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++
// 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
// 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
# 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#
// 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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