# Longest Subarray having strictly positive XOR

Given an array arr[] of N non-negative integers. The task is to find the length of the longest sub-array such that XOR of all the elements of this sub-array is strictly positive. If no such sub-array exists then print -1

Examples:

Input: arr[] = {1, 1, 1, 1}
Output: 3
Take sub-array[0:2] = {1, 1, 1}
Xor of this sub-array is equal to 1.

Input: arr[] = {0, 1, 5, 19}
Output: 4

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

Approach:

• If the XOR of the complete array is positive, then answer is equal to N.
• If all the elements are zeroes then the answer is -1 as it is impossible to get strictly positive XOR.
• Otherwise, let’s say that index of the first positive number is l and the last positive number is r.
• Now XOR of all the elements of the index range [l, r] must be zero as elements before l and after r are 0s which will not contribute to the XOR value and the XOR of the original array was 0.
• Consider the sub-arrays A1, A1, …, Ar-1 and Al+1, Al+2, …, AN.
• The first subarray would have XOR value equal to A[r] and second would have XOR value A[l] which are positive.
• Return the length of the larger sub-array among these two sub-arrays.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the length of the ` `// longest sub-array having positive XOR ` `int` `StrictlyPositiveXor(``int` `A[], ``int` `N) ` `{ ` ` `  `    ``// To store the XOR ` `    ``// of all the elements ` `    ``int` `allxor = 0; ` ` `  `    ``// To check if all the ` `    ``// elements of the array are 0s ` `    ``bool` `checkallzero = ``true``; ` ` `  `    ``for` `(``int` `i = 0; i < N; i += 1) { ` ` `  `        ``// Take XOR of all the elements ` `        ``allxor ^= A[i]; ` ` `  `        ``// If any positive value is found ` `        ``// the make the checkallzero false ` `        ``if` `(A[i] > 0) ` `            ``checkallzero = ``false``; ` `    ``} ` ` `  `    ``// If complete array is the answer ` `    ``if` `(allxor != 0) ` `        ``return` `N; ` ` `  `    ``// If all elements are equal to zero ` `    ``if` `(checkallzero) ` `        ``return` `-1; ` ` `  `    ``// Initialize l and r ` `    ``int` `l = N, r = -1; ` ` `  `    ``for` `(``int` `i = 0; i < N; i += 1) { ` ` `  `        ``// First positive value of the array ` `        ``if` `(A[i] > 0) { ` `            ``l = i + 1; ` `            ``break``; ` `        ``} ` `    ``} ` `    ``for` `(``int` `i = N - 1; i >= 0; i -= 1) { ` ` `  `        ``// Last positive value of the array ` `        ``if` `(A[i] > 0) { ` `            ``r = i + 1; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// Maximum length among ` `    ``// these two subarrays ` `    ``return` `max(N - l, r - 1); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `A[] = { 1, 0, 0, 1 }; ` ` `  `    ``int` `N = ``sizeof``(A) / ``sizeof``(A); ` ` `  `    ``cout << StrictlyPositiveXor(A, N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the length of the ` `// longest sub-array having positive XOR ` `static` `int` `StrictlyPositiveXor(``int` `[]A, ``int` `N) ` `{ ` ` `  `    ``// To store the XOR ` `    ``// of all the elements ` `    ``int` `allxor = ``0``; ` ` `  `    ``// To check if all the ` `    ``// elements of the array are 0s ` `    ``boolean` `checkallzero = ``true``; ` ` `  `    ``for` `(``int` `i = ``0``; i < N; i += ``1``)  ` `    ``{ ` ` `  `        ``// Take XOR of all the elements ` `        ``allxor ^= A[i]; ` ` `  `        ``// If any positive value is found ` `        ``// the make the checkallzero false ` `        ``if` `(A[i] > ``0``) ` `            ``checkallzero = ``false``; ` `    ``} ` ` `  `    ``// If complete array is the answer ` `    ``if` `(allxor != ``0``) ` `        ``return` `N; ` ` `  `    ``// If all elements are equal to zero ` `    ``if` `(checkallzero) ` `        ``return` `-``1``; ` ` `  `    ``// Initialize l and r ` `    ``int` `l = N, r = -``1``; ` ` `  `    ``for` `(``int` `i = ``0``; i < N; i += ``1``) ` `    ``{ ` ` `  `        ``// First positive value of the array ` `        ``if` `(A[i] > ``0``)  ` `        ``{ ` `            ``l = i + ``1``; ` `            ``break``; ` `        ``} ` `    ``} ` `    ``for` `(``int` `i = N - ``1``; i >= ``0``; i -= ``1``)  ` `    ``{ ` ` `  `        ``// Last positive value of the array ` `        ``if` `(A[i] > ``0``)  ` `        ``{ ` `            ``r = i + ``1``; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// Maximum length among ` `    ``// these two subarrays ` `    ``return` `Math.max(N - l, r - ``1``); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `A[] = { ``1``, ``0``, ``0``, ``1` `}; ` ` `  `    ``int` `N = A.length; ` ` `  `    ``System.out.print(StrictlyPositiveXor(A, N)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the length of the  ` `# longest sub-array having positive XOR  ` `def` `StrictlyPositiveXor(A, N) : ` ` `  `    ``# To store the XOR  ` `    ``# of all the elements  ` `    ``allxor ``=` `0``;  ` ` `  `    ``# To check if all the  ` `    ``# elements of the array are 0s  ` `    ``checkallzero ``=` `True``;  ` ` `  `    ``for` `i ``in` `range``(N) : ` ` `  `        ``# Take XOR of all the elements  ` `        ``allxor ^``=` `A[i];  ` ` `  `        ``# If any positive value is found  ` `        ``# the make the checkallzero false  ` `        ``if` `(A[i] > ``0``) : ` `            ``checkallzero ``=` `False``;  ` ` `  `    ``# If complete array is the answer  ` `    ``if` `(allxor !``=` `0``) : ` `        ``return` `N;  ` ` `  `    ``# If all elements are equal to zero  ` `    ``if` `(checkallzero) : ` `        ``return` `-``1``;  ` ` `  `    ``# Initialize l and r  ` `    ``l ``=` `N; r ``=` `-``1``;  ` ` `  `    ``for` `i ``in` `range``(N) :  ` ` `  `        ``# First positive value of the array  ` `        ``if` `(A[i] > ``0``) : ` `            ``l ``=` `i ``+` `1``;  ` `            ``break``;  ` `             `  `    ``for` `i ``in` `range``(N ``-` `1``, ``-``1``, ``-``1``) : ` ` `  `        ``# Last positive value of the array  ` `        ``if` `(A[i] > ``0``) : ` `            ``r ``=` `i ``+` `1``;  ` `            ``break``;  ` ` `  `    ``# Maximum length among  ` `    ``# these two subarrays  ` `    ``return` `max``(N ``-` `l, r ``-` `1``);  ` ` `  ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``A``=` `[ ``1``, ``0``, ``0``, ``1` `]; ` `    ``N ``=` `len``(A); ` `    ``print``(StrictlyPositiveXor(A, N));  ` ` `  `    ``# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the length of the ` `// longest sub-array having positive XOR ` `static` `int` `StrictlyPositiveXor(``int` `[]A, ``int` `N) ` `{ ` ` `  `    ``// To store the XOR ` `    ``// of all the elements ` `    ``int` `allxor = 0; ` ` `  `    ``// To check if all the ` `    ``// elements of the array are 0s ` `    ``bool` `checkallzero = ``true``; ` ` `  `    ``for` `(``int` `i = 0; i < N; i += 1)  ` `    ``{ ` ` `  `        ``// Take XOR of all the elements ` `        ``allxor ^= A[i]; ` ` `  `        ``// If any positive value is found ` `        ``// the make the checkallzero false ` `        ``if` `(A[i] > 0) ` `            ``checkallzero = ``false``; ` `    ``} ` ` `  `    ``// If complete array is the answer ` `    ``if` `(allxor != 0) ` `        ``return` `N; ` ` `  `    ``// If all elements are equal to zero ` `    ``if` `(checkallzero) ` `        ``return` `-1; ` ` `  `    ``// Initialize l and r ` `    ``int` `l = N, r = -1; ` ` `  `    ``for` `(``int` `i = 0; i < N; i += 1) ` `    ``{ ` ` `  `        ``// First positive value of the array ` `        ``if` `(A[i] > 0)  ` `        ``{ ` `            ``l = i + 1; ` `            ``break``; ` `        ``} ` `    ``} ` `    ``for` `(``int` `i = N - 1; i >= 0; i -= 1)  ` `    ``{ ` ` `  `        ``// Last positive value of the array ` `        ``if` `(A[i] > 0)  ` `        ``{ ` `            ``r = i + 1; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// Maximum length among ` `    ``// these two subarrays ` `    ``return` `Math.Max(N - l, r - 1); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `[]A = { 1, 0, 0, 1 }; ` ` `  `    ``int` `N = A.Length; ` ` `  `    ``Console.WriteLine(StrictlyPositiveXor(A, N)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

Output:

```3
```

Time complexity: O(N)

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