给定大小为N的数组arr [] ,任务是找到给定数组所有可能的无序对的按位XOR。
例子:
Input: arr[] = {1, 5, 3, 7}
Output: 0
Explanation:
All possible unordered pairs are (1, 5), (1, 3), (1, 7), (5, 3), (5, 7), (3, 7)
Bitwise XOR of all possible pairs are = 1 ^ 5 ^ 1 ^3 ^ 1 ^ 7 ^ 5 ^ 3 ^ 5^ 7 ^ 3 ^ 7 = 0
Therefore, the required output is 0.
Input: arr[] = {1, 2, 3, 4}
Output: 4
天真的方法:想法是遍历数组并生成给定数组的所有可能的对。最后,打印给定数组的这些对中存在的每个元素的按位XOR。请按照以下步骤解决问题:
- 初始化一个变量,例如totalXOR ,以存储这些对中每个元素的按位XOR。
- 遍历给定数组,并从给定数组生成所有可能的对(arr [i],arr [j]) 。
- 对于每对(arr [i],arr [j]) ,更新totalXOR =(totalXOR ^ arr [i] ^ arr [j])的值。
- 最后,打印totalXOR的值。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to get bitwise XOR
// of all possible pairs of
// the given array
int TotalXorPair(int arr[], int N)
{
// Stores bitwise XOR
// of all possible pairs
int totalXOR = 0;
// Generate all possible pairs
// and calculate bitwise XOR
// of all possible pairs
for (int i = 0; i < N; i++) {
for (int j = i + 1; j < N;
j++) {
// Calculate bitwise XOR
// of each pair
totalXOR ^= arr[i]
^ arr[j];
}
}
return totalXOR;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << TotalXorPair(arr, N);
}
Java
// Java program to implement
// the above approach
class GFG{
// Function to get bitwise XOR
// of all possible pairs of
// the given array
public static int TotalXorPair(int arr[],
int N)
{
// Stores bitwise XOR
// of all possible pairs
int totalXOR = 0;
// Generate all possible pairs
// and calculate bitwise XOR
// of all possible pairs
for (int i = 0; i < N; i++)
{
for (int j = i + 1; j < N; j++)
{
// Calculate bitwise XOR
// of each pair
totalXOR ^= arr[i] ^ arr[j];
}
}
return totalXOR;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {1, 2, 3, 4};
int N = arr.length;
System.out.print(TotalXorPair(arr, N));
}
}
// This code is contributed by divyeshrabadiya07
Python3
# Python3 program to implement
# the above approach
# Function to get bitwise XOR
# of all possible pairs of
# the given array
def TotalXorPair(arr, N):
# Stores bitwise XOR
# of all possible pairs
totalXOR = 0;
# Generate all possible pairs
# and calculate bitwise XOR
# of all possible pairs
for i in range(0, N):
for j in range(i + 1, N):
# Calculate bitwise XOR
# of each pair
totalXOR ^= arr[i] ^ arr[j];
return totalXOR;
# Driver code
if __name__ == '__main__':
arr = [1, 2, 3, 4];
N = len(arr);
print(TotalXorPair(arr, N));
# This code is contributed by shikhasingrajput
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to get bitwise XOR
// of all possible pairs of
// the given array
public static int TotalXorPair(int []arr,
int N)
{
// Stores bitwise XOR
// of all possible pairs
int totalXOR = 0;
// Generate all possible pairs
// and calculate bitwise XOR
// of all possible pairs
for(int i = 0; i < N; i++)
{
for(int j = i + 1; j < N; j++)
{
// Calculate bitwise XOR
// of each pair
totalXOR ^= arr[i] ^ arr[j];
}
}
return totalXOR;
}
// Driver code
public static void Main(String[] args)
{
int []arr = {1, 2, 3, 4};
int N = arr.Length;
Console.Write(TotalXorPair(arr, N));
}
}
// This code is contributed by Princi Singh
Javascript
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to get bitwise XOR
// of all possible pairs of
// the given array
int TotalXorPair(int arr[], int N)
{
// Stores bitwise XOR
// of all possible pairs
int totalXOR = 0;
// Check if N is odd
if (N % 2 != 0) {
return 0;
}
// If N is even then calculate
// bitwise XOR of all elements
// of the given array.
