Numbers having difference with digit sum more than s

You are given two positive integer value n and s. You have to find the total number of such integer from 1 to n such that the difference of integer and its digit sum is greater than given s.

Examples :

Input : n = 20, s = 5
Output :11
Explanation : Integer from 1 to 9 have 
diff(integer - digitSum) = 0 but for 10 to 
20 they have diff(value - digitSum) > 5

Input : n = 20, s = 20
Output : 0
Explanation : Integer from 1 to 20 have diff
(integer - digitSum) >  5



The very first and basic approach to solve this question is to check for all integer starting from 1 to n and for each check whether integer minus digit sum is greater than s or not. This will become very time costly because we have to traverse 1 to n and for each integer we also have to calculate the digit sum.

Before moving to better approach lets have some key analysis about this questions and its features:

  • For the largest possible integer (say long long int i.e. 10^18), the maximum possible digit sum is 9*18 (when all of digits are nine) = 162. This means in any case all the integer greater than s + 162 satisfy the condition of integer – digitSum > s.
  • All integer less than s can not satisfy the given condition for sure.
  • All the integers within a tens range (0-9, 10-19…100-109) does have same value of integer minus digitSum.

Using above three key features we can shorten our approach and time complexity in a manner where we have to iterate only over s to s+163 integers. Beside checking for all integer within range we only check for each 10th integer (e.g 150, 160, 170..).
Algorithm:

// if n < s then return 0
if n<s 
    return 0
else

    // iterate for s to min(n, s+163)
    for i=s to i min(n, s+163)

        // return n-i+1
        if (i-digitSum)>s
            return (n-i+1)

// if no such integer found return 0
return 0

C++

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// Program to find number of integer such that
// integer - digSum  > s
#include <bits/stdc++.h>
using namespace std;
  
// function for digit sum
int digitSum(long long int n) {
  int digSum = 0;
  while (n) {
    digSum += n % 10;
    n /= 10;
  }
  return digSum;
}
  
// function to calculate count of integer s.t.
// integer - digSum > s
  
long long int countInteger(long long int n, 
                          long long int s) {
  
  // if n < s no integer possible
  if (n < s)
    return 0;
  
  // iterate for s range and then calculate 
  // total count of such integer if starting 
  // integer is found
  for (long long int i = s; i <= min(n, s + 163); i++)
    if ((i - digitSum(i)) > s)
      return (n - i + 1);
  
  // if no integer found return 0
  return 0;
}
  
// driver program
int main() {
  long long int n = 1000, s = 100;
  cout << countInteger(n, s);
  return 0;
}

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Java

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// Java Program to find number of integer 
// such that integer - digSum > s
import java.io.*;
  
class GFG
{
    // function for digit sum
    static int digitSum(long n) 
    {
        int digSum = 0;
        while (n > 0
        {
            digSum += n % 10;
            n /= 10;
        }
        return digSum;
    }
  
    // function to calculate count of integer s.t.
    // integer - digSum > s
    public static long countInteger(long n, long s) 
    {
        // if n < s no integer possible
        if (n < s)
            return 0;
      
        // iterate for s range and then calculate 
        // total count of such integer if starting 
        // integer is found
        for (long i = s; i <= Math.min(n, s + 163); i++)
            if ((i - digitSum(i)) > s)
                return (n - i + 1);
      
        // if no integer found return 0
        return 0;
    }
  
    // Driver program
    public static void main(String args[])
    {
            long n = 1000, s = 100;
            System.out.println(countInteger(n, s));
    }
}
  
// This code is contributed by Anshika Goyal.

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Python3

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# Program to find number
# of integer such that
# integer - digSum  > s
  
# function for digit sum
def digitSum(n):
  
    digSum = 0
  
    while (n>0):
        digSum += n % 10
        n //= 10
    
    return digSum
   
# function to calculate
# count of integer s.t.
# integer - digSum > s
   
def countInteger(n, s):
      
    # if n < s no integer possible
    if (n < s):
        return 0
   
    # iterate for s range
    # and then calculate 
    # total count of such
    # integer if starting 
    # integer is found
    for i in range(s,min(n, s + 163)+1):
        if ((i - digitSum(i)) > s):
            return (n - i + 1)
   
    # if no integer found return 0
    return 0
  
# driver code
n = 1000
s = 100
  
print(countInteger(n, s))
  
# This code is contributed
# by Anant Agarwal.

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C#

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// C# Program to find number of integer 
// such that integer - digSum > s
using System;
  
class GFG
{
    // function for digit sum
    static long digitSum(long n) 
    {
        long digSum = 0;
          
        while (n > 0) 
        {
            digSum += n % 10;
            n /= 10;
        }
        return digSum;
    }
  
    // function to calculate count of integer s.t.
    // integer - digSum > s
    public static long countInteger(long n, long s) 
    {
        // if n < s no integer possible
        if (n < s)
            return 0;
      
        // iterate for s range and then calculate 
        // total count of such integer if starting 
        // integer is found
        for (long i = s; i <= Math.Min(n, s + 163); i++)
            if ((i - digitSum(i)) > s)
                return (n - i + 1);
      
        // if no integer found return 0
        return 0;
    }
  
    // Driver program
    public static void Main()
    {
            long n = 1000, s = 100;
            Console.WriteLine(countInteger(n, s));
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// Program to find number of integer 
// such that integer - digSum > s
  
// function for digit sum
function digitSum( $n
{
$digSum = 0;
while ($n
{
    $digSum += $n % 10;
    $n /= 10;
}
return $digSum;
}
  
// function to calculate count of 
// integer s.t. integer - digSum > s
  
function countInteger( $n, $s
{
  
// if n < s no integer possible
if ($n < $s)
    return 0;
  
// iterate for s range and then  
// calculate total count of such 
// integer if starting integer is found
for ( $i = $s; $i <= min($n, $s + 163); $i++)
    if (($i - digitSum($i)) > $s)
    return ($n - $i + 1);
  
// if no integer found return 0
return 0;
}
  
// Driver Code
$n = 1000; $s = 100;
echo countInteger($n, $s);
  
// This code is contributed by anuj_67.
?>

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Output :

891


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