Find the Largest number with given number of digits and sum of digits

Given an integer s and d, The task is to find the largest number with given digit sum s and the number of digits d.

Examples:

Input: s = 9, d = 2
Output: 90

Input: s = 20, d = 3
Output: 992

Naive Approach:

Consider all m digit numbers and keep a max variable to store the maximum number with m digits and digit sum as s. 

Time complexity: O(10^m).
Auxiliary Space: O(1)

Find the Largest number with the given number of digits and sum of digits Greedy approach

Below is the idea to solve the problem:

The idea is to one by one fill all digits from leftmost to rightmost compare the remaining sum with 9 if the remaining sum is more than or equal to 9, 9 at the current position, else put the remaining sum. Since digits are filled from left to right, the highest digits will be placed on the left side, hence get the largest number y.

Illustration: 

Follow the below steps to Implement the idea:

• If s is zero 
  • if m=1 print 0
  • Else no such number is possible.
• If s > 9*m then no such number is possible.
• Run a for loop from 0 to m-1
  • If s>=9 subtract 9 from s and print 9.
  • Else print s and set s to 0.

Below is the Implementation of the above approach:

C++

// C++ program to find the largest number that can be
// formed from given sum of digits and number of digits.
#include <iostream>
using namespace std;

// Prints the smallest possible number with digit sum 's'
// and 'm' number of digits.
void findLargest(int m, int s)
{
    // If sum of digits is 0, then a number is possible
    // only if number of digits is 1.
    if (s == 0) {
        (m == 1) ? cout << "Largest number is " << 0
                 : cout << "Not possible";
        return;
    }

    // Sum greater than the maximum possible sum.
    if (s > 9 * m) {
        cout << "Not possible";
        return;
    }

    // Create an array to store digits of result
    int res[m];

    // Fill from most significant digit to least
    // significant digit.
    for (int i = 0; i < m; i++) {
        // Fill 9 first to make the number largest
        if (s >= 9) {
            res[i] = 9;
            s -= 9;
        }

        // If remaining sum becomes less than 9, then
        // fill the remaining sum
        else {
            res[i] = s;
            s = 0;
        }
    }

    cout << "Largest number is ";
    for (int i = 0; i < m; i++)
        cout << res[i];
}

// Driver code
int main()
{
    int s = 9, m = 2;
    findLargest(m, s);
    return 0;
}

C

// C program to find the largest number that can be
// formed from given sum of digits and number of digits.
#include <stdio.h>

// Prints the smallest possible number with digit sum 's'
// and 'm' number of digits.
void findLargest(int m, int s)
{
    // If sum of digits is 0, then a number is possible
    // only if number of digits is 1.
    if (s == 0) {
        (m == 1) ? printf("Largest number is 0")
                 : printf("Not possible");
        return;
    }

    // Sum greater than the maximum possible sum.
    if (s > 9 * m) {
        printf("Not possible");
        return;
    }

    // Create an array to store digits of result
    int res[m];

    // Fill from most significant digit to least
    // significant digit.
    for (int i = 0; i < m; i++) {
        // Fill 9 first to make the number largest
        if (s >= 9) {
            res[i] = 9;
            s -= 9;
        }

        // If remaining sum becomes less than 9, then
        // fill the remaining sum
        else {
            res[i] = s;
            s = 0;
        }
    }

    printf("Largest number is ");
    for (int i = 0; i < m; i++)
        printf("%d", res[i]);
}

// Driver code
int main()
{
    int s = 9, m = 2;
    findLargest(m, s);
    return 0;
}

// This code is contributed by Sania Kumari Gupta

Java

// Java program to find the largest number that can be
// formed from given sum of digits and number of digits

class GFG
{
    // Function to print the largest possible number with digit sum 's'
    // and 'm' number of digits
    static void findLargest(int m, int s)
    {
        // If sum of digits is 0, then a number is possible
        // only if number of digits is 1
        if (s == 0)
        {
            System.out.print(m == 1 ? "Largest number is 0" : "Not possible");

            return ;
        }

        // Sum greater than the maximum possible sum
        if (s > 9*m)
        {
            System.out.println("Not possible");
            return ;
        }

        // Create an array to store digits of result
        int[] res = new int[m];

        // Fill from most significant digit to least
        // significant digit
        for (int i=0; i<m; i++)
        {
            // Fill 9 first to make the number largest
            if (s >= 9)
            {
                res[i] = 9;
                s -= 9;
            }

            // If remaining sum becomes less than 9, then
            // fill the remaining sum
            else
            {
                res[i] = s;
                s = 0;
            }
        }

        System.out.print("Largest number is ");
        for (int i=0; i<m; i++)
            System.out.print(res[i]);
    }

    // driver program
    public static void main (String[] args)
    {
        int s = 9, m = 2;
        findLargest(m, s);
    }
}

