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-1and 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)
Auxiliary Space: O(1)