给定范围内 x 的计数，使得 x、(x+1) 和 (x+2)、(x+3) 的按位异或相等

给定一个整数N ，任务是计算[0, 2 ^N -1]范围内的整数个数（比如x ），使得x⊕(x+1) = (x+2)⊕(x+3) . [其中⊕代表按位异或]

例子：

Input: N = 1
Output: 1
Explanation: Only 0 is the valid x, as, 0 ⊕ 1 = 2 ⊕ 3 = 1

Input: N = 3
Output: 4

朴素方法：解决给定问题的简单方法是生成 x 在[0, 2 ^N -1]范围内的所有可能值，并检查它们是否满足给定标准，即x⊕(x+1) = (x+ 2)⊕(x+3) 。

按照下面提到的步骤来实现这个想法：

• 从i = 0 迭代到 N 。
  • 检查(i ⊕ (i+1)) = ((i+2) ⊕ (i+3))
  • 如果满足条件，则增加此类数字的计数。
• 返回最终计数作为答案。

下面是上述方法的实现：

Python3

# Python3 code to implement the approach
  
# Function to find number of integers 
# satisfying the condition
def countOfvalues(N):
          
    # Stores the resultant count of pairs
    count = 0
  
    for i in range(0, 2**N):
  
       # Iterate over the range
        if i ^ (i + 1) == (i + 2) ^ (i + 3):
            count += 1
  
    return count
  
  
# Driver Code
if __name__ == '__main__':
    N = 3
      
    # Function call
    print(countOfvalues(N))

输出

4

时间复杂度： O(2 ^N )
辅助空间：O(1)

Efficient Approach：这个问题可以基于以下数学思想高效解决：

• If x is such that it is a even number then x+2 is also a even number and both (x+1) and (x+3) will be odd numbers. 
• Now two consecutive even and odd number has only a single bit difference only on their LSB position.
• So the bitwise XOR of (x and x+1) and (x+2 and x+3) both will be 1, when x is even.
• Therefore all the even numbers in the given range [0, 2^N−1] is a possible value of x.

So total number of such values are (2^N – 1)/2 = 2^N-1

按照下图更好地可视化这个想法：

插图：

Consider N = 3. So the numbers are in range [0, 7]

All even numbers in the range are 0, 2, 4 and 6

=> When x = 0: 0 ⊕ 1 = 1 and 2 ⊕ 3 = 1. Therefore the relation holds
=> When x = 2: 2 ⊕ 3 = 1 and 4 ⊕ 5 = 1. Therefore the relation holds
=> When x = 4. 4 ⊕ 5 = 1 and 6 ⊕ 7 = 1. Therefore the relation holds
=> When x = 6: 6 ⊕ 7 = 1 and 8 ⊕ 9 = 1. Therefore the relation holds.

Now for the odd numbers if it is done:

=> When x = 1: 1 ⊕ 2 = 3 and 3 ⊕ 4 = 7. Therefore the relation does not hold.
=> When x = 3: 3 ⊕ 4 = 7 and 5 ⊕ 6 = 3. Therefore the relation does not hold.
=> When x = 5: 5 ⊕ 6 = 3 and 7 ⊕ 8 = 15. Therefore the relation does not hold.
=> When x = 7: 7 ⊕ 8 = 15 and 9 ⊕ 10 = 3. Therefore the relation does not hold.

So total possible values are 4 = 2^3-1

因此得到答案计算2 ^N-1的值 对于给定的N

下面是上述方法的实现。

Python3

# Python code to implement the above approach
  
# Function to calculate 
# the number of possible values
def countOfValues (N):
    return pow(2, N - 1)
  
# Driver code
if __name__ == '__main__':
    N = 3
      
    # Function call
    res = countOfValues(N)
    print(res)

输出

4

时间复杂度：O(1)
辅助空间：O(1)