# Find the node whose xor with x gives maximum value

Given a tree, and the weights of all the nodes and an integer x, the task is to find a node i such that weight[i] xor x is maximum.
Examples:

Input:

x = 15
Output:
Node 1: 5 xor 15 = 10
Node 2: 10 xor 15 = 5
Node 3: 11 xor 15 = 4
Node 4: 8 xor 15 = 7
Node 5: 6 xor 15 = 9

Approach: Perform dfs on the tree and keep track of the node whose weighted xor with x gives the maximum value.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `int` `maximum = INT_MIN, x, ans;`   `vector<``int``> graph[100];` `vector<``int``> weight(100);`   `// Function to perform dfs to find` `// the maximum xored value` `void` `dfs(``int` `node, ``int` `parent)` `{` `    ``// If current value is less than` `    ``// the current maximum` `    ``if` `(maximum < (weight[node] ^ x)) {` `        ``maximum = weight[node] ^ x;` `        ``ans = node;` `    ``}` `    ``for` `(``int` `to : graph[node]) {` `        ``if` `(to == parent)` `            ``continue``;` `        ``dfs(to, node);` `    ``}` `}`   `// Driver code` `int` `main()` `{` `    ``x = 15;`   `    ``// Weights of the node` `    ``weight[1] = 5;` `    ``weight[2] = 10;` `    ``weight[3] = 11;` `    ``weight[4] = 8;` `    ``weight[5] = 6;`   `    ``// 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);`   `    ``cout << ans;`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.util.*;`   `class` `GFG ` `{`   `    ``static` `int` `maximum = Integer.MIN_VALUE, x, ans;`   `    ``@SuppressWarnings``(``"unchecked"``)` `    ``static` `Vector[] graph = ``new` `Vector[``100``];` `    ``static` `int``[] weight = ``new` `int``[``100``];`   `    ``// This block is executed even before main() function` `    ``// This is necessary otherwise this program will` `    ``// throw "NullPointerException"` `    ``static` `    ``{` `        ``for` `(``int` `i = ``0``; i < ``100``; i++)` `            ``graph[i] = ``new` `Vector<>();` `    ``}`   `    ``// Function to perform dfs to find` `    ``// the maximum xored value` `    ``static` `void` `dfs(``int` `node, ``int` `parent) ` `    ``{`   `        ``// If current value is less than` `        ``// the current maximum` `        ``if` `(maximum < (weight[node] ^ x)) ` `        ``{` `            ``maximum = weight[node] ^ x;` `            ``ans = node;` `        ``}` `        ``for` `(``int` `to : graph[node]) ` `        ``{` `            ``if` `(to == parent)` `                ``continue``;` `            ``dfs(to, node);` `        ``}` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``x = ``15``;`   `        ``// Weights of the node` `        ``weight[``1``] = ``5``;` `        ``weight[``2``] = ``10``;` `        ``weight[``3``] = ``11``;` `        ``weight[``4``] = ``8``;` `        ``weight[``5``] = ``6``;`   `        ``// Edges of the tree` `        ``graph[``1``].add(``2``);` `        ``graph[``2``].add(``3``);` `        ``graph[``2``].add(``4``);` `        ``graph[``1``].add(``5``);`   `        ``dfs(``1``, ``1``);`   `        ``System.out.println(ans);` `    ``}` `}`   `// This code is contributed by` `// sanjeev2552`

## Python3

 `# Python3 implementation of the approach` `import` `sys` `maximum ``=` `-``sys.maxsize ``-` `1` `graph ``=` `[[``0` `for` `i ``in` `range``(``100``)]` `            ``for` `j ``in` `range``(``100``)]` `weight ``=` `[``0` `for` `i ``in` `range``(``100``)]` `ans ``=` `[]`   `# Function to perform dfs to find` `# the maximum xored value` `def` `dfs(node, parent):` `    ``global` `maximum` `    `  `    ``# If current value is less than` `    ``# the current maximum` `    ``if` `(maximum < (weight[node] ^ x)):` `        ``maximum ``=` `weight[node] ^ x` `        ``ans.append(node)` `        `  `    ``for` `to ``in` `graph[node]:` `        ``if` `(to ``=``=` `parent):` `            ``continue` `        ``dfs(to, node)` `        `  `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``x ``=` `15`   `    ``# Weights of the node` `    ``weight[``1``] ``=` `5` `    ``weight[``2``] ``=` `10` `    ``weight[``3``] ``=` `11` `    ``weight[``4``] ``=` `8` `    ``weight[``5``] ``=` `6`   `    ``# Edges of the tree` `    ``graph[``1``].append(``2``)` `    ``graph[``2``].append(``3``)` `    ``graph[``2``].append(``4``)` `    ``graph[``1``].append(``5``)`   `    ``dfs(``1``, ``1``)`   `    ``print``(ans[``0``])` `    `  `# This code is contributed by` `# Surendra_Gangwar`

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; `   `class` `GFG ` `{ `   `static` `int` `maximum = ``int``.MinValue, x, ` `ans = ``int``.MaxValue; `   `static` `List> graph = ``new` `List>(); ` `static` `List<``int``> weight = ``new` `List<``int``>(); `     `// Function to perform dfs to find ` `// the maximum value ` `static` `void` `dfs(``int` `node, ``int` `parent) ` `{ ` `    ``// If current value is less than` `    ``// the current maximum` `    ``if` `(maximum < (weight[node] ^ x)) ` `    ``{` `        ``maximum = weight[node] ^ x;` `        ``ans = node;` `    ``} ` `        `  `    ``for` `(``int` `i = 0; i < graph[node].Count; i++) ` `    ``{ ` `        ``if` `(graph[node][i] == parent) ` `            ``continue``; ` `        ``dfs(graph[node][i], node); ` `    ``} ` `} `   `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``x = 15; `   `    ``// Weights of the node ` `    ``weight.Add(0); ` `    ``weight.Add(5); ` `    ``weight.Add(10); ` `    ``weight.Add(11);; ` `    ``weight.Add(8); ` `    ``weight.Add(6); ` `    `  `    ``for``(``int` `i = 0; i < 100; i++) ` `    ``graph.Add(``new` `List<``int``>()); `   `    ``// Edges of the tree ` `    ``graph[1].Add(2); ` `    ``graph[2].Add(3); ` `    ``graph[2].Add(4); ` `    ``graph[1].Add(5); `   `    ``dfs(1, 1); ` `    ``Console.Write( ans); ` `} ` `} `   `// This code is contributed by SHUBHAMSINGH10 `

## Javascript

 ``

Output:

`1`

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(1).
Any extra space is not required, so the space complexity is constant.

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