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Count numbers in range such that digits in it and it’s product with q are unequal
  • Difficulty Level : Easy
  • Last Updated : 26 Oct, 2020

Given a range of numbers [l, r] and an integer q. The task is to count all such number in the given range such that any digit of the number does not match with any digit in its product with the given number q.

Examples

Input : l = 10, r = 12, q = 2
Output : 1
10*2 = 20 which has 0 as same digit
12*2 = 24 which as 2 as same digit
11*2 = 22 no same digit

Input : l = 5, r = 15, q = 2
Output : 9

Source : Goldman Sachs Interview set 46
The idea is to run a loop from l to r to generate all numbers in the range and convert each such number n and it’s product with q, i.e. n*q to strings using to_string() method and then check if any character in string2 is present in string1 or not using basic string hashing.

Below is the implementation of the above approach: 

C++




// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if all of the digits
// in a number and it's product with q
// are unequal or not
bool checkIfUnequal(int n, int q)
{
 
    // convert first number into string
    string s1 = to_string(n);
    int a[26] = { 0 };
 
    // Insert elements from 1st number
    // to hash
    for (int i = 0; i < s1.size(); i++)
        a[s1[i] - '0']++;
 
    // Calculate corresponding product
    int prod = n * q;
 
    // Convert the product to string
    string s2 = to_string(prod);
 
    // Using the hash check if any digit of
    // product matches with the digits of
    // input number
    for (int i = 0; i < s2.size(); i++)
    {
        // If yes, return false
        if (a[s2[i] - '0'])
            return false;
    }
 
    // Return true
    return true;
}
 
// Function to count numbers in the range [l, r]
// such that all of the digits of the number and
// it's product with q are unequal
int countInRange(int l, int r, int q)
{
    int count = 0;
 
    for (int i = l; i <= r; i++) {
        // check for every number between l and r
        if (checkIfUnequal(i, q))
            count++;
    }
 
    return count;
}
 
// Driver Code
int main()
{
 
    int l = 10, r = 12, q = 2;
 
    // Function Call
    cout << countInRange(l, r, q);
    return 0;
}

Java




// Java program for above approach
class GfG {
 
    // Function to check if all of the digits
    // in a number and it's product with q
    // are unequal or not
    static boolean checkIfUnequal(int n, int q)
    {
 
        // convert first number into string
        String s1 = Integer.toString(n);
        int a[] = new int[26];
 
        // Insert elements from 1st number
        // to hash
        for (int i = 0; i < s1.length(); i++)
            a[s1.charAt(i) - '0']++;
 
        // Calculate corresponding product
        int prod = n * q;
 
        // Convert the product to string
        String s2 = Integer.toString(prod);
 
        // Using the hash check if any digit of
        // product matches with the digits of
        // input number
        for (int i = 0; i < s2.length(); i++)
        {
            // If yes, return false
            if (a[s2.charAt(i) - '0'] > 0)
                return false;
        }
        // else, return true
        return true;
    }
 
    // Function to count numbers in the range [l, r]
    // such that all of the digits of the number and
    // it's product with q are unequal
    static int countInRange(int l, int r, int q)
    {
        int count = 0;
 
        for (int i = l; i <= r; i++) {
 
            // check for every number between l and r
            if (checkIfUnequal(i, q))
                count++;
        }
 
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int l = 10, r = 12, q = 2;
 
        // Function Call
        System.out.println(countInRange(l, r, q));
    }
}

Python3




# Python 3 program for above approach
 
# Function to check if all of the digits
# in a number and it's product with q
# are unequal or not
def checkIfUnequal(n, q):
     
    # convert first number into string
    s1 = str(n)
    a = [0 for i in range(26)]
 
    # Insert elements from 1st number
    # to hash
    for i in range(0, len(s1), 1):
        a[ord(s1[i]) - ord('0')] += 1
 
    # Calculate corresponding product
    prod = n * q
 
    # Convert the product to string
    s2 = str(prod)
 
    # Using the hash check if any digit of
    # product matches with the digits of
    # input number
    for i in range(0, len(s2), 1):
         
        # If yes, return false
        if (a[ord(s2[i]) - ord('0')]):
            return False
 
    # Return true
    return True
 
# Function to count numbers in the range [l, r]
# such that all of the digits of the number and
# it's product with q are unequal
def countInRange(l, r, q):
    count = 0
 
    for i in range(l, r + 1, 1):
         
        # check for every number between l and r
        if (checkIfUnequal(i, q)):
            count += 1
     
    return count
 
# Driver Code
if __name__ == '__main__':
    l = 10
    r = 12
    q = 2
 
    # Function call
    print(countInRange(l, r, q))
 
# This code is contributed by
# Sahil_Shelangia

C#




// C# program for above approach
using System;
 
class GfG {
 
    // Function to check if all of the digits
    // in a number and it's product with q
    // are unequal or not
    static bool checkIfUnequal(int n, int q)
    {
 
        // convert first number into string
        string s1 = n.ToString();
        int[] a = new int[26];
 
        // Insert elements from 1st number
        // to hash
        for (int i = 0; i < s1.Length; i++)
            a[s1[i] - '0']++;
 
        // Calculate corresponding product
        int prod = n * q;
 
        // Convert the product to string
        string s2 = prod.ToString();
 
        // Using the hash check if any digit of
        // product matches with the digits of
        // input number
        for (int i = 0; i < s2.Length; i++)
        {
            // If yes, return false
            if (a[s2[i] - '0'])
                return false;
        }
 
        // Else, return true
        return true;
    }
 
    // Function to count numbers in the range [l, r]
    // such that all of the digits of the number and
    // it's product with q are unequal
    static int countInRange(int l, int r, int q)
    {
        int count = 0;
 
        for (int i = l; i <= r; i++)
        {
            // check for every number between l and r
            if (checkIfUnequal(i, q))
                count++;
        }
 
        return count;
    }
 
    // Driver Code
    public static void Main()
    {
 
        int l = 10, r = 12, q = 2;
 
        // Function call
        Console.WriteLine(countInRange(l, r, q));
    }
}
 
// This code is contributed bt Archana_kumari

PHP




// PHP program for above code
 
// Function to check if all of the digits
// in a number and it's product with q
// are unequal or not
function checkIfUnequal($n, $q)
{
    // convert first number into string
    $s1 = strval($n);
    $a = array_fill(0, 26, NULL);
 
    // Insert elements from 1st number
    // to hash
    for ($i = 0; $i < strlen($s1); $i++)
        $a[ord($s1[$i]) - ord('0')]++;
 
    // Calculate corresponding product
    $prod = $n * $q;
 
    // Convert the product to string
    $s2 = strval($prod);
 
    // Using the hash check if any digit of
    // product matches with the digits of
    // input number
    for ($i = 0; $i < strlen($s2); $i++)
    {
        
       // If yes, return false
       if ($a[ord($s2[$i]) - ord('0')])
            return false;
    }
 
    // Else, return true
    return true;
}
 
// Function to count numbers in the range
// [l, r] such that all of the digits of the
// number and it's product with q are unequal
function countInRange($l, $r, $q)
{
    $count = 0;
 
    for ($i = $l; $i <= $r; $i++)
    {
        // check for every number between l and r
        if (checkIfUnequal($i, $q))
            $count++;
    }
 
    return $count;
}
 
// Driver Code
$l = 10;
$r = 12;
$q = 2;
 
// Function call
echo countInRange($l, $r, $q);
 
// This code is contributed by ita_c
?>
Output
1

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