Check if number can be made prime by deleting a single digit

Given an integer N, the task is to check if it is possible to make N prime by deleting any single digit from N.

Examples:

Input: N = 610
Output: Yes
Explanation: 
Deleting 0 from 610, we get 61 which is prime. 

Input: N = 68 
Output: No
 

 

Approach: The idea is to convert N to a string. Now iterate for every digit of string and Delete character at index i from string and then convert the string after deleting character at index i to an integer, Now check if this integer is a prime, then return true. Otherwise, finally return false.
 



Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation to check if a number
// becomes prime by deleting any digit
 
#include <bits/stdc++.h>
using namespace std;
   
// Function to check if N is prime
bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
   
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
   
    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
   
    return true;
}
 
// Function to delete character at index i
// from given string str
string deleteIth(string str, int i)
{
    // Deletes character at position 4
    str.erase(str.begin() + i);
   
    return str;
}
 
// Function to check if a number
// becomes prime by deleting any digit
bool isPrimePossible(int N)
{
    // Converting the number to string
    string s = to_string(N);
   
    // length of string
    int l = s.length();
   
    // number should not be
    // of single digit
    if (l < 2)
        return false;
   
    // Loop to find all numbers
    // after deleting a single digit
    for (int i = 0; i < l ; i++) {
       
        // Deleting ith character
        // from the string
        string str = deleteIth(s, i);
         
        // converting string to int
        int num = stoi(str);
  
        if (isPrime(num))
            return true;
    }
    return false;
}
   
// Driver Code
int main()
{
    int N = 610;
    isPrimePossible(N) ? cout << "Yes"
                     : cout << "No";
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation to check if a number
// becomes prime by deleting any digit
import java.util.*;
class GFG{
 
  // Function to check if N is prime
  static boolean isPrime(int n)
  {
    // Corner cases
    if (n <= 1)
      return false;
    if (n <= 3)
      return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
      return false;
 
    for (int i = 5; i * i <= n; i = i + 6)
      if (n % i == 0 || n % (i + 2) == 0)
        return false;
 
    return true;
  }
 
  // Function to delete character at index i
  // from given String str
  static String deleteIth(String str, int i)
  {
    // Deletes character at position 4
    str = str.substring(0, i) +
            str.substring(i + 1);
 
    return str;
  }
 
  // Function to check if a number
  // becomes prime by deleting any digit
  static boolean isPrimePossible(int N)
  {
    // Converting the number to String
    String s = String.valueOf(N);
 
    // length of String
    int l = s.length();
 
    // number should not be
    // of single digit
    if (l < 2)
      return false;
 
    // Loop to find all numbers
    // after deleting a single digit
    for (int i = 0; i < l; i++)
    {
 
      // Deleting ith character
      // from the String
      String str = deleteIth(s, i);
 
      // converting String to int
      int num = Integer.valueOf(str);
 
      if (isPrime(num))
        return true;
    }
    return false;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 610;
    if (isPrimePossible(N))
      System.out.print("Yes");
    else
      System.out.print("No");
  }
}
 
// This code is contributed by Rajput-Ji

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation to check if a number
# becomes prime by deleting any digit
 
# Function to check if N is prime
#from builtins import range
def isPrime(n):
   
    # Corner cases
    if (n <= 1):
        return False;
    if (n <= 3):
        return True;
 
    # This is checked so that we can skip
    # middle five numbers in below loop
    if (n % 2 == 0 or n % 3 == 0):
        return False;
 
    for i in range(5, int(n**1/2), 6):
        if (n % i == 0 or n % (i + 2) == 0):
            return False;
 
    return True;
 
# Function to delete character at index i
# from given String str
def deleteIth(str, i):
   
    # Deletes character at position 4
    str = str[0:i] + str[i + 1:];
 
    return str;
 
# Function to check if a number
# becomes prime by deleting any digit
def isPrimePossible(N):
   
    # Converting the number to String
    s = str(N);
 
    # length of String
    l = len(s);
 
    # number should not be
    # of single digit
    if (l < 2):
        return False;
 
    # Loop to find all numbers
    # after deleting a single digit
    for i in range(l):
 
        # Deleting ith character
        # from the String
        str1 = deleteIth(s, i);
 
        # converting String to int
        num = int(str1);
 
        if (isPrime(num)):
            return True;
 
    return False;
 
# Driver Code
if __name__ == '__main__':
    N = 610;
    if (isPrimePossible(N)):
        print("Yes");
    else:
        print("No");
 
# This code is contributed by Rajput-Ji

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation to check if a number
// becomes prime by deleting any digit
using System;
 
class GFG{
 
// Function to check if N is prime
static bool isPrime(int n)
{
     
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for(int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}
 
// Function to delete character at index i
// from given String str
static String deleteIth(String str, int i)
{
     
    // Deletes character at position 4
    str = str.Substring(0, i) +
          str.Substring(i + 1);
 
    return str;
}
 
// Function to check if a number
// becomes prime by deleting any digit
static bool isPrimePossible(int N)
{
     
    // Converting the number to String
    String s = String.Join("", N);
 
    // length of String
    int l = s.Length;
 
    // number should not be
    // of single digit
    if (l < 2)
        return false;
 
    // Loop to find all numbers
    // after deleting a single digit
    for(int i = 0; i < l; i++)
    {
 
        // Deleting ith character
        // from the String
        String str = deleteIth(s, i);
 
        // converting String to int
        int num = Int32.Parse(str);
 
        if (isPrime(num))
            return true;
    }
    return false;
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 610;
     
    if (isPrimePossible(N))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by Rajput-Ji

chevron_right


Output

Yes



Time Complexity: O(D)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : Rajput-Ji