可以删除的最大位数，以便剩余整数是辅音

给定一个表示N位整数的字符串S ，任务是找到可以删除的最大位数，以便从辅音整数中删除剩余的位数。

请注意，0 和 1 也被视为非质数。

例子：

Input: S = “237”
Output: 1
Explanation: In the given integer, S[1] can be removed. Hence, S = “27”, which is the smallest possible integer that is non-prime. Therefore, the maximum number of digit that can be removed is 1.

Input: S = “35”
Output: -1
Explanation: It is not possible to create a non-prime integer using the above steps.

方法：给定的问题可以通过以下观察来解决：所有具有3位或更多位的字符串都包含长度最多为2位的数字子序列，使得该子序列表示非素数整数。使用此观察，可以使用以下步骤解决给定问题：

• 创建一个数组prime[] ，它存储给定整数是否为[0, 100)范围内的所有整数的素数。可以使用 Eratosthenes 筛高效地创建该阵列。
• 迭代给定的字符串str[]并检查是否存在任何非素数的1位字符串，即{0, 1, 4, 6, 8, 9} 。
• 如果不存在单个数字字符串，请检查给定字符串的长度是否 <= 2 。在这种情况下，如果S表示的整数是素数，则返回-1 ，否则返回0 。
• 否则，返回N – 2 ，这将是所需的答案。

下面是上述方法的实现：

C++

// C++ program of the above approach
#include 
using namespace std;
 
// Stores if integer representing
// the index is prime of not
bool prime[100];
 
// Function to calculate prime[]
// using sieve of eratosthenes
void sieve()
{
    // Set all integers as prime
    for (int i = 0; i < 100; i++) {
        prime[i] = true;
    }
 
    // Since 0 and 1 are considered
    // as the non prime integers
    prime[0] = false;
    prime[1] = false;
    for (int i = 2; i < 100; i++) {
        if (!prime[i])
            continue;
        for (int j = 2 * i; j < 100; j += i) {
            prime[j] = false;
        }
    }
}
 
// Function to find the maximum count of
// digits that can be removed such that
// the remaining integer is non-prime
int maxCount(string S)
{
    // Loop to iterate over all
    // digits in string S
    for (int i = 0; i < S.size(); i++) {
        // If a non-prime single
        // digit integer is found
        if (!prime[S[i] - '0']) {
            return S.length() - 1;
        }
    }
 
    // If the length of string
    // is at most 2
    if (S.length() <= 2) {
 
        // If S represents a
        // prime integer
        if (prime[stoi(S)])
            return -1;
        else
            return 0;
    }
    else {
 
        // Return Answer
        return S.length() - 2;
    }
}
 
// Driver Code
int main()
{
    string S = "237";
 
    sieve();
    cout << maxCount(S);
    return 0;
}

Java

// JAVA program of the above approach
import java.util.*;
class GFG
{
   
    // Stores if integer representing
    // the index is prime of not
    private static boolean[] prime = new boolean[100];
 
    // Function to calculate prime[]
    // using sieve of eratosthenes
    public static void sieve()
    {
        // Set all integers as prime
        for (int i = 0; i < 100; i++) {
            prime[i] = true;
        }
 
        // Since 0 and 1 are considered
        // as the non prime integers
        prime[0] = false;
        prime[1] = false;
        for (int i = 2; i < 100; i++) {
            if (!prime[i])
                continue;
            for (int j = 2 * i; j < 100; j += i) {
                prime[j] = false;
            }
        }
    }
 
    // Function to find the maximum count of
    // digits that can be removed such that
    // the remaining integer is non-prime
    public static int maxCount(String S)
    {
       
        // Loop to iterate over all
        // digits in string S
        for (int i = 0; i < S.length(); i++)
        {
           
            // If a non-prime single
            // digit integer is found
            if (!prime[S.charAt(i) - '0']) {
                return S.length() - 1;
            }
        }
 
        // If the length of string
        // is at most 2
        if (S.length() <= 2) {
 
            // If S represents a
            // prime integer
            if (prime[Integer.parseInt(S)])
                return -1;
            else
                return 0;
        }
        else {
 
