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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // Program to find number of integer such that // integer - digSum  > s #include 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; }

Java

 // 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.

Python3

 # 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.

C#

 // 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.

PHP

 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. ?>

Output :

891

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