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++ 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;
} |
// 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 |
# 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 |
// 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 |
<script> // JavaScript implementation to check if a number // becomes prime by deleting any digit // Function to check if N is prime function 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 || n % 3 == 0)
return false ;
for (let 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
function deleteIth(str, i)
{
// Deletes character at position 4
str = str.substr(0, i) +
str.substr(i + 1);
return str;
}
// Function to check if a number
// becomes prime by deleting any digit
function isPrimePossible(N)
{
// Converting the number to String
let s = N.toString();
// length of String
let 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 (let i = 0; i < l; i++)
{
// Deleting ith character
// from the String
let str = deleteIth(s, i);
// converting String to let
let num = parseInt(str);
if (isPrime(num))
return true ;
}
return false ;
}
// Driver Code let N = 610;
if (isPrimePossible(N))
document.write( "Yes" );
else
document.write( "No" );
</script> |
Yes
Time Complexity: O(D)
Auxiliary Space: O(1)