for (int i = 0; i < N; i++) {
totalXOR ^= arr[i];
}
return totalXOR;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << TotalXorPair(arr, N);
}
Java
// Java program to implement
// the above approach
class GFG{
// Function to get bitwise XOR
// of all possible pairs of
// the given array
public static int TotalXorPair(int arr[],
int N)
{
// Stores bitwise XOR
// of all possible pairs
int totalXOR = 0;
// Check if N is odd
if( N % 2 != 0 )
{
return 0;
}
// If N is even then calculate
// bitwise XOR of all elements
// of the array
for(int i = 0; i < N; i++)
{
totalXOR ^= arr[i];
}
return totalXOR;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {1, 2, 3, 4};
int N = arr.length;
System.out.print(TotalXorPair(arr, N));
}
}
// This code is contributed by math_lover
Python3
# Python3 program to implement
# the above appraoch
# Function to get bitwise XOR
# of all possible pairs of
# the given array
def TotalXorPair(arr, N):
# Stores bitwise XOR
# of all possible pairs
totalXOR = 0
# Check if N is odd
if (N % 2 != 0):
return 0
# If N is even then calculate
# bitwise XOR of all elements
# of the given array.
for i in range(N):
totalXOR ^= arr[i]
return totalXOR
# Driver code
if __name__ == '__main__':
arr = [ 1, 2, 3, 4 ]
N = len(arr)
print(TotalXorPair(arr, N))
# This code is contributed by Shivam Singh
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to get bitwise XOR
// of all possible pairs of
// the given array
static int TotalXorPair(int []arr,
int N)
{
// Stores bitwise XOR
// of all possible pairs
int totalXOR = 0;
// Check if N is odd
if (N % 2 != 0)
{
return 0;
}
// If N is even then calculate
// bitwise XOR of all elements
// of the array
for(int i = 0; i < N; i++)
{
totalXOR ^= arr[i];
}
return totalXOR;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4 };
int N = arr.Length;
Console.Write(TotalXorPair(arr, N));
}
}
// This code is contributed by doreamon_
Javascript
4
时间复杂度: O(N 2 )
辅助空间: O(1)
高效方法:要优化上述方法,请遵循以下观察结果:
Property of Bitwise XOR:
a ^ a ^ a …….( Even times ) = 0
a ^ a ^ a …….( Odd times ) = a
Each element of the given array occurs exactly (N – 1) times in all possible pairs.
Therefore, if N is even, then Bitwise XOR of all possible pairs are equal to bitwise XOR of all the array elements.
Otherwise, bitwise XOR of all possible pairs are equal to 0.
请按照以下步骤解决问题:
- 如果N为奇数,则打印0 。
- 如果N是偶数,则打印给定数组的所有元素的按位XOR值。
下面是上述方法的实现
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to get bitwise XOR
// of all possible pairs of
// the given array
int TotalXorPair(int arr[], int N)
{
// Stores bitwise XOR
// of all possible pairs
int totalXOR = 0;
// Check if N is odd
if (N % 2 != 0) {
return 0;
}
// If N is even then calculate
// bitwise XOR of all elements
// of the given array.
for (int i = 0; i < N; i++) {
totalXOR ^= arr[i];
}
return totalXOR;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << TotalXorPair(arr, N);
}
Java
// Java program to implement
// the above approach
class GFG{
// Function to get bitwise XOR
// of all possible pairs of
// the given array
public static int TotalXorPair(int arr[],
int N)
{
// Stores bitwise XOR
// of all possible pairs
int totalXOR = 0;
// Check if N is odd
if( N % 2 != 0 )
{
return 0;
}
// If N is even then calculate
// bitwise XOR of all elements
// of the array
for(int i = 0; i < N; i++)
{
totalXOR ^= arr[i];
}
return totalXOR;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {1, 2, 3, 4};
int N = arr.length;
System.out.print(TotalXorPair(arr, N));
}
}
// This code is contributed by math_lover
Python3
# Python3 program to implement
# the above appraoch
# Function to get bitwise XOR
# of all possible pairs of
# the given array
def TotalXorPair(arr, N):
# Stores bitwise XOR
# of all possible pairs
totalXOR = 0
# Check if N is odd
if (N % 2 != 0):
return 0
# If N is even then calculate
# bitwise XOR of all elements
# of the given array.
for i in range(N):
totalXOR ^= arr[i]
return totalXOR
# Driver code
if __name__ == '__main__':
arr = [ 1, 2, 3, 4 ]
N = len(arr)
print(TotalXorPair(arr, N))
# This code is contributed by Shivam Singh
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to get bitwise XOR
// of all possible pairs of
// the given array
static int TotalXorPair(int []arr,
int N)
{
// Stores bitwise XOR
// of all possible pairs
int totalXOR = 0;
// Check if N is odd
if (N % 2 != 0)
{
return 0;
}
// If N is even then calculate
// bitwise XOR of all elements
// of the array
for(int i = 0; i < N; i++)
{
totalXOR ^= arr[i];
}
return totalXOR;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4 };
int N = arr.Length;
Console.Write(TotalXorPair(arr, N));
}
}
// This code is contributed by doreamon_
Java脚本
4
时间复杂度: O(N)
辅助空间: O(1)