# Find maximum xor of k elements in an array

Given an array arr[] of N integers and an integer K. The task is to find the maximum xor subset of size K of the given array.

Examples:

Input: arr[] = {2, 5, 4, 1, 3, 7, 6, 8}, K = 3
Output: 15
We obtain 15 by selecting 4, 5, 6, 8

Input: arr[] = {3, 4, 7, 7, 9}, K = 3
Output: 14

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

Naive approach: Iterate over all subsets of size K of the array and find maximum xor among them.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the maximum xor for a ` `// subset of size k from the given array ` `int` `Max_Xor(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` ` `  `    ``// Initialize result ` `    ``int` `maxXor = INT_MIN; ` ` `  `    ``// Traverse all subsets of the array ` `    ``for` `(``int` `i = 0; i < (1 << n); i++) { ` ` `  `        ``// __builtin_popcount() returns the number ` `        ``// of sets bits in an integer ` `        ``if` `(__builtin_popcount(i) == k) { ` ` `  `            ``// Initialize current xor as 0 ` `            ``int` `cur_xor = 0; ` `            ``for` `(``int` `j = 0; j < n; j++) { ` ` `  `                ``// If jth bit is set in i then include ` `                ``// jth element in the current xor ` `                ``if` `(i & (1 << j)) ` `                    ``cur_xor = cur_xor ^ arr[j]; ` `            ``} ` ` `  `            ``// Update maximum xor so far ` `            ``maxXor = max(maxXor, cur_xor); ` `        ``} ` `    ``} ` `    ``return` `maxXor; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 5, 4, 1, 3, 7, 6, 8 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` `    ``int` `k = 3; ` ` `  `    ``cout << Max_Xor(arr, n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the maximum xor for a ` `// subset of size k from the given array ` `static` `int` `Max_Xor(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` ` `  `    ``// Initialize result ` `    ``int` `maxXor = Integer.MIN_VALUE; ` ` `  `    ``// Traverse all subsets of the array ` `    ``for` `(``int` `i = ``0``; i < (``1` `<< n); i++)  ` `    ``{ ` ` `  `        ``// __builtin_popcount() returns the number ` `        ``// of sets bits in an integer ` `        ``if` `(Integer.bitCount(i) == k) ` `        ``{ ` ` `  `            ``// Initialize current xor as 0 ` `            ``int` `cur_xor = ``0``; ` `            ``for` `(``int` `j = ``0``; j < n; j++) ` `            ``{ ` ` `  `                ``// If jth bit is set in i then include ` `                ``// jth element in the current xor ` `                ``if` `((i & (``1` `<< j)) == ``0``) ` `                    ``cur_xor = cur_xor ^ arr[j]; ` `            ``} ` ` `  `            ``// Update maximum xor so far ` `            ``maxXor = Math.max(maxXor, cur_xor); ` `        ``} ` `    ``} ` `    ``return` `maxXor; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``2``, ``5``, ``4``, ``1``, ``3``, ``7``, ``6``, ``8` `}; ` `    ``int` `n = arr.length; ` `    ``int` `k = ``3``; ` ` `  `    ``System.out.println(Max_Xor(arr, n, k)); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python implementation of the approach ` ` `  `MAX` `=` `10000` `MAX_ELEMENT ``=` `50` ` `  `dp ``=``[[[``-``1` `for` `i ``in` `range``(``MAX``)]  ` `    ``for` `j ``in` `range``(MAX_ELEMENT)]  ` `    ``for` `k ``in` `range``(MAX_ELEMENT)] ` ` `  `# Function to return the maximum xor for a ` `# subset of size j from the given array ` `def` `Max_Xor(arr, i, j, mask, n): ` `    ``if` `(i >``=` `n): ` `         `  `        ``# If the subset is complete then return ` `        ``# the xor value of the selected elements ` `        ``if` `(j ``=``=` `0``): ` `            ``return` `mask ` `        ``else``: ` `            ``return` `0` `     `  `    ``# Return if already calculated for some ` `    ``# mask and j at the i'th index ` `    ``if` `(dp[i][j][mask] !``=` `-``1``): ` `        ``return` `dp[i][j][mask] ` `     `  `    ``# Initialize answer to 0 ` `    ``ans ``=` `0` `     `  `    ``# If we can still include elements in our subset ` `    ``# include the i'th element ` `    ``if` `(j > ``0``): ` `        ``ans ``=` `Max_Xor(arr, i ``+` `1``, j ``-` `1``, mask ^ arr[i], n) ` `         `  `    ``# Exclude the i'th element ` `    ``# ans store the max of both operations ` `    ``ans ``=` `max``(ans, Max_Xor(arr, i ``+` `1``, j, mask, n)) ` `    ``dp[i][j][mask] ``=` `ans ` `    ``return` `ans ` ` `  `# Driver code ` `arr ``=` `[``2``, ``5``, ``4``, ``1``, ``3``, ``7``, ``6``, ``8``] ` `n ``=` `len``(arr) ` `k ``=` `3` ` `  `print``(Max_Xor(arr, ``0``, k, ``0``, n)) ` ` `  `# This code is contributed by shubhamsingh10 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` ` `  `// Function to return the maximum xor for a ` `// subset of size k from the given array ` `static` `int` `Max_Xor(``int` `[]arr, ``int` `n, ``int` `k) ` `{ ` ` `  `    ``// Initialize result ` `    ``int` `maxXor = ``int``.MinValue; ` ` `  `    ``// Traverse all subsets of the array ` `    ``for` `(``int` `i = 0; i < (1 << n); i++)  ` `    ``{ ` ` `  `        ``// __builtin_popcount() returns the number ` `        ``// of sets bits in an integer ` `        ``if` `(bitCount(i) == k) ` `        ``{ ` ` `  `            ``// Initialize current xor as 0 ` `            ``int` `cur_xor = 0; ` `            ``for` `(``int` `j = 0; j < n; j++) ` `            ``{ ` ` `  `                ``// If jth bit is set in i then include ` `                ``// jth element in the current xor ` `                ``if` `((i & (1 << j)) == 0) ` `                    ``cur_xor = cur_xor ^ arr[j]; ` `            ``} ` ` `  `            ``// Update maximum xor so far ` `            ``maxXor = Math.Max(maxXor, cur_xor); ` `        ``} ` `    ``} ` `    ``return` `maxXor; ` `} ` ` `  `static` `int` `bitCount(``long` `x) ` `{ ` `    ``int` `setBits = 0; ` `    ``while` `(x != 0) ` `    ``{ ` `        ``x = x & (x - 1); ` `        ``setBits++; ` `    ``} ` `    ``return` `setBits; ` `}  ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 2, 5, 4, 1, 3, 7, 6, 8 }; ` `    ``int` `n = arr.Length; ` `    ``int` `k = 3; ` ` `  `    ``Console.WriteLine(Max_Xor(arr, n, k)); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:

```15
```

