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# Split an array into subarrays with maximum Bitwise XOR of their respective Bitwise OR values

• Last Updated : 08 Jul, 2021

Given an array arr[] consisting of N integers, the task is to find the maximum Bitwise XOR of Bitwise OR of every subarray after splitting the array into subarrays(possible zero subarrays).

Examples:

Input: arr[] = {1, 5, 7}, N = 3
Output: 7
Explanation:
The given array can be expressed as the 1 subarray i.e., {1, 5, 7}.
The Bitwise XOR of the Bitwise OR of the formed subarray is 7, which is the maximum possible value.

Input: arr[] = {1, 2}, N = 2
Output: 3

Naive Approach: The simplest approach to solve the given above problem is to generate all possible combinations of breaking of subarrays using recursion and at each recursive call, find the maximum value of Bitwise XOR of Bitwise OR of all possible formed subarray and print it.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Recursive function to find all the``// possible breaking of arrays into``// subarrays and find the maximum``// Bitwise XOR``int` `maxXORUtil(``int` `arr[], ``int` `N,``               ``int` `xrr, ``int` `orr)``{``    ``// If the value of N is 0``    ``if` `(N == 0)``        ``return` `xrr ^ orr;` `    ``// Stores the result if the new``    ``// group is formed with the first``    ``// element as arr[i]``    ``int` `x = maxXORUtil(arr, N - 1,``                       ``xrr ^ orr,``                       ``arr[N - 1]);` `    ``// Stores if the result if the``    ``// arr[i] is included in the``    ``// last group``    ``int` `y = maxXORUtil(arr, N - 1,``                       ``xrr, orr | arr[N - 1]);` `    ``// Returns the maximum of``    ``// x and y``    ``return` `max(x, y);``}` `// Function to find the maximum possible``// Bitwise XOR of all possible values of``// the array after breaking the arrays``// into subarrays``int` `maximumXOR(``int` `arr[], ``int` `N)``{``    ``// Return the result``    ``return` `maxXORUtil(arr, N, 0, 0);``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 5, 7 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << maximumXOR(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``public` `class` `GFG{` `    ``// Recursive function to find all the``    ``// possible breaking of arrays into``    ``// subarrays and find the maximum``    ``// Bitwise XOR``    ``static` `int` `maxXORUtil(``int` `arr[], ``int` `N, ``int` `xrr,``                          ``int` `orr)``    ``{``      ` `        ``// If the value of N is 0``        ``if` `(N == ``0``)``            ``return` `xrr ^ orr;` `        ``// Stores the result if the new``        ``// group is formed with the first``        ``// element as arr[i]``        ``int` `x``            ``= maxXORUtil(arr, N - ``1``, xrr ^ orr, arr[N - ``1``]);` `        ``// Stores if the result if the``        ``// arr[i] is included in the``        ``// last group``        ``int` `y``            ``= maxXORUtil(arr, N - ``1``, xrr, orr | arr[N - ``1``]);` `        ``// Returns the maximum of``        ``// x and y``        ``return` `Math.max(x, y);``    ``}` `    ``// Function to find the maximum possible``    ``// Bitwise XOR of all possible values of``    ``// the array after breaking the arrays``    ``// into subarrays``    ``static` `int` `maximumXOR(``int` `arr[], ``int` `N)``    ``{``      ` `        ``// Return the result``        ``return` `maxXORUtil(arr, N, ``0``, ``0``);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``1``, ``5``, ``7` `};``        ``int` `N = arr.length;``        ``System.out.println(maximumXOR(arr, N));``    ``}``}` `// This code is contributed by abhinavjain194`

