``100` `x ``=` `15` `n ``=` `5` ` `  `# Weights of the node ` `weight[``1``] ``=` `-``1` `weight[``2``] ``=` `5` `weight[``3``] ``=` `-``1` `weight[``4``] ``=` `3` `weight[``5``] ``=` `-``2` ` `  `# Edges of the tree ` `graph[``1``].append(``2``) ` `graph[``2``].append(``3``) ` `graph[``2``].append(``4``) ` `graph[``1``].append(``5``) ` ` `  `dfs(``1``, ``1``) ` `findMaxX(n, x) ` ` `  `print``(ans) ` ` `  `# This code is contributed by SHUBHAMSINGH10 `

## C#

 `     `  `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG ` `{ ` `    ``static` `int` `ans = 0, maxi = ``int``.MinValue; ` ` `  `    ``static` `List<``int``>[] graph = ``new` `List<``int``>; ` `    ``static` `int``[] weight = ``new` `int``; ` ` `  `    ``// Function to perform dfs and update the tree ` `    ``// such that every node's weight is the sum of ` `    ``// the weights of all the nodes in the sub-tree ` `    ``// of the current node including itself ` `    ``static` `void` `dfs(``int` `node, ``int` `parent) ` `    ``{ ` `        ``foreach` `(``int` `to ``in` `graph[node]) ` `        ``{ ` `            ``if` `(to == parent) ` `                ``continue``; ` `            ``dfs(to, node); ` ` `  `            ``// Calculating the weighted ` `            ``// sum of the subtree ` `            ``weight[node] += weight[to]; ` `        ``} ` `    ``} ` ` `  `    ``// Function to find the node ` `    ``// having maximum sub-tree sum XOR x ` `    ``static` `void` `findMaxX(``int` `n, ``int` `x) ` `    ``{ ` ` `  `        ``// For every node ` `        ``for` `(``int` `i = 1; i <= n; i++) ` `        ``{ ` ` `  `            ``// If current node's weight XOR x ` `            ``// is maximum so far ` `            ``if` `(maxi < (weight[i] ^ x)) ` `            ``{ ` `                ``maxi = (weight[i] ^ x); ` `                ``ans = i; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int` `x = 15; ` `        ``int` `n = 5; ` `        ``for` `(``int` `i = 0; i < 100; i++) ` `            ``graph[i] = ``new` `List<``int``>(); ` `         `  `        ``// Weights of the node ` `        ``weight = -1; ` `        ``weight = 5; ` `        ``weight = -1; ` `        ``weight = 3; ` `        ``weight = -2; ` ` `  `        ``// Edges of the tree ` `        ``graph.Add(2); ` `        ``graph.Add(3); ` `        ``graph.Add(4); ` `        ``graph.Add(5); ` ` `  `        ``dfs(1, 1); ` `        ``findMaxX(n, x); ` ` `  `        ``Console.Write(ans); ` `    ``} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```4
```

Complexity Analysis:

• Time Complexity : O(N).
In dfs, every node of the tree is processed once and hence the complexity due to the dfs is O(N) if there are total N nodes in the tree. Therefore, the time complexity is O(N).
• Auxiliary Space : O(n).
Recursion Stack.

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