Count of prime digits of a Number which divides the number
Last Updated :
20 Dec, 2022
Given an integer N, the task is to count the number of digits in N which is a prime number, and also divides the number.
Examples:
Input: N = 12
Output: 1
Explanation:
Digits of the number = {1, 2}
But, only 2 is prime number that divides N.
Input: N = 1032
Output: 2
Explanation:
Digits of the number = {1, 0, 3, 2}
3 and 2 divides the number and are also prime.
Naive Approach: The idea is to find all the digits of the number. For each digit, check if is prime or not. If yes, then check if it divides the number or not. If both the cases are true, then increment the count by 1. The final count is the required answer.
Efficient Approach: Since only 2, 3, 5, and 7 are the prime single-digit numbers, therefore for each digit, check if divides the number and if is 2, 3, 5 or 7. If both the cases are true, then increment the count by 1. The final count is the required answer.
Below is the implementation of this approach:
C++
#include <bits/stdc++.h>
using namespace std;
int countDigit( int n)
{
bool prime[10];
memset (prime, false , sizeof (prime));
prime[2] = prime[3] = true ;
prime[5] = prime[7] = true ;
int temp = n, count = 0;
while (temp != 0) {
int d = temp % 10;
temp /= 10;
if (d > 0 && n % d == 0 && prime[d])
count++;
}
return count;
}
int main()
{
int n = 1032;
cout << countDigit(n) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
static int countDigit( int n)
{
boolean prime[] = new boolean [ 10 ];
for ( int i = 0 ; i < 10 ; i++)
prime[i] = false ;
prime[ 2 ] = prime[ 3 ] = true ;
prime[ 5 ] = prime[ 7 ] = true ;
int temp = n, count = 0 ;
while (temp != 0 ) {
int d = temp % 10 ;
temp /= 10 ;
if (d > 0 && n % d == 0 && prime[d] == true )
count++;
}
return count;
}
public static void main (String[] args)
{
int n = 1032 ;
System.out.println(countDigit(n)) ;
}
}
|
Python3
def countDigit(n):
prime = [ False ] * 10
prime[ 2 ] = True
prime[ 3 ] = True ;
prime[ 5 ] = True
prime[ 7 ] = True ;
temp = n
count = 0 ;
while (temp ! = 0 ):
d = temp % 10 ;
temp / / = 10 ;
if (d > 0 and n % d = = 0 and prime[d]):
count + = 1
return count
n = 1032
print (countDigit(n))
|
C#
using System;
class GFG {
static int countDigit( int n)
{
bool []prime = new bool [10];
for ( int i = 0; i < 10; i++)
prime[i] = false ;
prime[2] = prime[3] = true ;
prime[5] = prime[7] = true ;
int temp = n, count = 0;
while (temp != 0) {
int d = temp % 10;
temp /= 10;
if (d > 0 && n % d == 0 && prime[d] == true )
count++;
}
return count;
}
public static void Main ( string [] args)
{
int n = 1032;
Console.WriteLine(countDigit(n)) ;
}
}
|
Javascript
<script>
function countDigit(n)
{
var prime = Array(10).fill( false );
prime[2] = prime[3] = true ;
prime[5] = prime[7] = true ;
var temp = n, count = 0;
while (temp != 0) {
var d = temp % 10;
temp = parseInt(temp/10);
if (d > 0 && n % d == 0 && prime[d])
count++;
}
return count;
}
n = 1032;
document.write(countDigit(n));
</script>
|
Time Complexity: O(log10n)
Auxiliary Space: O(prime)
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