# Find smallest number with given number of digits and sum of digits under given constraints

Given two integers S and D, the task is to find the number having D number of digits and the sum of its digits as S such that the difference between the maximum and the minimum digit in the number is as minimum as possible. If multiple such numbers are possible, print the smallest number.

Examples:

Input: S = 25, D = 4
Output: 6667
The difference between maximum digit 7 and minimum digit 6 is 1.

Input: S = 27, D = 3
Output: 999

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

Approach:

• Finding smallest number for given number of digits and sum is already discussed in this article.
• In this article, the idea is to minimize the difference between the maximum and minimum digit in the required number. Therefore, the sum s should be evenly distributed among d digits.
• If the sum is evenly distributed then the difference can be at most 1. The difference is zero when sum s is divisible by d. In that case, each of the digits has the same value equal to s/d.
• The difference is one when sum s is not divisible by d. In that case, after each digit is assigned value s/d, s%d sum value is still left to be distributed.
• As the smallest number is required, this remaining value is evenly distributed among last s%d digits of the number, i.e., last s%d digits in the number are incremented by one.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the number having ` `// sum of digits as s and d number of ` `// digits such that the difference between ` `// the maximum and the minimum digit ` `// the minimum possible ` `string findNumber(``int` `s, ``int` `d) ` `{ ` `    ``// To store the final number ` `    ``string num = ``""``; ` ` `  `    ``// To store the value that is evenly ` `    ``// distributed among all the digits ` `    ``int` `val = s / d; ` ` `  `    ``// To store the remaining sum that still ` `    ``// remains to be distributed among d digits ` `    ``int` `rem = s % d; ` ` `  `    ``int` `i; ` ` `  `    ``// rem stores the value that still remains ` `    ``// to be distributed ` `    ``// To keep the difference of digits minimum ` `    ``// last rem digits are incremented by 1 ` `    ``for` `(i = 1; i <= d - rem; i++) { ` `        ``num = num + to_string(val); ` `    ``} ` ` `  `    ``// In the last rem digits one is added to ` `    ``// the value obtained by equal distribution ` `    ``if` `(rem) { ` `        ``val++; ` `        ``for` `(i = d - rem + 1; i <= d; i++) { ` `            ``num = num + to_string(val); ` `        ``} ` `    ``} ` ` `  `    ``return` `num; ` `} ` ` `  `// Driver function ` `int` `main() ` `{ ` `    ``int` `s = 25, d = 4; ` ` `  `    ``cout << findNumber(s, d); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to find the number having ` `// sum of digits as s and d number of ` `// digits such that the difference between ` `// the maximum and the minimum digit ` `// the minimum possible ` `static` `String findNumber(``int` `s, ``int` `d) ` `{ ` `    ``// To store the final number ` `    ``String num = ``""``; ` ` `  `    ``// To store the value that is evenly ` `    ``// distributed among all the digits ` `    ``int` `val = s / d; ` ` `  `    ``// To store the remaining sum that still ` `    ``// remains to be distributed among d digits ` `    ``int` `rem = s % d; ` ` `  `    ``int` `i; ` ` `  `    ``// rem stores the value that still remains ` `    ``// to be distributed ` `    ``// To keep the difference of digits minimum ` `    ``// last rem digits are incremented by 1 ` `    ``for` `(i = ``1``; i <= d - rem; i++) ` `    ``{ ` `        ``num = num + String.valueOf(val); ` `    ``} ` ` `  `    ``// In the last rem digits one is added to ` `    ``// the value obtained by equal distribution ` `    ``if` `(rem > ``0``)  ` `    ``{ ` `        ``val++; ` `        ``for` `(i = d - rem + ``1``; i <= d; i++) ` `        ``{ ` `            ``num = num + String.valueOf(val); ` `        ``} ` `    ``} ` `    ``return` `num; ` `} ` ` `  `// Driver function ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `s = ``25``, d = ``4``; ` ` `  `    ``System.out.print(findNumber(s, d)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to find the number having  ` `# sum of digits as s and d number of  ` `# digits such that the difference between  ` `# the maximum and the minimum digit  ` `# the minimum possible  ` `def` `findNumber(s, d) : ` ` `  `    ``# To store the final number  ` `    ``num ``=` `"";  ` ` `  `    ``# To store the value that is evenly  ` `    ``# distributed among all the digits  ` `    ``val ``=` `s ``/``/` `d;  ` ` `  `    ``# To store the remaining sum that still  ` `    ``# remains to be distributed among d digits  ` `    ``rem ``=` `s ``%` `d;  ` ` `  `    ``# rem stores the value that still remains  ` `    ``# to be distributed  ` `    ``# To keep the difference of digits minimum  ` `    ``# last rem digits are incremented by 1  ` `    ``for` `i ``in` `range``(``1``, d ``-` `rem ``+` `1``) : ` `        ``num ``=` `num ``+` `str``(val);  ` ` `  `    ``# In the last rem digits one is added to  ` `    ``# the value obtained by equal distribution  ` `    ``if` `(rem) : ` `        ``val ``+``=` `1``;  ` `        ``for` `i ``in` `range``(d ``-` `rem ``+` `1``, d ``+` `1``) : ` `            ``num ``=` `num ``+` `str``(val);  ` ` `  `    ``return` `num;  ` ` `  `# Driver function  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``s ``=` `25``; d ``=` `4``;  ` ` `  `    ``print``(findNumber(s, d));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to find the number having ` `    ``// sum of digits as s and d number of ` `    ``// digits such that the difference between ` `    ``// the maximum and the minimum digit ` `    ``// the minimum possible ` `    ``static` `String findNumber(``int` `s, ``int` `d) ` `    ``{ ` `        ``// To store the readonly number ` `        ``String num = ``""``; ` ` `  `        ``// To store the value that is evenly ` `        ``// distributed among all the digits ` `        ``int` `val = s / d; ` ` `  `        ``// To store the remaining sum that still ` `        ``// remains to be distributed among d digits ` `        ``int` `rem = s % d; ` ` `  `        ``int` `i; ` ` `  `        ``// rem stores the value that still remains ` `        ``// to be distributed ` `        ``// To keep the difference of digits minimum ` `        ``// last rem digits are incremented by 1 ` `        ``for` `(i = 1; i <= d - rem; i++)  ` `        ``{ ` `            ``num = num + String.Join(``""``, val); ` `        ``} ` ` `  `        ``// In the last rem digits one is added to ` `        ``// the value obtained by equal distribution ` `        ``if` `(rem > 0) ` `        ``{ ` `            ``val++; ` `            ``for` `(i = d - rem + 1; i <= d; i++) ` `            ``{ ` `                ``num = num + String.Join(``""``, val); ` `            ``} ` `        ``} ` `        ``return` `num; ` `    ``} ` ` `  `    ``// Driver function ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int` `s = 25, d = 4; ` ` `  `        ``Console.Write(findNumber(s, d)); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```6667
```

Time Complexity: O(d)
Auxiliary Space: O(1)

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Improved By : 29AjayKumar, AnkitRai01