数组的最小大小，MEX 为 A，数组元素的 XOR 为 B

给定两个整数A和B ，任务是找到数组的MEX的最小可能大小 是A和按位异或 所有数组元素中的B 。

例子：

Input: A = 1, B = 1
Output: 3
Explanation: The array which can satisfy the given condition is {0, 3, 2} which is of minimum possible size i.e, 3.

Input: A = 1, B = 999
Output: 2

方法：给定的问题可以通过以下观察来解决：

• 由于数组的 MEX 应为A ，因为必须包含[0, A – 1]范围内的所有元素。
• 为了使所有数组元素的异或为B ，必须在数组中添加几个整数，可以分为 3 种情况。假设变量currXOR在[0, A – 1]范围内存储 XOR 的值，可以使用本文中讨论的方法计算该值。
  • 案例 1，其中currXor = B 。在这种情况下，不需要添加整数。
  • 案例 2，其中currXor^B = A 。在这种情况下，由于A不能添加到数组中，因此必须添加2 个整数，使得它们的XOR为A 。
  • 在所有其他情况下，必须添加1 个整数才能使给定数组的XOR为B 。

因此，上述问题可以通过对从0到A-1的所有数字进行异或并检查上述情况来解决。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the minimum size
// of array with given MEX and XOR
int minimumSizeArr(int A, int B)
{
    int currXor = 0;
 
    // Find the XOR of values from
    // 0 to A-1
    int reminder = (A - 1) % 4;
 
    // If A is a multiple of 4
    if (reminder == 0)
        currXor = A - 1;
 
    // If A % 4 gives remainder 1
    else if (reminder == 1)
        currXor = 1;
 
    // If A % 4 gives remainder 2
    else if (reminder == 2)
        currXor = A;
 
    // Initializing array size by A
    int minSize = A;
 
    // If XOR of all values of array
    // is equal to B
    if (currXor == B)
        return minSize;
 
    // If the required integer to be
    // added is equal to A
    else if (currXor ^ B == A)
        return minSize + 2;
 
    // Else any integer can be added
    else
        return minSize + 1;
}
 
// Driver Code
int main()
{
    int A = 1;
    int B = 999;
    cout << minimumSizeArr(A, B);
 
    return 0;
}

Java

// Java program for the above approach
public class GFG {
     
    // Function to find the minimum size
    // of array with given MEX and XOR
    static int minimumSizeArr(int A, int B)
    {
        int currXor = 0;
     
        // Find the XOR of values from
        // 0 to A-1
        int reminder = (A - 1) % 4;
     
        // If A is a multiple of 4
        if (reminder == 0)
            currXor = A - 1;
     
        // If A % 4 gives remainder 1
        else if (reminder == 1)
            currXor = 1;
     
        // If A % 4 gives remainder 2
        else if (reminder == 2)
            currXor = A;
     
        // Initializing array size by A
        int minSize = A;
     
        // If XOR of all values of array
        // is equal to B
        if (currXor == B)
            return minSize;
     
        // If the required integer to be
        // added is equal to A
        else if ((currXor ^ B) == A)
            return minSize + 2;
     
        // Else any integer can be added
        else
            return minSize + 1;
    }
     
    // Driver Code
    public static void main (String[] args) {
         
        int A = 1;
        int B = 999;
        System.out.println(minimumSizeArr(A, B));       
    }
}
 
// This code is contributed by AnkThon

Python3

# python program for the above approach
 
# Function to find the minimum size
# of array with given MEX and XOR
def minimumSizeArr(A, B):
 
    currXor = 0
 
    # Find the XOR of values from
    # 0 to A-1
    reminder = (A - 1) % 4
 
    # If A is a multiple of 4
    if (reminder == 0):
        currXor = A - 1
 
    # If A % 4 gives remainder 1
    elif (reminder == 1):
        currXor = 1
 
    # If A % 4 gives remainder 2
    elif (reminder == 2):
        currXor = A
 
    # Initializing array size by A
    minSize = A
 
    # If XOR of all values of array
    # is equal to B
    if (currXor == B):
        return minSize
 
    # If the required integer to be
    # added is equal to A
    elif (currXor ^ B == A):
        return minSize + 2
 
    # Else any integer can be added
    else:
        return minSize + 1
 
# Driver Code
if __name__ == "__main__":
 
    A = 1
    B = 999
    print(minimumSizeArr(A, B))
 
# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
public class GFG {
     
    // Function to find the minimum size
    // of array with given MEX and XOR
    static int minimumSizeArr(int A, int B)
    {
        int currXor = 0;
     
        // Find the XOR of values from
        // 0 to A-1
        int reminder = (A - 1) % 4;
     
        // If A is a multiple of 4
        if (reminder == 0)
            currXor = A - 1;
     
        // If A % 4 gives remainder 1
        else if (reminder == 1)
            currXor = 1;
     
        // If A % 4 gives remainder 2
        else if (reminder == 2)
            currXor = A;
     
        // Initializing array size by A
        int minSize = A;
     
        // If XOR of all values of array
        // is equal to B
        if (currXor == B)
            return minSize;
     
        // If the required integer to be
        // added is equal to A
        else if ((currXor ^ B) == A)
            return minSize + 2;
     
        // Else any integer can be added
        else
            return minSize + 1;
    }
     
    // Driver Code
    public static void Main () {
         
        int A = 1;
        int B = 999;
        Console.WriteLine(minimumSizeArr(A, B));       
    }
}
 
// This code is contributed by ihritik

Javascript

// Javascript program for the above approach
 
// Function to find the minimum size
// of array with given MEX and XOR
function minimumSizeArr(A, B)
{
  let currXor = 0;
 
  // Find the XOR of values from
  // 0 to A-1
  let reminder = (A - 1) % 4;
 
  // If A is a multiple of 4
  if (reminder == 0) currXor = A - 1;
   
  // If A % 4 gives remainder 1
  else if (reminder == 1) currXor = 1;
   
  // If A % 4 gives remainder 2
  else if (reminder == 2) currXor = A;
 
  // Initializing array size by A
  let minSize = A;
 
  // If XOR of all values of array
  // is equal to B
  if (currXor == B) return minSize;
  // If the required integer to be
  // added is equal to A
  else if (currXor ^ (B == A)) return minSize + 2;
  // Else any integer can be added
  else return minSize + 1;
}
 
// Driver Code
 
let A = 1;
let B = 999;
document.write(minimumSizeArr(A, B));
 
// This code is contributed by saurabh_jaiswal.

输出：

2

时间复杂度： O(1)
辅助空间： O(1)