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, …, A_r-1 and A_l+1, A_l+2, …, A_N.
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 <bits/stdc++.h> 
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[0]); 
  
    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:

Time complexity: O(N)

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes

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

C++

Java

Python3

C#

Recommended Posts: