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

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  • Last Updated : 01 Jun, 2021

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++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

PHP




<?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
?>

Javascript




<script>
 
    // JavaScript implementation of above approach
     
    // Function to calculate the
    // sum of digits
    function sumOfDigits(n)
    {
        let sum = 0;
       
        while (n > 0)
        {
            sum += n % 10;
            n = parseInt(n / 10, 10);
        }
       
        return sum;
    }
       
    // Function to count the numbers
    function countDigits(num, limit)
    {
        let left = 1, right = num, result = 0;
       
        // use binary search
        while (left <= right) {
       
            let mid = parseInt((right + left) / 2, 10);
       
            // 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;
    }
     
    // Get the value of N and L
    let N = 1546300, L = 30651;
 
    // Count the numbers
    document.write(countDigits(N, L));
     
</script>
Output: 
1515631

 

Time Complexity: O(log N) 
Auxiliary Space: O(1)
 


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