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Longest Subarray having strictly positive XOR

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  • Difficulty Level : Medium
  • Last Updated : 27 Apr, 2021
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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:
Take sub-array[0:2] = {1, 1, 1} 
Xor of this sub-array is equal to 1.
Input: arr[] = {0, 1, 5, 19} 
Output:
 

 

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 an XOR value A[l] which is 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..

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the length of the
// longest sub-array having positive XOR
function StrictlyPositiveXor(A, N)
{
 
    // To store the XOR
    // of all the elements
    let allxor = 0;
 
    // To check if all the
    // elements of the array are 0s
    let checkallzero = true;
 
    for (let 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
    let l = N, r = -1;
 
    for (let i = 0; i < N; i += 1) {
 
        // First positive value of the array
        if (A[i] > 0) {
            l = i + 1;
            break;
        }
    }
    for (let 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
 
    let A = [ 1, 0, 0, 1 ];
 
    let N = A.length;
 
    document.write(StrictlyPositiveXor(A, N));
 
</script>

Output: 

3

 

Time complexity: O(N)
 


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