通过删除或附加任何数字来最小化将 N 更改为 2 的幂的步骤

给定一个整数N ，任务是使用以下步骤找到将数字N更改为2的完美幂所需的最小步骤数：

• 从数字中删除任何一位数字d 。
• 在数字N的末尾添加任何数字d 。

例子：

Input: N = 1092
Output: 2
Explanation:
Following are the steps followed:

1. Removing digit 9 from the number N(= 1092) modifies it to 102.
2. Adding digit 4 to the end of number N(= 102) modifies it to 1024.

After the above operations, the number N is converted into perfect power of 2 and the number of moves required is 2, which is the minimum. Therefore, print 2.

Input: N = 4444
Output: 3

方法：给定的问题可以使用贪心方法来解决，这个想法是存储所有 2 的幂且小于10 ²⁰的数字，并检查每个数字将给定数字转换为 20 的最小步骤2.按照以下步骤解决问题：

• 初始化一个数组并存储所有2的幂且小于10 ²⁰的数字。
• 将答案变量最好初始化为长度 + 1作为所需的最大步数。
• 使用变量x遍历[0, len)范围，其中len是数组的长度，并执行以下步骤：
  • 初始化变量，例如position为0 。
  • 迭代范围[0, len) ，其中len是使用变量i的数字的长度，如果位置小于len(x)并且x[position]等于num[i] ，则增加一个位置1 。
  • 将best的值更新为best或len(x) + len(num) – 2*position的最小值。
• 执行上述步骤后，打印best的值作为结果。

下面是上述方法的实现：

Java

// Java Program to implement
// the above approach
import java.util.*;
 
class GFG {
 
  // Function to find the minimum number of
  // steps required to reduce N to perfect
  // power of 2 by performing given steps
  static void findMinimumSteps(int N) {
    String num = String.valueOf(N);
    int c = 1;
 
    Stack a = new Stack<>();
 
    // Stores all the perfect power of 2
    while (true) {
      if (c > (int)Math.pow(10, 20)|| c>10 ) {
        break;
      }
      a.add(String.valueOf(c));
      c = c * 2;
    }
 
    // Maximum number of steps required
    int best = num.length()+1;
 
    // Iterate for each perfect power of 2
    String x;
    int i = 0;
    while (!a.isEmpty()) {
      x = a.pop();
      int position = 0;
 
      // Comparing with all numbers
      for (i = 0; i < num.length(); i++) {
 
        if (position < x.length() && x.charAt(position) == num.charAt(i))
          position += 1;
      }
 
      // Update the minimum number of
      // steps required
      best = (int) (Math.min(best, x.length() + num.length() - 2 * position));
 
    }
 
    // Print the result
    System.out.print(best-1);
 
  }
 
  // Driver Code
  public static void main(String[] args) {
    int N = 1092;
 
    // Function Call
    findMinimumSteps(N);
  }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python program for the above approach
 
# Function to find the minimum number of
# steps required to reduce N to perfect
# power of 2 by performing given steps
def findMinimumSteps(N):
 
    num = str(N)
    c = 1
    a = []
 
    # Stores all the perfect power of 2
    while True:
        if (c > 10 ** 20):
            break
        a.append(str(c))
        c = c * 2
 
    # Maximum number of steps required
    best = len(num) + 1
     
    # Iterate for each perfect power of 2
    for x in a:
        position = 0
         
        # Comparing with all numbers
        for i in range(len(num)):
           
            if position < len(x) and x[position] == num[i]:
                position += 1
                 
        # Update the minimum number of
        # steps required
        best = min(best, len(x) + len(num) - 2 * position)
         
    # Print the result
    print(best)
 
# Driver Code
N = 1092
 
# Function Call
findMinimumSteps(N)

Javascript

 
        // JavaScript Program to implement
        // the above approach
 
        // Function to find the minimum number of
        // steps required to reduce N to perfect
        // power of 2 by performing given steps
        function findMinimumSteps(N)
        {
            num = N.toString()
            c = 1
            a = []
 
            // Stores all the perfect power of 2
            while (1) {
                if (c > Math.pow(10, 20)) {
                    break;
                }
                a.push(c.toString())
                c = c * 2
            }
             
            // Maximum number of steps required
            best = num.length + 1
 
            // Iterate for each perfect power of 2
            for (x of a) {
                position = 0
 
                // Comparing with all numbers
                for (let i = 0; i < num.length; i++) {
 
                    if (position < x.length && x[position] == num[i])
                        position += 1
                }
                 
                // Update the minimum number of
                // steps required
                best = Math.min(best, x.length + num.length - 2 * position)
            }
             
            // Print the result
            document.write(best)
        }
         
        // Driver Code
        let N = 1092
 
        // Function Call
        findMinimumSteps(N)
 
    // This code is contributed by Potta Lokesh
    

输出：

2

时间复杂度： O(N)
辅助空间： O(1)