Efficient approach: The problem can be solved using dynamic programming. Create a dp table dp[i][j][mask] which stores the maximum xor possible at the ith index (with or without including it) and j denotes the number of remaining elements we can include in our subset of K elements. Mask is the xor of all the elements selected till the ith index.
Note: This approach will only work for smaller arrays due to space requirements for the dp array.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `#define MAX 10000 ` `#define MAX_ELEMENT 50 ` ` `  `int` `dp[MAX_ELEMENT][MAX_ELEMENT][MAX]; ` ` `  `// Function to return the maximum xor for a ` `// subset of size j from the given array ` `int` `Max_Xor(``int` `arr[], ``int` `i, ``int` `j, ``int` `mask, ``int` `n) ` `{ ` `    ``if` `(i >= n) { ` ` `  `        ``// If the subset is complete then return ` `        ``// the xor value of the selected elements ` `        ``if` `(j == 0) ` `            ``return` `mask; ` `        ``else` `            ``return` `0; ` `    ``} ` ` `  `    ``// Return if already calculated for some ` `    ``// mask and j at the i'th index ` `    ``if` `(dp[i][j][mask] != -1) ` `        ``return` `dp[i][j][mask]; ` ` `  `    ``// Initialize answer to 0 ` `    ``int` `ans = 0; ` ` `  `    ``// If we can still include elements in our subset ` `    ``// include the i'th element ` `    ``if` `(j > 0) ` `        ``ans = Max_Xor(arr, i + 1, j - 1, mask ^ arr[i], n); ` ` `  `    ``// Exclude the i'th element ` `    ``// ans store the max of both operations ` `    ``ans = max(ans, Max_Xor(arr, i + 1, j, mask, n)); ` ` `  `    ``return` `dp[i][j][mask] = ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 5, 4, 1, 3, 7, 6, 8 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` `    ``int` `k = 3; ` ` `  `    ``memset``(dp, -1, ``sizeof``(dp)); ` ` `  `    ``cout << Max_Xor(arr, 0, k, 0, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `static` `int` `MAX = ``10000``; ` `static` `int` `MAX_ELEMENT = ``50``; ` ` `  `static` `int` `[][][] dp = ``new` `int``[MAX_ELEMENT][MAX_ELEMENT][MAX]; ` ` `  `// Function to return the maximum xor for a ` `// subset of size j from the given array ` `static` `int` `Max_Xor(``int` `arr[], ``int` `i,  ` `                   ``int` `j, ``int` `mask, ``int` `n) ` `{ ` `    ``if` `(i >= n)  ` `    ``{ ` ` `  `        ``// If the subset is complete then return ` `        ``// the xor value of the selected elements ` `        ``if` `(j == ``0``) ` `            ``return` `mask; ` `        ``else` `            ``return` `0``; ` `    ``} ` ` `  `    ``// Return if already calculated for some ` `    ``// mask and j at the i'th index ` `    ``if` `(dp[i][j][mask] != -``1``) ` `        ``return` `dp[i][j][mask]; ` ` `  `    ``// Initialize answer to 0 ` `    ``int` `ans = ``0``; ` ` `  `    ``// If we can still include elements in our subset ` `    ``// include the i'th element ` `    ``if` `(j > ``0``) ` `        ``ans = Max_Xor(arr, i + ``1``, j - ``1``,  ` `                      ``mask ^ arr[i], n); ` ` `  `    ``// Exclude the i'th element ` `    ``// ans store the max of both operations ` `    ``ans = Math.max(ans, Max_Xor(arr, i + ``1``, j,  ` `                                ``mask, n)); ` ` `  `    ``return` `dp[i][j][mask] = ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``2``, ``5``, ``4``, ``1``, ``3``, ``7``, ``6``, ``8` `}; ` `    ``int` `n = arr.length; ` `    ``int` `k = ``3``; ` ` `  `    ``for``(``int` `i = ``0``; i < MAX_ELEMENT; i++) ` `    ``{ ` `        ``for``(``int` `j = ``0``; j < MAX_ELEMENT; j++) ` `        ``{ ` `            ``for``(``int` `l = ``0``; l < MAX; l++) ` `            ``dp[i][j][l] = -``1``; ` `        ``} ` `    ``} ` ` `  `    ``System.out.println(Max_Xor(arr, ``0``, k, ``0``, n)); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