// Contributed by Pramod Kumar

Python3

# Python 3 program to find
# the largest number that
# can be formed from given
# sum of digits and number
# of digits.


# Prints the smallest
# possible number with digit
# sum 's' and 'm' number of
# digits.
def findLargest( m, s) :

    # If sum of digits is 0,
    # then a number is possible
    # only if number of digits
    # is 1.
    if (s == 0) :

        if(m == 1) :
            print("Largest number is " , "0",end = "")
        else :
            print("Not possible",end = "")

        return

    # Sum greater than the
    # maximum possible sum.
    if (s > 9 * m) :
        print("Not possible",end = "")
        return

    # Create an array to
    # store digits of
    # result
    res = [0] * m

    # Fill from most significant
    # digit to least significant
    # digit.
    for i in range(0, m) :

        # Fill 9 first to make
        # the number largest
        if (s >= 9) :
            res[i] = 9
            s = s - 9

        # If remaining sum
        # becomes less than
        # 9, then fill the
        # remaining sum
        else :
            res[i] = s
            s = 0


    print( "Largest number is ",end = "")

    for i in range(0, m) :
        print(res[i],end = "")

# Driver code
s = 9
m = 2
findLargest(m, s)

# This code is contributed by Nikita Tiwari.

C#

// C# program to find the
// largest number that can
// be formed from given sum
// of digits and number of digits
using System;

class GFG
{

    // Function to print the
    // largest possible number
    // with digit sum 's' and
    // 'm' number of digits
    static void findLargest(int m, int s)
    {
        // If sum of digits is 0,
        // then a number is possible
        // only if number of digits is 1
        if (s == 0)
        {
            Console.Write(m == 1 ?
                   "Largest number is 0" :
                          "Not possible");

            return ;
        }

        // Sum greater than the
        // maximum possible sum
        if (s > 9 * m)
        {
            Console.WriteLine("Not possible");
            return ;
        }

        // Create an array to
        // store digits of result
        int []res = new int[m];

        // Fill from most significant
        // digit to least significant digit
        for (int i = 0; i < m; i++)
        {
            // Fill 9 first to make
            // the number largest
            if (s >= 9)
            {
                res[i] = 9;
                s -= 9;
            }

            // If remaining sum becomes
            // less than 9, then
            // fill the remaining sum
            else
            {
                res[i] = s;
                s = 0;
            }
        }

        Console.Write("Largest number is ");
        for (int i = 0; i < m; i++)
            Console.Write(res[i]);
    }

    // Driver Code
    static public void Main ()
    {
        int s = 9, m = 2;
        findLargest(m, s);
    }
}

// This code is Contributed by ajit

PHP

<?php
// PHP program to find the largest
// number that can be formed from
// given sum of digits and number
// of digits.

// Prints the smallest possible
// number with digit sum 's'
// and 'm' number of digits.
function findLargest($m, $s)
{
    // If sum of digits is 0, then
    // a number is possible only if
    // number of digits is 1.
    if ($s == 0)
    {
        if(($m == 1) == true)
            echo "Largest number is " , 0;
        else
            echo "Not possible";
        return ;
    }

    // Sum greater than the
    // maximum possible sum.
    if ($s > 9 * $m)
    {
        echo "Not possible";
        return ;
    }

    // Create an array to store
    // digits of result Fill from
    // most significant digit to
    // least significant digit.
    for ($i = 0; $i < $m; $i++)
    {
        // Fill 9 first to make
        // the number largest
        if ($s >= 9)
        {
            $res[$i] = 9;
            $s -= 9;
        }

        // If remaining sum becomes
        // less than 9, then fill
        // the remaining sum
        else
        {
            $res[$i] = $s;
            $s = 0;
        }
    }

    echo "Largest number is ";
    for ($i = 0; $i < $m; $i++)
        echo $res[$i];
}

// Driver code
$s = 9; $m = 2;
findLargest($m, $s);

// This code is contributed by m_kit
?>

Javascript

<script>
// Javascript program to find the largest number that can be
// formed from given sum of digits and number of digits.

// Prints the smallest possible number with digit sum 's'
// and 'm' number of digits.
function findLargest(m, s)
{
    // If sum of digits is 0, then a number is possible
    // only if number of digits is 1.
    if (s == 0)
    {
        (m == 1)? document.write("Largest number is " + 0)
                    : document.write("Not possible");
        return ;
    }

    // Sum greater than the maximum possible sum.
    if (s > 9*m)
    {
        document.write("Not possible");
        return ;
    }

    // Create an array to store digits of result
    let res = new Array(m);

    // Fill from most significant digit to least
    // significant digit.
    for (let i=0; i<m; i++)
    {
        // Fill 9 first to make the number largest
        if (s >= 9)
        {
            res[i] = 9;
            s -= 9;
        }

        // If remaining sum becomes less than 9, then
        // fill the remaining sum
        else
        {
            res[i] = s;
            s = 0;
        }
    }

    document.write("Largest number is ");
    for (let i=0; i<m; i++)
        document.write(res[i]);
}

// Driver code
    let s = 9, m = 2;
    findLargest(m, s);

// This code is contributed by Mayank Tyagi

</script>

Output

Largest number is 90

Time Complexity of this solution is O(m).
Auxiliary Space: O(m), where m is the given integer.

