# 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 = 4Output:6667

The difference between maximum digit 7 and minimum digit 6 is 1.Input:S = 27, D = 3Output:999

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the

DSA Self Paced Courseat a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please referComplete Interview Preparation Course.In case you wish to attend

live classeswith experts, please referDSA Live Classes for Working ProfessionalsandCompetitive Programming Live for Students.

**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 <bits/stdc++.h>` `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` |

## Javascript

`<script>` `// Javascript 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` `function` `findNumber(s, d)` `{` ` ` `// To store the final number` ` ` `var` `num = [] ;` ` ` `// To store the value that is evenly` ` ` `// distributed among all the digits` ` ` `var` `val = parseInt(s / d);` ` ` `// To store the remaining sum that still` ` ` `// remains to be distributed among d digits` ` ` `var` `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` `(` `var` `i = 1; i <= d - rem; i++)` ` ` `{` ` ` ` ` `// num = num.concat(toString(val));` ` ` `num.push(val.toString());` ` ` `}` ` ` `// In the last rem digits one is added to` ` ` `// the value obtained by equal distribution` ` ` `if` `(rem != 0)` ` ` `{` ` ` `val++;` ` ` `for` `(` `var` `i = d - rem + 1; i <= d; i++)` ` ` `{` ` ` `// num = num + toString(val);` ` ` `num.push(val.toString());` ` ` `}` ` ` `}` ` ` `return` `num;` `}` `var` `s = 25, d = 4;` `var` `n=findNumber(s, d);` `for` `(` `var` `i = 0; i < n.length; i++)` `{` ` ` `document.write(n[i]);` `}` `// This code is contributed by SoumikMondal` `</script>` |

**Output:**

6667

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