Counting numbers whose difference from reverse is a product of k

Given two number l and r. Count the total numbers between l and r which when subtracted from their respective reverse, the difference is a product of k.

Examples:

Input : 20 23 6
Output : 2
20 and 22 are the two numbers.
|20-2| = 18 which is a product of 6
|22-22| = 0 which is a product of 6

Input : 35 45 5
Output : 2
38 and 44 are the two numbers.
|38-83| = 45 which is a product of 5
|44-44| = 0 which is a product of 5



Approach: For each number between the given range check if the difference between the number and its reverse number is divisible by k, increment count if yes.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to Count the numbers
// within a given range in which when
// you subtract a number from its
// reverse, the difference is a product
// of k
#include <iostream>
using namespace std;
  
bool isRevDiffDivisible(int x, int k)
    // function to check if the number
    // and its reverse have their 
    // absolute difference divisible by k
    int n = x;
    int m = 0;
    int flag;
    while (x > 0) 
    {   
        // reverse the number
        m = m * 10 + x % 10;
        x /= 10;
    }
      
    return (abs(n - m) % k == 0);
}
  
int countNumbers(int l, int r, int k)
{
    int count = 0; 
    for (int i = l; i <= r; i++)    
        if (isRevDiffDivisible(i, k))        
            ++count; 
    return count;
}
  
// Driver code
int main()
{
    int l = 20, r = 23, k = 6;
    cout << countNumbers(l, r, k) << endl; 
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to Count the 
// numbers within a given range 
// in which when you subtract 
// a number from its reverse, 
// the difference is a product of k
import java.io.*;
import java.math.*;
  
class GFG {
      
    static boolean isRevDiffDivisible(int x, int k)
    
        // function to check if the number
        // and its reverse have their 
        // absolute difference divisible by k
        int n = x;
        int m = 0;
        int flag;
        while (x > 0
        
            // reverse the number
            m = m * 10 + x % 10;
            x /= 10;
        }
          
        return (Math.abs(n - m) % k == 0);
    }
      
    static int countNumbers(int l, int r, int k)
    {
        int count = 0
        for (int i = l; i <= r; i++) 
            if (isRevDiffDivisible(i, k))     
                ++count; 
        return count;
    }
      
    // Driver code
    public static void main(String args[])
    {
        int l = 35, r = 45, k = 5;
        System.out.println(countNumbers(l, r, k));
    }
}
  
// This code is contributed 
// by Nikita Tiwari.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 program to Count the numbers
# within a given range in which when you
# subtract a number from its reverse,
# the difference is a product of k
  
def isRevDiffDivisible(x, k) :
    # function to check if the number
    # and its reverse have their 
    # absolute difference divisible by k
    n = x; m = 0
    while (x > 0) :
  
        # Reverse the number
        m = m * 10 + x % 10
        x = x // 10
          
    return (abs(n - m) % k == 0)
  
def countNumbers(l, r, k) :
    count = 0
    for i in range(l, r + 1) :
  
        if (isRevDiffDivisible(i, k)) :
            count = count + 1
      
    return count
      
      
# Driver code
l = 20; r = 23; k = 6
print(countNumbers(l, r, k))
  
  
# This code is contributed by Nikita Tiwari.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to Count the 
// numbers within a given range 
// in which when you subtract 
// a number from its reverse, 
// the difference is a product of k
using System;
  
class GFG {
      
    static bool isRevDiffDivisible(int x, int k)
    
        // function to check if the number
        // and its reverse have their 
        // absolute difference divisible by k
        int n = x;
        int m = 0;
          
        // int flag;
        while (x > 0) 
        
            // reverse the number
            m = m * 10 + x % 10;
            x /= 10;
        }
          
        return (Math.Abs(n - m) % k == 0);
    }
      
    static int countNumbers(int l, int r, int k)
    {
        int count = 0; 
          
        for (int i = l; i <= r; i++) 
            if (isRevDiffDivisible(i, k)) 
                ++count; 
        return count;
    }
      
    // Driver code
    public static void Main()
    {
        int l = 35, r = 45, k = 5;
        Console.WriteLine(countNumbers(l, r, k));
    }
}
  
// This code is contributed 
// by vt_m.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to Count the 
// numbers within a given 
// range in which when you
// subtract a number from 
// its reverse, the difference 
// is a product of k
function isRevDiffDivisible($x, $k)
    // function to check if
    // the number and its
    // reverse have their 
    // absolute difference 
    // divisible by k
    $n = $x;
    $m = 0;
    $flag;
    while ($x > 0) 
    
        // reverse the number
        $m = $m * 10 + $x % 10;
        $x = (int)$x / 10;
    }
      
    return (abs($n - $m) %
                $k == 0);
}
  
function countNumbers($l, $r, $k)
{
    $count = 0; 
    for ($i = $l; $i <= $r; $i++) 
        if (isRevDiffDivisible($i, $k)) 
            ++$count
    return $count;
}
  
// Driver code
$l = 20; $r = 23; $k = 6;
echo countNumbers($l, $r, $k); 
  
// This code is contributed by mits.
?>

chevron_right



Output:

2


My Personal Notes arrow_drop_up

Budding Web DeveloperKeen learnerAverage CoderDancer&Social Activist

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 : Mithun Kumar