将 N 减少到 0 所需执行的给定操作的最小数量

给定一个整数N ，任务是使用以下操作任意次数将N减少到 0 的最小操作次数：

更改N的二进制表示中最右边的 (0 ^th ) 位。
如果第( ⁱ -1)位设置为 1，并且第( ⁱ -2)到第⁰位设置为 0，则更改N的二进制表示中的第ⁱ位。

例子：

Input: N = 3
Output: 2
Explanation: The binary representation of 3 is “11”.
“11″ → “01” with the 2nd operation since the 0th bit is 1.
“01″ → “00″ with the 1st operation.

Input: N = 6
Output: 4
Explanation: The binary representation of 6 is “110”.
“110” → “010” with the 2nd operation since the 1st bit is 1 and 0th through 0th bits are 0.
“010” → “011” with the 1st operation.
“011” → “001” with the 2nd operation since the 0th bit is 1.
“001” → “000” with the 1st operation.

编程需要懂一点英语

方法：这个想法基于以下观察：

1 → 0 需要 1 次操作。
2 → 0 即10 → 11 → 01 → 0 需要 3 次操作。
4 → 0 即100 → 101 → 111 → 110 → 010 → 11 → 01 →0 需要 7 次操作。
因此，它可以推广到任何 2 的幂，即 2 ^k需要 2 ^(k+1) -1 次操作。
如果a→b需要k次操作，b→a也需要k次操作。

从 2 ⁿ到 2 ^{(n-1) 的}中间数包含 2 ⁿ到 2 (n- ¹ ) 之间的所有数字，这表明任何给定的非负整数都可以转换为 0。设 f(n) 是一个函数找到所需的最小操作数。递归关系为：
f(n) = f(2 ^k ) – f(n xor 2 ^k )，其中 k = n 的最高有效位的下一位。

这个想法是使用递归。请按照以下步骤解决问题：

创建一个以数字N作为参数的递归函数。
- 如果N的值等于0 ，则返回0 。
- 否则，找到小于或等于N的 2 的最大幂并将其存储在变量X中。
- 将通过递归调用(X^(X/2)^N)的函数返回的值存储在变量S中。
- 将X的值添加到S并返回。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find minimum
// operations required to convert
// N to 0
int minimumOneBitOperations(int n, int res = 0)
{
    // Base Case
    if (n == 0)
        return res;
 
    // Store the highest power of 2
    // less than or equal to n
    int b = 1;
    while ((b << 1) <= n)
        b = b << 1;
 
    // Return the result
    return minimumOneBitOperations(
        (b >> 1) ^ b ^ n, res + b);
}
 
// Driver Code
int main()
{
    // Given Input
    int N = 6;
 
    // Function call
    cout << minimumOneBitOperations(N);
 
    return 0;
}

Java

// Java program for the above approach
class GFG {
 
    // Function to find minimum
    // operations required to convert
    // N to 0
    public static int minimumOneBitOperations(int n, int res)
    {
       
        // Base Case
        if (n == 0)
            return res;
 
        // Store the highest power of 2
        // less than or equal to n
        int b = 1;
        while ((b << 1) <= n)
            b = b << 1;
 
        // Return the result
        return minimumOneBitOperations((b >> 1) ^ b ^ n, res + b);
    }
 
    // Driver Code
    public static void main(String args[])
    {
       
        // Given Input
        int N = 6;
 
        // Function call
        System.out.println(minimumOneBitOperations(N, 0));
    }
}
 
// This code is contributed by gfgking.

Python3

# Python program for the above approach
 
# Function to find minimum
# operations required to convert
# N to 0
def minimumOneBitOperations(n, res):
   
  # Base case
    if n == 0:
        return res
       
     # Store the highest power of 2
    # less than or equal to n 
    b = 1
    while (b << 1) <= n:
        b = b << 1
         
    # Return the result
    return minimumOneBitOperations((b >> 1) ^ b ^ n, res + b)
 
# Driver code
N = 6
print(minimumOneBitOperations(N, 0))
 
# This code is contributed by Parth Manchanda

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find minimum
// operations required to convert
// N to 0
static int minimumOneBitOperations(int n, int res)
{
    // Base Case
    if (n == 0)
        return res;
 
    // Store the highest power of 2
    // less than or equal to n
    int b = 1;
    while ((b << 1) <= n)
        b = b << 1;
 
    // Return the result
    return minimumOneBitOperations(
        (b >> 1) ^ b ^ n, res + b);
}
 
// Driver Code
public static void Main()
{
    // Given Input
    int N = 6;
 
    // Function call
    Console.Write(minimumOneBitOperations(N,0));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

输出：

时间复杂度： O(log(N))
辅助空间： O(1)