给定N个非负整数的数组arr [] 。任务是找到最长子数组的长度,以使该子数组所有元素的XOR严格为正。如果不存在这样的子数组,则打印-1
例子:
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
方法:
- 如果完整数组的XOR为正,则答案等于N。
- 如果所有元素均为零,则答案为-1 ,因为不可能获得严格的正XOR。
- 否则,假设第一个正数的索引为l ,最后一个正数的索引为r 。
- 现在,索引范围[l,r]的所有元素的XOR必须为零,因为l之前和r之后的元素均为0 ,这不会对XOR值起作用,并且原始数组的XOR为0 。
- 考虑子阵列A 1 ,A 1 ,…,A r-1和A l + 1 ,A l + 2 ,…,A N。
- 第一个子数组的XOR值等于A [r] ,第二个子数组的XOR值A [l]为正。
- 返回这两个子数组中较大的子数组的长度。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
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
输出:
3
时间复杂度: O(N)