# 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: 1
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

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

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; ` `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 = 5; ` `    ``weight = 10; ` `    ``weight = 11; ` `    ``weight = 8; ` `    ``weight = 6; ` ` `  `    ``// Edges of the tree ` `    ``graph.push_back(2); ` `    ``graph.push_back(3); ` `    ``graph.push_back(4); ` `    ``graph.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.Add(2);  ` `    ``graph.Add(3);  ` `    ``graph.Add(4);  ` `    ``graph.Add(5);  ` ` `  `    ``dfs(1, 1);  ` `    ``Console.Write( ans);  ` `}  ` `}  ` ` `  `// This code is contributed by SHUBHAMSINGH10  `

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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