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

Last Updated : 21 Sep, 2022

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

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# 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>