Count digits in given number N which divide N

Given a number N which may be 10^5 digits long, the task is to count all the digits in N which divide N. Divisibility by 0 is not allowed. If any digit in N which is repeated divides N, then all repetitions of that digit should be counted i. e., N = 122324, here 2 divides N and it appears 3 times. So count for digit 2 will be 3.

Examples:

Input  : N = "35"
Output : 1
There are two digits in N and 1 of them
(5)divides it.

Input  : N = "122324"
Output : 5
N is divisible by 1, 2 and 4



How to check divisibility of a digit for large N stored as string?
The idea is to use distributive property of mod operation.
(x+y)%a = ((x%a) + (y%a)) % a.

// This function returns true if digit divides N, 
// else false
bool divisible(string N, int digit)
{
    int ans = 0;
    for (int i = 0; i < N.length(); i++)
    {     
        // (N[i]-'0') gives the digit value and
        // form the number       
        ans  = (ans*10 + (N[i]-'0'));

        // We use distributive property of mod here. 
        ans %= digit;
    }
    return (ans == 0);
}

A simple solution for this problem is to read number in string form and one by one check divisibility by each digit which appears in N. Time complexity for this approach will be O(N2).

An efficient solution for this problem is to use an extra array divide[] of size 10. Since we have only 10 digits so run a loop from 1 to 9 and check divisibility of N with each digit from 1 to 9. If any digit divides N then mark true in divide[] array at digit as index. Now traverse the number string and increment result if divide[i] is true for current digit i.

C++

// C++ program to find number of digits in N that
// divide N.
#include<bits/stdc++.h>
using namespace std;
  
// Utility function to check divisibility by digit
bool divisible(string N, int digit)
{
    int ans = 0;
    for (int i = 0; i < N.length(); i++)
    {
        // (N[i]-'0') gives the digit value and
        // form the number
        ans  = (ans*10 + (N[i]-'0'));
        ans %= digit;
    }
    return (ans == 0);
}
  
// Function to count digits which appears in N and
// divide N
// divide[10]  --> array which tells that particular
//                 digit divides N or not
// count[10]   --> counts frequency of digits which
//                 divide N
int allDigits(string N)
{
    // We initialize all digits of N as not divisble
    // by N.
    bool divide[10] = {false};
    divide[1] = true// 1 divides all numbers
  
    // start checking divisibilty of N by digits 2 to 9
    for (int digit=2; digit<=9; digit++)
    {
        // if digit divides N then mark it as true
        if (divisible(N, digit))
            divide[digit] = true;
    }
  
    // Now traverse the number string to find and increment
    // result whenever a digit divides N.
    int result = 0;
    for (int i=0; i<N.length(); i++)
    {
        if (divide[N[i]-'0'] == true)
            result++;
    }
  
    return result;
}
  
// Driver program to run the case
int main()
{
    string N = "122324";
    cout << allDigits(N);
    return 0;
}

C#

// C# program to find number of digits 
// in N that divide N.
using System;
  
class GFG {
      
// Utility function to 
// check divisibility by digit
static bool divisible(string N, int digit)
{
    int ans = 0;
    for (int i = 0; i < N.Length; i++)
    {
          
        // (N[i]-'0') gives the digit value and
        // form the number
        ans = (ans * 10 + (N[i] - '0'));
        ans %= digit;
    }
    return (ans == 0);
}
  
// Function to count digits which 
// appears in N and divide N
// divide[10] --> array which 
// tells that particular
// digit divides N or not
// count[10] --> counts 
// frequency of digits which
// divide N
static int allDigits(string N)
{
      
    // We initialize all digits
    // of N as not divisble by N
    bool[] divide = new bool[10];
      
    for (int i = 0; i < divide.Length; i++) 
    {
        divide[i] = false;
    }
      
    // 1 divides all numbers
    divide[1] = true
  
    // start checking divisibilty
    // of N by digits 2 to 9
    for (int digit = 2; digit <= 9; digit++)
    {
          
        // if digit divides N 
        // then mark it as true
        if (divisible(N, digit))
            divide[digit] = true;
    }
  
    // Now traverse the number
    // string to find and increment
    // result whenever a digit divides N.
    int result = 0;
      
    for (int i = 0; i < N.Length; i++)
    {
        if (divide[N[i] - '0'] == true)
            result++;
    }
  
    return result;
}
  
// Driver Code
public static void Main()
{
    string N = "122324";
    Console.Write(allDigits(N));
}
}
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)

PHP

<?php
// PHP program to find number of 
// digits in N that divide N.
  
// Utility function to check 
// divisibility by digit
function divisible($N, $digit)
{
    $ans = 0;
    for ($i = 0; $i < strlen($N); $i++)
    {
        // (N[i]-'0') gives the digit 
        // value and form the number
        $ans = ($ans * 10 + (int)($N[$i] - '0'));
        $ans %= $digit;
    }
    return ($ans == 0);
}
  
// Function to count digits which 
// appears in N and divide N
// divide[10] --> array which tells 
// that particular digit divides N or not
// count[10] --> counts frequency of 
//                 digits which divide N
function allDigits($N)
{
    // We initialize all digits of N 
    // as not divisble by N.
    $divide = array_fill(0, 10, false);
    $divide[1] = true; // 1 divides all numbers
  
    // start checking divisibilty of
    // N by digits 2 to 9
    for ($digit = 2; $digit <= 9; $digit++)
    {
        // if digit divides N then
        // mark it as true
        if (divisible($N, $digit))
            $divide[$digit] = true;
    }
  
    // Now traverse the number string to 
    // find and increment result whenever
    // a digit divides N.
    $result = 0;
    for ($i = 0; $i < strlen($N); $i++)
    {
        if ($divide[(int)($N[$i] - '0')] == true)
            $result++;
    }
  
    return $result;
}
  
// Driver Code
$N = "122324";
echo allDigits($N);
  
// This code is contributed by mits
?>


Output :

5

Time Complexity: O(n)
Auxiliary space: O(1)

This article is contributed by Shashank Mishra ( Gullu ). 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.

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Improved By : Mithun Kumar, Abby_akku



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