最小清除数量为10的幂次幂可除以K

给定两个正整数N和K。找到可以从数字N中删除的最小位数，这样在删除之后该数字可以被10 ^K整除，或者如果不可能，则打印-1。

例子：

Input : N = 10904025, K = 2
Output : 3
Explanation : We can remove the digits 4, 2 and 5 such that the number 
becomes 10900 which is divisible by 102.

Input : N = 1000, K = 5
Output : 3
Explanation : We can remove the digits 1 and any two zeroes such that the
number becomes 0 which is divisible by 105

Input : N = 23985, K = 2
Output : -1

方法：这个想法是在保留一个计数器的同时开始从最后一位开始遍历数字。如果当前数字不为零，则递增计数器变量，否则递减变量K。当K变为零时，返回计数器作为答案。遍历整数后，检查K的当前值是否为零。如果为零，则将计数器作为答案返回，否则将答案作为N – 1中的位数返回，因为我们需要将整数减小为单个零，该整数可以被任何数字整除。另外，如果给定的数字不包含任何零，则返回-1作为答案。

下面是上述方法的实现。

C++

// CPP Program to count the number
// of digits that can be removed such
// that number is divisible by 10^K
#include 
using namespace std;
  
// function to return the required
// number of digits to be removed
int countDigitsToBeRemoved(int N, int K)
{
    // Converting the given number
    // into string
    string s = to_string(N);
  
    // variable to store number of
    // digits to be removed
    int res = 0;
  
    // variable to denote if atleast
    // one zero has been found
    int f_zero = 0;
    for (int i = s.size() - 1; i >= 0; i--) {
        if (K == 0)
            return res;
        if (s[i] == '0') {
  
            // zero found
            f_zero = 1;
            K--;
        }
        else
            res++;
    }
  
    // return size - 1 if K is not zero and
    // atleast one zero is present, otherwise
    // result
    if (!K)
        return res;
    else if (f_zero)
        return s.size() - 1;
    return -1;
}
  
// Driver Code to test above function
int main()
{
    int N = 10904025, K = 2;
    cout << countDigitsToBeRemoved(N, K) << endl;
  
    N = 1000, K = 5;
    cout << countDigitsToBeRemoved(N, K) << endl;
  
    N = 23985, K = 2;
    cout << countDigitsToBeRemoved(N, K) << endl;
    return 0;
}

Java

// Java Program to count the number 
// of digits that can be removed such 
// that number is divisible by 10^K 
  
public class GFG{
      
    // function to return the required 
    // number of digits to be removed 
    static int countDigitsToBeRemoved(int N, int K) 
    { 
        // Converting the given number 
        // into string 
        String s = Integer.toString(N); 
      
        // variable to store number of 
        // digits to be removed 
        int res = 0; 
      
        // variable to denote if atleast 
        // one zero has been found 
        int f_zero = 0; 
        for (int i = s.length() - 1; i >= 0; i--) { 
            if (K == 0) 
                return res; 
            if (s.charAt(i) == '0') { 
      
                // zero found 
                f_zero = 1; 
                K--; 
            } 
            else
                res++; 
        } 
      
        // return size - 1 if K is not zero and 
        // atleast one zero is present, otherwise 
        // result 
        if (K == 0) 
            return res; 
        else if (f_zero == 1) 
            return s.length() - 1; 
        return -1; 
    } 
      
    // Driver Code to test above function 
    public static void main(String []args)
    { 
        int N = 10904025;
        int K = 2; 
        System.out.println(countDigitsToBeRemoved(N, K)) ; 
      
        N = 1000 ;
        K = 5; 
        System.out.println(countDigitsToBeRemoved(N, K))  ;
      
        N = 23985;
        K = 2; 
        System.out.println(countDigitsToBeRemoved(N, K)) ; 
    } 
  
    // This code is contributed by Ryuga
    }

Python3

# Python3 Program to count the number
# of digits that can be removed such
# that number is divisible by 10^K
  
# function to return the required
# number of digits to be removed
def countDigitsToBeRemoved(N, K):
  
    # Converting the given number
    # into string
    s = str(N);
  
    # variable to store number of
    # digits to be removed
    res = 0;
  
    # variable to denote if atleast
    # one zero has been found
    f_zero = 0;
    for i in range(len(s) - 1, -1, -1):
        if (K == 0):
            return res;
        if (s[i] == '0'):
  
            # zero found
            f_zero = 1;
            K -= 1;
        else:
            res += 1;
  
    # return size - 1 if K is not zero and
    # atleast one zero is present, otherwise
    # result
    if (K == 0):
        return res;
    elif (f_zero > 0):
        return len(s) - 1;
    return -1;
  
# Driver Code
N = 10904025;
K = 2;
print(countDigitsToBeRemoved(N, K));
  
N = 1000;
K = 5;
print(countDigitsToBeRemoved(N, K));
  
N = 23985;
K = 2;
print(countDigitsToBeRemoved(N, K));
      
# This code is contributed by mits

C#

// C# Program to count the number 
// of digits that can be removed such 
// that number is divisible by 10^K 
   
using System;
public class GFG{
       
    // function to return the required 
    // number of digits to be removed 
    static int countDigitsToBeRemoved(int N, int K) 
    { 
        // Converting the given number 
        // into string 
        string s = Convert.ToString(N); 
       
        // variable to store number of 
        // digits to be removed 
        int res = 0; 
       
        // variable to denote if atleast 
        // one zero has been found 
        int f_zero = 0; 
        for (int i = s.Length - 1; i >= 0; i--) { 
            if (K == 0) 
                return res; 
            if (s[i] == '0') { 
       
                // zero found 
                f_zero = 1; 
                K--; 
            } 
            else
                res++; 
        } 
       
        // return size - 1 if K is not zero and 
        // atleast one zero is present, otherwise 
        // result 
        if (K == 0) 
            return res; 
        else if (f_zero == 1) 
            return s.Length - 1; 
        return -1; 
    } 
       
    // Driver Code to test above function 
    public static void Main()
    { 
        int N = 10904025;
        int K = 2; 
        Console.Write(countDigitsToBeRemoved(N, K)+"\n") ; 
       
        N = 1000 ;
        K = 5; 
        Console.Write(countDigitsToBeRemoved(N, K)+"\n")  ;
       
        N = 23985;
        K = 2; 
        Console.Write(countDigitsToBeRemoved(N, K)+"\n") ; 
    } 
   
     
    }

PHP

= 0; $i--) {
        if ($K == 0)
            return $res;
        if ($s[$i] == '0') {
  
            // zero found
            $f_zero = 1;
            $K--;
        }
        else
            $res++;
    }
  
    // return size - 1 if K is not zero and
    // atleast one zero is present, otherwise
    // result
    if (!$K)
        return $res;
    else if ($f_zero)
        return strlen($s) - 1;
    return -1;
}
  
// Driver Code to test above function
  
    $N = 10904025;
    $K = 2;
    echo countDigitsToBeRemoved($N, $K)."\n";
  
    $N = 1000;
    $K = 5;
    echo countDigitsToBeRemoved($N, $K)."\n";
  
    $N = 23985;
    $K = 2;
    echo countDigitsToBeRemoved($N, $K);
      
// This code is contributed by mits
?>

输出：

3
3
-1

时间复杂度：给定数字中的位数。