通过删除或附加任何数字来最小化将 N 转换为 K 的幂的操作

给定一个数字N ，任务是找到将给定整数转换为K的任意幂的最小操作数，其中在每个操作中，可以删除任何数字，或者可以将任何数字附加到后面整数。

例子：

Input: N = 247, K = 3
Output: 1
Explanation: In the 1st operation, digit 4 can be removed. Hence, N = 27 which is a power of K.

Input: N = 5, K = 2
Output: 2
Explanation: In the 1st operation, digit 5 can be removed and in the 2nd operation, digit 2 can be appended in the end. Hence, N = 2 which is a power of K.

方法：给定问题可以通过将K的所有幂存储在一个向量中并计算将给定整数N转换为当前幂所需的操作数来解决。以下是要遵循的步骤：

• 将K的所有幂存储在向量幂中。
• 现在遍历向量幂并计算将给定整数N转换为当前幂所需的操作数，可按如下方式完成：
  • 将给定的整数转换为字符串s1和s2 。
  • 初始化两个变量i = 0和j = 0 。
  • 如果s1[i]等于s2[j] ，则增加i和j 。否则，只增加j 。
  • 所需的操作数将为S1.length + S2.length – (2 * i) 。
• 在所有值上保持所需操作的最小值，这是所需的答案。

下面是上述方法的实现：

C++

// C++ code for the above approach
#include 
using namespace std;
#define int long long
 
// Function to find all powers of
// K in the range [1, 10^18]
vector findPowers(int K)
{
    // Stores the powers of K
    vector powers;
    int val = 1;
 
    // Loop to iterate over all
    // powers in the range
    while (val <= 1e18) {
        powers.push_back(val);
        val *= K;
    }
 
    // Return answer
    return powers;
}
 
// Function to find minimum operations to
// convert given integer into a power of K
int minOperations(int N, int K)
{
    // Store all the powers of K
    vector powers = findPowers(K);
 
    // Stores the final result
    int res = INT_MAX;
 
    // Loop to iterate through all
    // the powers of K
    for (int x = 0; x < powers.size(); x++) {
 
        // Convert integers to strings
        // for easier comparison
        string s1 = to_string(powers[x]);
        string s2 = to_string(N);
 
        int i = 0, j = 0;
 
        // Loop to calculate operations
        // required to convert s1 to s2
        while (i < s1.size() && j < s2.size()) {
 
            // Count no of equal character
            if (s1[i] == s2[j]) {
                i++;
            }
 
            // Increment j by 1
            j++;
        }
 
        // Update res
        res = min(res,
                  (int)(s1.size()
                        + s2.size() - 2 * i));
    }
 
    // Return answer;
    return res;
}
 
// Driver Code
int32_t main()
{
    int N = 247;
    int K = 3;
    cout << minOperations(N, K);
 
    return 0;
}

Java

// Java code for the above approach
import java.util.*;
 
class GFG{
 
// Function to find all powers of
// K in the range [1, 10^18]
static Vector findPowers(int K)
{
    // Stores the powers of K
    Vector powers = new Vector();
    int val = 1;
 
    // Loop to iterate over all
    // powers in the range
    while (val <= (int)(9999999999L)) {
        powers.add(val);
        val *= K;
    }
 
    // Return answer
    return powers;
}
 
// Function to find minimum operations to
// convert given integer into a power of K
static int minOperations(int N, int K)
{
    // Store all the powers of K
    Vector powers = findPowers(K);
 
    // Stores the final result
    int res = Integer.MAX_VALUE;
 
    // Loop to iterate through all
    // the powers of K
    for (int x = 0; x < powers.size(); x++) {
 
        // Convert integers to Strings
        // for easier comparison
        String s1 = String.valueOf(powers.get(x));
        String s2 = String.valueOf(N);
 
        int i = 0, j = 0;
 
        // Loop to calculate operations
        // required to convert s1 to s2
        while (i < s1.length() && j < s2.length()) {
 
            // Count no of equal character
            if (s1.charAt(i) == s2.charAt(j)) {
                i++;
            }
 
            // Increment j by 1
            j++;
        }
 
        // Update res
        res = Math.min(res,
                  (int)(s1.length()
                        + s2.length() - 2 * i));
    }
 
    // Return answer;
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 247;
    int K = 3;
    System.out.print(minOperations(N, K));
 
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python 3 code for the above approach
import sys
 