Approach: Greedy Algorithm

• Create an empty string to store the result
• If d is 1, append s to the result and return it
• Loop from the leftmost digit to the rightmost digit
  a. If the remaining sum of digits is greater than or equal to 9, append 9 to the result and subtract 9 from the remaining sum of digits
  b. If the remaining sum of digits is less than 9, append the remaining sum of digits to the result and fill the remaining digits with 0s
• Return the result

C++

#include <iostream>
#include <string>

using namespace std;

int largest_number(int s, int d) {
    if (s == 0) {
        return 0;
    }
    if (s > 9 * d) {
        return -1;
    }
    string result = "";
    for (int i = 0; i < d; i++) {
        if (s >= 9) {
            result += '9';
            s -= 9;
        } else {
            result += to_string(s);
            s = 0;
        }
        if (s == 0 && i < d-1) {
            result += string(d-i-1, '0');
            break;
        }
    }
    return stoi(result);
}

int main() {
    // Test case 1
    cout << largest_number(9, 2) << endl; // Output: 90

    // Test case 2
    cout << largest_number(20, 3) << endl; // Output: 992

    return 0;
}

Java

import java.util.*;

public class Main {

    public static int largest_number(int s, int d)
    {
        // If s is 0, then the largest number is 0.
        if (s == 0) {
            return 0;
        }
        // If s is greater than 9 times d, then it is
        // impossible to form a d-digit number whose sum of
        // digits is s.
        if (s > 9 * d) {
            return -1;
        }
        // Initialize an empty string to store the result.
        String result = "";
        // Loop through each digit of the number.
        for (int i = 0; i < d; i++) {
            // If s is greater than or equal to 9, then add
            // 9 to the result and subtract 9 from s.
            if (s >= 9) {
                result += '9';
                s -= 9;
            }
            // Otherwise, add s to the result and set s to
            // 0.
            else {
                result += Integer.toString(s);
                s = 0;
            }
            // If s is 0 and there are still digits left to
            // fill, then fill the remaining digits with 0s
            // and break out of the loop.
            if (s == 0 && i < d - 1) {
                result += String.join(
                    "",
                    Collections.nCopies(d - i - 1, "0"));
                break;
            }
        }
        // Convert the result to an integer and return it.
        return Integer.parseInt(result);
    }

    public static void main(String[] args)
    {
        // Test case 1
        System.out.println(
            largest_number(9, 2)); // Output: 90

        // Test case 2
        System.out.println(
            largest_number(20, 3)); // Output: 992
    }
}

Python3

def largest_number(s, d):
    if s == 0:
        return 0
    if s > 9 * d:
        return -1
    result = ''
    for i in range(d):
        if s >= 9:
            result += '9'
            s -= 9
        else:
            result += str(s)
            s = 0
        if s == 0 and i < d-1:
            result += '0' * (d-i-1)
            break
    return int(result)

# Test case 1
print(largest_number(9, 2)) # Output: 90

# Test case 2
print(largest_number(20, 3)) # Output: 992

C#

using System;

class Program {
    static int LargestNumber(int s, int d) {
        if (s == 0) {
            return 0;
        }
        if (s > 9 * d) {
            return -1;
        }
        string result = "";
        for (int i = 0; i < d; i++) {
            if (s >= 9) {
                result += "9";
                s -= 9;
            } else {
                result += s.ToString();
                s = 0;
            }
            if (s == 0 && i < d - 1) {
                result += new string('0', d - i - 1);
                break;
            }
        }
        return int.Parse(result);
    }

    static void Main(string[] args) {
        // Test case 1
        Console.WriteLine(LargestNumber(9, 2)); // Output: 90

        // Test case 2
        Console.WriteLine(LargestNumber(20, 3)); // Output: 992
    }
}

Javascript

function largestNumber(s, d) {
if (s == 0) {
return 0;
}
if (s > 9 * d) {
return -1;
}
let result = "";
for (let i = 0; i < d; i++) {
if (s >= 9) {
result += "9";
s -= 9;
} else {
result += s.toString();
s = 0;
}
if (s == 0 && i < d - 1) {
result += "0".repeat(d - i - 1);
break;
}
}
return parseInt(result);
}

// Test cases
console.log(largestNumber(9, 2)); // Output: 90
console.log(largestNumber(20, 3)); // Output: 992

Output

90
992

Time complexity: O(d),
Auxiliary space: O(d)

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