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

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++

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// 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;
}

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Java

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// 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..

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Python3

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# 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

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C#

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// 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..

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Output:

3

Time complexity: O(N)



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