# Function to find all powers of
# K in the range [1, 10^18]
def findPowers(K):
 
    # Stores the powers of K
    powers = []
    val = 1
 
    # Loop to iterate over all
    # powers in the range
    while (val <= 1e18):
        powers.append(val)
        val *= K
 
    # Return answer
    return powers
 
# Function to find minimum operations to
# convert given integer into a power of K
 
 
def minOperations(N,  K):
 
    # Store all the powers of K
    powers = findPowers(K)
 
    # Stores the final result
    res = sys.maxsize
 
    # Loop to iterate through all
    # the powers of K
    for x in range(len(powers)):
 
        # Convert integers to strings
        # for easier comparison
        s1 = str(powers[x])
        s2 = str(N)
 
        i = 0
        j = 0
 
        # Loop to calculate operations
        # required to convert s1 to s2
        while (i < len(s1) and j < len(s2)):
 
            # Count no of equal character
            if (s1[i] == s2[j]):
                i += 1
 
            # Increment j by 1
            j += 1
 
        # Update res
        res = min(res,
                  (int)(len(s1)
                        + len(s2) - 2 * i))
 
    # Return answer;
    return res
 
# Driver Code
if __name__ == "__main__":
 
    N = 247
    K = 3
    print(minOperations(N, K))
 
    # This code is contributed by ukasp.

C#

// C# code for the above approach
using System;
using System.Collections.Generic;
 
public class GFG{
 
  // Function to find all powers of
  // K in the range [1, 10^18]
  static List findPowers(int K)
  {
 
    // Stores the powers of K
    List powers = new List();
    int val = 1;
 
    // Loop to iterate over all
    // powers in the range
    while (val <= ((int)(999999999))) {
      powers.Add(val);
      val *= K;
    }
 
    // Return answer
    return powers;
  }
 
  // Function to find minimum operations to
  // convert given integer into a power of K
  static int minOperations(int N, int K)
  {
 
    // Store all the powers of K
    List powers = findPowers(K);
 
    // Stores the readonly result
    int res = int.MaxValue;
 
    // Loop to iterate through all
    // the powers of K
    for (int x = 0; x < powers.Count; x++) {
 
      // Convert integers to Strings
      // for easier comparison
      String s1 = String.Join("",powers[x]);
      String s2 = String.Join("",N);
 
      int i = 0, j = 0;
 
      // Loop to calculate operations
      // required to convert s1 to s2
      while (i < s1.Length && j < s2.Length) {
 
        // Count no of equal character
        if (s1[i] == s2[j]) {
          i++;
        }
 
        // Increment j by 1
        j++;
      }
 
      // Update res
      res = Math.Min(res,
                     (int)(s1.Length
                           + s2.Length - 2 * i));
    }
 
    // Return answer;
    return res;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int N = 247;
    int K = 3;
    Console.Write(minOperations(N, K));
 
  }
}
 
// This code is contributed by shikhasingrajput

Javascript

// Python 3 code for the above approach
 
// Function to find all powers of
// K in the range [1, 10^18]
function findPowers(K){
 
    // Stores the powers of K
    var powers = []
    var val = 1
 
    // Loop to iterate over all
    // powers in the range
    while (val <= 1e18){
        powers.push(val)
        val *= K
    }
    // Return answer
    return powers
}
 
// Function to find minimum operations to
// convert given integer into a power of K
function minOperations(N,  K){
 
    // Store all the powers of K
    var powers = findPowers(K)
 
    // Stores the final result
    var res = 99999999999;
 
    // Loop to iterate through all
    // the powers of K
    for(var x = 0; x < powers.length;x++){
 
        // Convert integers to strings
        // for easier comparison
        var s1 = (powers[x].toString());
        var s2 = (N.toString());
 
        var i = 0
        var j = 0
 
        // Loop to calculate operations
        // required to convert s1 to s2
        while (i < s1.length && j < s2.length){
 
            // Count no of equal character
            if (s1[i] == s2[j]){
                i += 1
            }
            // Increment j by 1
            j += 1
        }
        // Update res
        res = Math.min(res,
                  (s1.length
                        + s2.length - 2 * i))
      }
 
    // Return answer;
    return res
}
 
// Driver Code
    N = 247
    K = 3
    document.write(minOperations(N, K));
 
// This code is contributed by 29AjayKumar

输出

1

时间复杂度： O((log N) ² )
辅助空间： O(1)