最小化将 N 减少到 0 所需的给定翻转

给定一个整数N ，任务是通过执行以下操作的最少次数将N的值减少到0 ：

• 翻转N的二进制表示中最右边(^第0^位)的位。
• 如果(I – 1)^位被设置，然后翻转^第i位和选自(i – 2)清除所有位到^第0^个比特。

例子：

Input: N = 3 
Output: 2 
Explanation: 
The binary representation of N (= 3) is 11 
Since 0^th bit in binary representation of N(= 3) is set, flipping the 1^st bit of binary representation of N modifies N to 1(01). 
Flipping the rightmost bit of binary representation of N(=1) modifies N to 0(00). 
Therefore, the required output is 2

Input: N = 4 
Output: 7

方法：该问题可以基于以下观察来解决：

1 -> 0 => 1 
10 -> 11 -> 01 -> 00 => 2 + 1 = 3 
100 -> 101 -> 111 -> 110 -> 010 -> … => 4 + 2 + 1 = 7 
1000 -> 1001 -> 1011 -> 1010 -> 1110 -> 1111 -> 1101 -> 1100 -> 0100 -> … => 8 + 7 = 15 
Therefore, for N = 2^N total (2^{(N + 1)} – 1) operations required.
If N is not a power of 2, then the recurrence relation is: 
MinOp(N) = MinOp((1 << cntBit) – 1) – MinOp(N – (1 << (cntBit – 1)))
cntBit = total count of bits in binary representation of N. 
MinOp(N) denotes minimum count of operations required to reduce N to 0. 
 

请按照以下步骤解决问题：

• 使用log ₂ (N) + 1计算N 的二进制表示中的位数。
• 使用上述递推关系并计算将N减少到0所需的最小操作数。

下面是上述方法的实现。

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to find the minimum count of
// operations required to Reduce N to 0
int MinOp(int N)
{
 
    if (N <= 1)
        return N;
 
    // Stores count of
    // bits in N
    int bit = log2(N) + 1;
 
    // Recurrence relation
    return ((1 << bit) - 1)
           - MinOp(N - (1 << (bit - 1)));
}
 
// Driver Code
int main()
{
 
    int N = 4;
    cout << MinOp(N);
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{
     
// Function to find the minimum count of
// operations required to Reduce N to 0
public static int MinOp(int N)
{
    if (N <= 1)
        return N;
   
    // Stores count of
    // bits in N
    int bit = (int)(Math.log(N) /
                    Math.log(2)) + 1;
   
    // Recurrence relation
    return ((1 << bit) - 1) - MinOp(
        N - (1 << (bit - 1)));
}
 
// Driver code
public static void main(String[] args)
{
    int N = 4;
     
    System.out.println(MinOp(N));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python program to implement
# the above approach
 
# Function to find the minimum count of
# operations required to Reduce N to 0
import math
def MinOp(N):
    if (N <= 1):
        return N;
 
    # Stores count of
    # bits in N
    bit = (int)(math.log(N) / math.log(2)) + 1;
 
    # Recurrence relation
    return ((1 << bit) - 1) - MinOp(N - (1 << (bit - 1)));
 
# Driver code
if __name__ == '__main__':
    N = 4;
 
    print(MinOp(N));
 
# This code is contributed by 29AjayKumar

C#

// C# program to implement
// the above approach 
using System;
 
class GFG{
     
// Function to find the minimum count of
// operations required to Reduce N to 0
public static int MinOp(int N)
{
    if (N <= 1)
        return N;
         
    // Stores count of
    // bits in N
    int bit = (int)(Math.Log(N) /
                    Math.Log(2)) + 1;
                     
    // Recurrence relation
    return ((1 << bit) - 1) - MinOp(
        N - (1 << (bit - 1)));
}
  
// Driver code
public static void Main()
{
    int N = 4;
     
    Console.WriteLine(MinOp(N));
}
}
 
// This code is contributed by sanjoy_62

Javascript

// javascript program to implement
// the above approach
 
// Function to find the minimum count of
// operations required to Reduce N to 0
function MinOp(N)
{
    if (N <= 1)
        return N;
    
    // Stores count of
    // bits in N
    let bit = (Math.log(N) /
                    Math.log(2)) + 1;
    
    // Recurrence relation
    return ((1 << bit) - 1) - MinOp(
        N - (1 << (bit - 1)));
}
 
// Driver code
    let N = 4;
    document.write(MinOp(N));
     
    // This code is contributed by souravghosh0416.

输出：

7

时间复杂度： O(log(N))
辅助空间： O(1)

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