            // Return Answer
            return S.length() - 2;
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String S = "237";
 
        sieve();
        System.out.print(maxCount(S));
    }
}
 
// This code is contributed by Taranpreet

Python3

# Stores if integer representing
# the index is prime of not
prime = []
 
# Function to calculate prime[]
# using sieve of eratosthenes
def sieve():
    # Set all integers as prime
    for i in range(100):
        prime.append(True)
 
    # Since 0 and 1 are considered
    # as the non prime integers
    prime[0] = False
    prime[1] = False
    for i in range(2, 100):
        if (not prime[i]):
            continue
        for j in range(2*i, 100, i):
            prime[j] = False
 
 
# Function to find the maximum count of
# digits that can be removed such that
# the remaining integer is non-prime
def maxCount(S):
 
    # Loop to iterate over all
    # digits in string S
    N = len(S)
    for i in range(N):
        # If a non-prime single
        # digit integer is found
        if (not prime[int(S[i])]):
            return N - 1
 
    # If the length of string
    # is at most 2
    if (N <= 2):
 
        # If S represents a
        # prime integer
        if (prime[int(S)]):
            return -1
        else:
            return 0
    else:
        # Return Answer
        return N - 2
 
 
# driver code
S = "237"
sieve()
print(maxCount(S))
 
 
# This code is contributed by Palak Gupta

C#

using System;
 
public class GFG
{
 
  // Stores if integer representing
  // the index is prime of not
  private static bool[] prime = new bool[100];
 
  // Function to calculate prime[]
  // using sieve of eratosthenes
  public static void sieve()
  {
    // Set all integers as prime
    for (int i = 0; i < 100; i++) {
      prime[i] = true;
    }
 
    // Since 0 and 1 are considered
    // as the non prime integers
    prime[0] = false;
    prime[1] = false;
    for (int i = 2; i < 100; i++) {
      if (!prime[i])
        continue;
      for (int j = 2 * i; j < 100; j += i) {
        prime[j] = false;
      }
    }
  }
 
  // Function to find the maximum count of
  // digits that can be removed such that
  // the remaining integer is non-prime
  public static int maxCount(string S)
  {
 
    // Loop to iterate over all
    // digits in string S
    for (int i = 0; i < S.Length; i++)
    {
 
      // If a non-prime single
      // digit integer is found
      if (!prime[S[i] - '0']) {
        return S.Length - 1;
      }
    }
 
    // If the length of string
    // is at most 2
    if (S.Length <= 2) {
 
      // If S represents a
      // prime integer
      if (prime[Int32.Parse(S)])
        return -1;
      else
        return 0;
    }
    else {
 
      // Return Answer
      return S.Length - 2;
    }
  }
 
  // Driver code
  static public void Main (){
 
    string S = "237";
 
    sieve();
    Console.Write(maxCount(S));
  }
}
 
// This code is contributed by sanjoy_62.

Javascript

       // JavaScript code for the above approach
 
       // Stores if integer representing
       // the index is prime of not
       let prime = new Array(100);
 
       // Function to calculate prime[]
       // using sieve of eratosthenes
       function sieve()
       {
        
           // Set all integers as prime
           for (let i = 0; i < 100; i++) {
               prime[i] = true;
           }
 
           // Since 0 and 1 are considered
           // as the non prime integers
           prime[0] = false;
           prime[1] = false;
           for (let i = 2; i < 100; i++) {
               if (!prime[i])
                   continue;
               for (let j = 2 * i; j < 100; j += i) {
                   prime[j] = false;
               }
           }
       }
 
       // Function to find the maximum count of
       // digits that can be removed such that
       // the remaining integer is non-prime
       function maxCount(S)
       {
        
           // Loop to iterate over all
           // digits in string S
           for (let i = 0; i < S.length; i++)
           {
            
               // If a non-prime single
               // digit integer is found
               if (!prime[S[i].charCodeAt(0) - '0'.charCodeAt(0)]) {
                   return S.length - 1;
               }
           }
 
           // If the length of string
           // is at most 2
           if (S.length <= 2) {
 
               // If S represents a
               // prime integer
               if (prime[parseInt(S)])
                   return -1;
               else
                   return 0;
           }
           else {
 
               // Return Answer
               return S.length - 2;
           }
       }
 
       // Driver Code
       let S = "237";
 
       sieve();
       document.write(maxCount(S));
 
      // This code is contributed by Potta Lokesh
   

输出

1

时间复杂度： O(N)
辅助空间： O(1)