通过在每一步加 1 或加倍来最小化从 0 到 K 的步骤

给定一个正整数K ，任务是找到以下两种类型的最小操作次数，需要将 0 更改为 K：

• 向操作数加一
• 将操作数乘以 2。

例子：

Input: K = 1 
Output: 1 
Explanation: 
Step 1: 0 + 1 = 1 = K
Input: K = 4 
Output: 3 
Explanation: 
Step 1: 0 + 1 = 1, 
Step 2: 1 * 2 = 2, 
Step 3: 2 * 2 = 4 = K 
 

方法：

• 如果K是奇数，那么最后一步必须是给它加 1。
• 如果K是偶数，最后一步是乘以 2 以最小化步数。
• 创建一个 dp[] 表，该表存储在每个dp[i] 中，即达到i所需的最小步骤。

$dp[i] =\begin{cases} dp[i - 1] + 1 & \text{; if i is odd} \\ dp[\frac{i}{2}] + 1 & \text{; if i is even} \end{cases}$
 

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 minimum operations
int minOperation(int k)
{
    // vector dp is initialised
    // to store the steps
    vector dp(k + 1, 0);
 
    for (int i = 1; i <= k; i++) {
 
        dp[i] = dp[i - 1] + 1;
 
        // For all even numbers
        if (i % 2 == 0) {
            dp[i]
                = min(dp[i],
                      dp[i / 2] + 1);
        }
    }
    return dp[k];
}
 
// Driver Code
int main()
{
    int K = 12;
    cout << minOperation(k);
}

Java

// Java program to implement the above approach
class GFG{
     
// Function to find minimum operations
static int minOperation(int k)
{
     
    // dp is initialised
    // to store the steps
    int dp[] = new int[k + 1];
 
    for(int i = 1; i <= k; i++)
    {
       dp[i] = dp[i - 1] + 1;
        
       // For all even numbers
       if (i % 2 == 0)
       {
           dp[i] = Math.min(dp[i], dp[i / 2] + 1);
       }
    }
    return dp[k];
}
 
// Driver Code
public static void main (String []args)
{
    int K = 12;
    System.out.print( minOperation(K));
}
}
 
// This code is contributed by chitranayal

Python3

# Python3 program to implement the above approach
 
# Function to find minimum operations
def minOperation(k):
     
    # dp is initialised
    # to store the steps
    dp = [0] * (k + 1)
 
    for i in range(1, k + 1):
        dp[i] = dp[i - 1] + 1
 
        # For all even numbers
        if (i % 2 == 0):
            dp[i]= min(dp[i], dp[i // 2] + 1)
 
    return dp[k]
 
# Driver Code
if __name__ == '__main__':
     
    k = 12
     
    print(minOperation(k))
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement the above approach
using System;
class GFG{
     
// Function to find minimum operations
static int minOperation(int k)
{
     
    // dp is initialised
    // to store the steps
    int []dp = new int[k + 1];
 
    for(int i = 1; i <= k; i++)
    {
        dp[i] = dp[i - 1] + 1;
             
        // For all even numbers
        if (i % 2 == 0)
        {
            dp[i] = Math.Min(dp[i], dp[i / 2] + 1);
        }
    }
    return dp[k];
}
 
// Driver Code
public static void Main()
{
    int K = 12;
    Console.Write(minOperation(K));
}
}
 
// This code is contributed by Nidhi_Biet

Javascript

// Javascript implementation of the above approach
 
// Function to find minimum operations
function minOperation(k)
{
       
    // dp is initialised
    // to store the steps
    let dp = Array.from({length: k+1}, (_, i) => 0);
   
    for(let i = 1; i <= k; i++)
    {
       dp[i] = dp[i - 1] + 1;
          
       // For all even numbers
       if (i % 2 == 0)
       {
           dp[i] = Math.min(dp[i], dp[i / 2] + 1);
       }
    }
    return dp[k];
}
 
  // Driver Code   
    let K = 12;
    document.write( minOperation(K));
  
 // This code is contributed by target_2.

输出：

5