Count of Numbers such that difference between the number and sum of its digits not less than L

Given a Natural number N and a whole number L, the task is to find the count of numbers, smaller than or equal to N, such that the difference between the number and sum of its digits is not less than L.

Examples:

Input: N = 1500, L = 30
Output: 1461

Input: N = 1546300, L = 30651
Output: 1515631

Approach:
Our solution depends on a simple observation that if a number, say X, is such that the difference of X and sumOfDigits(X) is less than or equal to L, then X+1 is also a valid number. Hence, we have to find a minimum such X using binary search.

Implementation:

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to calculate the
// sum of digits
int sumOfDigits(long long n)
{
    long long sum = 0;
  
    while (n > 0) {
        sum += n % 10;
        n /= 10;
    }
  
    return sum;
}
  
// Function to count the numbers
long long countDigits(long long num, long long limit)
{
    long long left = 1, right = num, result = 0;
  
    // use binary search
    while (left <= right) {
  
        long long mid = (right + left) / 2;
  
        // Check if you are at a valid number or not
        if ((mid - sumOfDigits(mid)) >= limit) {
  
            // update the answer at each valid step
            result = num - mid + 1;
            right = mid - 1;
        }
        else {
  
            left = mid + 1;
        }
    }
  
    // return the answer
    return result;
}
  
// Driver code
int main()
{
    // Get the value of N and L
    long long N = 1546300, L = 30651;
  
    // Count the numbers
    cout << countDigits(N, L);
  
    return 0;
}

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Java














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// Java implementation of above approach 


  


class GFG 


{ 


    // Function to calculate the 


    // sum of digits 


    static long sumOfDigits( long n ) 


    { 


        long sum = 0; 


      


        while (n > 0)  


        { 


            sum += n % 10; 


            n /= 10; 


        } 


      


        return sum; 


    } 


      


    // Function to count the numbers 


    static long countDigits(long num, long limit) 


    { 


        long left = 1, right = num, result = 0; 


      


        // use binary search 


        while (left <= right) 


        { 


            long mid = (right + left) / 2; 


      


            // Check if you are at a valid number or not 


            if ((mid - sumOfDigits(mid)) >= limit) 


            { 


      


                // update the answer at each valid step 


                result = num - mid + 1; 


                right = mid - 1; 


            } 


            else


            { 


                left = mid + 1; 


            } 


        } 


      


        // return the answer 


        return result; 


    } 


      


    // Driver code 


    public static void main(String []args) 


    { 


        // Get the value of N and L 


        long N = 1546300, L = 30651; 


      


        // Count the numbers 


        System.out.println(countDigits(N, L)); 


    } 


} 


  


// This code is contributed by ihritik 



















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Python3














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# Python3 program for above approach 


  


# Function to calculate the sum of digits 


def sumOfDigits(n): 


    sum = 0


  


    while (n > 0): 


        sum += n % 10


        n = int(n / 10) 


  


    return sum


  


# Function to count the numbers 


def countDigits(num, limit): 


    left = 1


    right = num 


    result = 0


  


    # use binary search 


    while (left <= right): 


        mid = int((right + left) / 2) 


  


        # Check if you are at a valid number or not 


        if ((mid - sumOfDigits(mid)) >= limit): 


              


            # update the answer at each valid step 


            result = num - mid + 1


            right = mid - 1


          


        else: 


            left = mid + 1


          


    # return the answer 


    return result 


  


# Driver code 


if __name__ == '__main__': 


      


    # Get the value of N and L 


    N = 1546300


    L = 30651


  


    # Count the numbers 


    print(countDigits(N, L)) 


  


# This code is contributed by 


# Surendra_Gangwar 



















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C#














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// C# implementation of above approach 


using System; 


  


class GFG 


{ 


    // Function to calculate the 


    // sum of digits 


    static long sumOfDigits( long n ) 


    { 


        long sum = 0; 


      


        while (n > 0)  


        { 


            sum += n % 10; 


            n /= 10; 


        } 


      


        return sum; 


    } 


      


    // Function to count the numbers 


    static long countDigits(long num, long limit) 


    { 


        long left = 1, right = num, result = 0; 


      


        // use binary search 


        while (left <= right) { 


      


            long mid = (right + left) / 2; 


      


            // Check if you are at a valid number or not 


            if ((mid - sumOfDigits(mid)) >= limit)  


            { 


      


                // update the answer at each valid step 


                result = num - mid + 1; 


                right = mid - 1; 


            } 


            else 


            { 


      


                left = mid + 1; 


            } 


        } 


      


        // return the answer 


        return result; 


    } 


      


    // Driver code 


    public static void Main() 


    { 


        // Get the value of N and L 


        long N = 1546300, L = 30651; 


      


        // Count the numbers 


        Console.WriteLine(countDigits(N, L)); 


    } 


} 


  


// This code is contributed by ihritik 



















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PHP














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<?php 


// PHP implementation of the above approach 


      


// Function to calculate the  


// sum of digits  


function sumOfDigits($n)  


{  


    $sum = 0;  


  


    while ($n > 0) 


    {  


        $sum += $n % 10;  


        $n /= 10;  


    }  


  


    return $sum;  


}  


  


// Function to count the numbers  


function countDigits($num, $limit)  


{  


    $left = 1; 


    $right = $num; 


    $result = 0;  


  


    // use binary search  


    while ($left <= $right)  


    {  


  


        $mid = floor(($right + $left) / 2);  


  


        // Check if you are at a valid number or not  


        if (($mid - sumOfDigits($mid)) >= $limit)  


        {  


  


            // update the answer at each valid step  


            $result = $num - $mid + 1;  


            $right = $mid - 1;  


        }  


        else 


        {  


  


            $left = $mid + 1;  


        }  


    }  


  


    // return the answer  


    return $result;  


}  


  


// Driver code  


  


// Get the value of N and L  


$N = 1546300; 


$L = 30651;  


  


// Count the numbers  


echo countDigits($N, $L);  


  


// This code is contributed by Ryuga 


?> 



















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Output:


1515631





Time Complexity: O(log N)

Auxiliary Space: O(1)



          
          
          
          

            

    

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