# 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.

**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 <bits/stdc++.h>` `using` `namespace` `std;` `int` `ans = 0, maxi = INT_MIN;` `vector<` `int` `> graph[100];` `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] = -1;` ` ` `weight[2] = 5;` ` ` `weight[3] = -1;` ` ` `weight[4] = 3;` ` ` `weight[5] = -2;` ` ` `// Edges of the tree` ` ` `graph[1].push_back(2);` ` ` `graph[2].push_back(3);` ` ` `graph[2].push_back(4);` ` ` `graph[1].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<Integer>[] 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<Integer>();` ` ` ` ` `// 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` `>[100];` ` ` `static` `int` `[] weight = ` `new` `int` `[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)` ` ` `{` ` ` `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] = -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);` ` ` `Console.Write(ans);` ` ` `}` `}` `// This code is contributed by PrinciRaj1992` |

## Javascript

`<script>` `// Javascript implementation of the approach` ` ` `let maxi = Number.MIN_VALUE, x, ans;` ` ` `let graph = ` `new` `Array(100);` ` ` `let weight = ` `new` `Array(100);` ` ` `for` `(let i = 0; i < 100; i++)` ` ` `{` ` ` `graph[i] = [];` ` ` `weight[i] = 0;` ` ` `}` ` ` ` ` `// 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` ` ` `function` `dfs(node, parent)` ` ` `{` ` ` `for` `(let 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` ` ` `function` `findMaxX(n, x)` ` ` `{` ` ` ` ` `// For every node` ` ` `for` `(let 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` ` ` `x = 15;` ` ` `let 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].push(2);` ` ` `graph[2].push(3);` ` ` `graph[2].push(4);` ` ` `graph[1].push(5);` ` ` `dfs(1, 1);` ` ` `findMaxX(n, x);` ` ` `document.write(ans);` ` ` ` ` `// This code is contributed by unknown2108` `</script>` |

**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.