## Python3

 `# Python implementation of the approach ` ` `  `MAX` `=` `10000` `MAX_ELEMENT ``=` `50` ` `  `dp ``=``[[[``-``1` `for` `i ``in` `range``(``MAX``)] ``for` `j ``in` `range``(MAX_ELEMENT)] ``for` `k ``in` `range``(MAX_ELEMENT)] ` ` `  `# Function to return the maximum xor for a ` `# subset of size j from the given array ` `def` `Max_Xor(arr, i, j, mask, n): ` `    ``if` `(i >``=` `n): ` `         `  `        ``# If the subset is complete then return ` `        ``# the xor value of the selected elements ` `        ``if` `(j ``=``=` `0``): ` `            ``return` `mask ` `        ``else``: ` `            ``return` `0` `     `  `    ``# Return if already calculated for some ` `    ``# mask and j at the i'th index ` `    ``if` `(dp[i][j][mask] !``=` `-``1``): ` `        ``return` `dp[i][j][mask] ` `     `  `    ``# Initialize answer to 0 ` `    ``ans ``=` `0` `     `  `    ``# If we can still include elements in our subset ` `    ``# include the i'th element ` `    ``if` `(j > ``0``): ` `        ``ans ``=` `Max_Xor(arr, i ``+` `1``, j ``-` `1``, mask ^ arr[i], n) ` `         `  `    ``# Exclude the i'th element ` `    ``# ans store the max of both operations ` `    ``ans ``=` `max``(ans, Max_Xor(arr, i ``+` `1``, j, mask, n)) ` `    ``dp[i][j][mask] ``=` `ans ` `    ``return` `ans ` ` `  ` `  `# Driver code ` ` `  `arr ``=` `[``2``, ``5``, ``4``, ``1``, ``3``, ``7``, ``6``, ``8``] ` `n ``=` `len``(arr) ` `k ``=` `3` ` `  `print``(Max_Xor(arr, ``0``, k, ``0``, n)) ` ` `  `# This code is contributed by shubhamsingh10 `

Output:

```15
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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