## Python3

 `# C++ program for the above approach``# Recursive function to find all the``# possible breaking of arrays o``# subarrays and find the maximum``# Bitwise XOR``def` `maxXORUtil(arr, N, xrr, orr):` `    ``# If the value of N is 0``    ``if` `(N ``=``=` `0``):``        ``return` `xrr ^ orr` `    ``# Stores the result if the new``    ``# group is formed with the first``    ``# element as arr[i]``    ``x ``=` `maxXORUtil(arr, N ``-` `1``, xrr ^ orr, arr[N ``-` `1``])` `    ``# Stores if the result if the``    ``# arr[i] is included in the``    ``# last group``    ``y ``=` `maxXORUtil(arr, N ``-` `1``, xrr, orr | arr[N ``-` `1``])` `    ``# Returns the maximum of``    ``# x and y``    ``return` `max``(x, y)`  `# Function to find the maximum possible``# Bitwise XOR of all possible values of``# the array after breaking the arrays``# o subarrays``def` `maximumXOR(arr,  N):` `    ``# Return the result``    ``return` `maxXORUtil(arr, N, ``0``, ``0``)`  `# Driver Code``arr ``=`  `1``, ``5``, ``7``N ``=` `len``(arr)``print``(maximumXOR(arr, N))` `# this code is contributed by shivanisinghss2110`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG``{` `    ``// Recursive function to find all the``    ``// possible breaking of arrays into``    ``// subarrays and find the maximum``    ``// Bitwise XOR``    ``static` `int` `maxXORUtil(``int``[] arr, ``int` `N, ``int` `xrr,``                          ``int` `orr)``    ``{``      ` `        ``// If the value of N is 0``        ``if` `(N == 0)``            ``return` `xrr ^ orr;` `        ``// Stores the result if the new``        ``// group is formed with the first``        ``// element as arr[i]``        ``int` `x``            ``= maxXORUtil(arr, N - 1, xrr ^ orr, arr[N - 1]);` `        ``// Stores if the result if the``        ``// arr[i] is included in the``        ``// last group``        ``int` `y``            ``= maxXORUtil(arr, N - 1, xrr, orr | arr[N - 1]);` `        ``// Returns the maximum of``        ``// x and y``        ``return` `Math.Max(x, y);``    ``}` `    ``// Function to find the maximum possible``    ``// Bitwise XOR of all possible values of``    ``// the array after breaking the arrays``    ``// into subarrays``    ``static` `int` `maximumXOR(``int``[] arr, ``int` `N)``    ``{``      ` `        ``// Return the result``        ``return` `maxXORUtil(arr, N, 0, 0);``    ``}`  `// Driver code``static` `void` `Main()``{``    ``int``[] arr = { 1, 5, 7 };``        ``int` `N = arr.Length;``        ``Console.Write(maximumXOR(arr, N));``}``}` `// This code is contributed by sanjoy_62.`

## Javascript

 ``
Output:
`7`

Time Complexity: O(2N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by observing the relationship between the Bitwise XOR and Bitwise OR  i.e., the value of Bitwise XOR of N elements is at most the value of Bitwise OR of N elements. Therefore, to find the maximum value, the idea is to split the group into only 1 group of the whole array.

Hence, print the value of Bitwise OR of the array elements arr[] as the resultant maximum value.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the bitwise OR of``// array elements``int` `MaxXOR(``int` `arr[], ``int` `N)``{``    ``// Stores the resultant maximum``    ``// value of Bitwise XOR``    ``int` `res = 0;` `    ``// Traverse the array arr[]``    ``for` `(``int` `i = 0; i < N; i++) {``        ``res |= arr[i];``    ``}` `    ``// Return the maximum value res``    ``return` `res;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 5, 7 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << MaxXOR(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.lang.*;``import` `java.util.*;` `class` `GFG{``    ` `// Function to find the bitwise OR of``// array elements``static` `int` `MaxXOR(``int` `arr[], ``int` `N)``{``    ` `    ``// Stores the resultant maximum``    ``// value of Bitwise XOR``    ``int` `res = ``0``;` `    ``// Traverse the array arr[]``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``res |= arr[i];``    ``}``    ` `    ``// Return the maximum value res``    ``return` `res;``}` `public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``1``, ``5``, ``7` `};``    ``int` `N = arr.length;``    ` `    ``System.out.println(MaxXOR(arr, N));``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 program for the above approach` `# Function to find the bitwise OR of``# array elements``def` `MaxXOR(arr, N):``    ` `    ``# Stores the resultant maximum``    ``# value of Bitwise XOR``    ``res ``=` `0` `    ``# Traverse the array arr[]``    ``for` `i ``in` `range``(N):``        ``res |``=` `arr[i]` `    ``# Return the maximum value res``    ``return` `res` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``arr ``=` `[ ``1``, ``5``, ``7` `]``    ``N ``=` `len``(arr)``    ` `    ``print` `(MaxXOR(arr, N))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG``{``    ` `// Function to find the bitwise OR of``// array elements``static` `int` `MaxXOR(``int` `[]arr, ``int` `N)``{``    ` `    ``// Stores the resultant maximum``    ``// value of Bitwise XOR``    ``int` `res = 0;` `    ``// Traverse the array arr[]``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``res |= arr[i];``    ``}``    ` `    ``// Return the maximum value res``    ``return` `res;``}` `public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 1, 5, 7 };``    ``int` `N = arr.Length;``    ` `    ``Console.Write(MaxXOR(arr, N));``}``}` `// This code is contributed by shivanisinghss2110`

## Javascript

 ``
Output:
`7`

Time Complexity: O(N)
Auxiliary Space: O(1)

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