Smallest number k such that the product of digits of k is equal to n

Given a non-negative number n. The problem is to find the smallest number k such that the product of digits of k is equal to n. If no such number k can be formed then print “-1”.
Examples:

Input : 100
Output : 455
Explanation: 4*5*5 = 100 and 455 is the
smallest possible number.

Input : 26
Output : -1

Source: Asked in Amazon Interview

Approach: For each i = 9 to 2, repeatedly divide n by i until it cannot be further divided or the list of numbers from 9 to 2 gets finished. Also, in the process of division push each digit i onto the stack which divides n completely. After the above process gets completed check whether n == 1 or not. If not, then print “-1”, else form the number k using the digits from the stack containing the digits in the same sequence as popped from the stack. 

C++

// C++ implementation to find smallest number k such that
// the product of digits of k is equal to n
#include <bits/stdc++.h>

using namespace std;

// function to find smallest number k such that
// the product of digits of k is equal to n
long long int smallestNumber(int n)
{
    // if 'n' is a single digit number, then
    // it is the required number
    if (n >= 0 && n <= 9)
        return n;

    // stack the store the digits
    stack<int> digits;

    // repeatedly divide 'n' by the numbers
    // from 9 to 2 until all the numbers are
    // used or 'n' > 1
    for (int i=9; i>=2 && n > 1; i--)
    {
        while (n % i == 0)
        {
            // save the digit 'i' that divides 'n'
            // onto the stack
            digits.push(i);
            n = n / i;
        }
    }

    // if true, then no number 'k' can be formed
    if (n != 1)
        return -1;

    // pop digits from the stack 'digits'
    // and add them to 'k'
    long long int k = 0;
    while (!digits.empty())
    {
        k = k*10 + digits.top();
        digits.pop();
    }

    // required smallest number
    return k;
}

// Driver program to test above
int main()
{
    int n = 100;
    cout << smallestNumber(n);
    return 0;
}

Java

//Java implementation to find smallest number k such that
// the product of digits of k is equal to n
import java.util.Stack;

public class GFG {

// function to find smallest number k such that
// the product of digits of k is equal to n
    static long smallestNumber(int n) {
        // if 'n' is a single digit number, then
        // it is the required number
        if (n >= 0 && n <= 9) {
            return n;
        }

        // stack the store the digits
        Stack<Integer> digits = new Stack<>();

        // repeatedly divide 'n' by the numbers
        // from 9 to 2 until all the numbers are
        // used or 'n' > 1
        for (int i = 9; i >= 2 && n > 1; i--) {
            while (n % i == 0) {
                // save the digit 'i' that divides 'n'
                // onto the stack
                digits.push(i);
                n = n / i;
            }
        }

        // if true, then no number 'k' can be formed
        if (n != 1) {
            return -1;
        }

        // pop digits from the stack 'digits'
        // and add them to 'k'
        long k = 0;
        while (!digits.empty()) {
            k = k * 10 + digits.peek();
            digits.pop();
        }

        // required smallest number
        return k;
    }

// Driver program to test above
    static public void main(String[] args) {
        int n = 100;
        System.out.println(smallestNumber(n));
    }
}

/*This code is contributed by PrinciRaj1992*/

Python3

# Python3 implementation to find smallest
# number k such that the product of digits
# of k is equal to n
import math as mt

# function to find smallest number k such that
# the product of digits of k is equal to n
def smallestNumber(n):

    # if 'n' is a single digit number, then
    # it is the required number
    if (n >= 0 and n <= 9):
        return n

    # stack the store the digits
    digits = list()

    # repeatedly divide 'n' by the numbers
    # from 9 to 2 until all the numbers are
    # used or 'n' > 1
    for i in range(9,1, -1):

        while (n % i == 0):

            # save the digit 'i' that
            # divides 'n' onto the stack
            digits.append(i)
            n = n //i

    # if true, then no number 'k'
    # can be formed
    if (n != 1):
        return -1

    # pop digits from the stack 'digits'
    # and add them to 'k'
    k = 0
    while (len(digits) != 0):

        k = k * 10 + digits[-1]
        digits.pop()

    # required smallest number
    return k

# Driver Code
n = 100
print(smallestNumber(n))

# This code is contributed by
# Mohit kumar 29

C#

// C# implementation to find smallest number k such that
// the product of digits of k is equal to n
using System;
using System.Collections.Generic;
public class GFG {

// function to find smallest number k such that
// the product of digits of k is equal to n
    static long smallestNumber(int n) {
        // if 'n' is a single digit number, then
        // it is the required number
        if (n >= 0 && n <= 9) {
            return n;
        }

        // stack the store the digits
        Stack<int> digits = new Stack<int>();

        // repeatedly divide 'n' by the numbers
        // from 9 to 2 until all the numbers are
        // used or 'n' > 1
        for (int i = 9; i >= 2 && n > 1; i--) {
            while (n % i == 0) {
                // save the digit 'i' that divides 'n'
                // onto the stack
                digits.Push(i);
                n = n / i;
            }
        }

        // if true, then no number 'k' can be formed
        if (n != 1) {
            return -1;
        }

        // pop digits from the stack 'digits'
        // and add them to 'k'
        long k = 0;
        while (digits.Count!=0) {
            k = k * 10 + digits.Peek();
            digits.Pop();
        }

        // required smallest number
        return k;
    }

// Driver program to test above
    static public void Main() {
        int n = 100;
        Console.Write(smallestNumber(n));
    }
}

/*This code is contributed by Rajput-Ji*/

PHP

<?php
// PHP implementation to find smallest number k such that
// the product of digits of k is equal to n

// function to find smallest number k such that
// the product of digits of k is equal to n
function smallestNumber($n)
{
    // if 'n' is a single digit number, then
    // it is the required number
    if ($n >= 0 && $n <= 9)
        return $n;

    // stack the store the digits
    $digits = array();

    // repeatedly divide 'n' by the numbers
    // from 9 to 2 until all the numbers are
    // used or 'n' > 1
    for ($i = 9; $i >= 2 && $n > 1; $i--)
    {
        while ($n % $i == 0)
        {
            // save the digit 'i' that divides 'n'
            // onto the stack
            array_push($digits,$i);
            $n =(int)( $n / $i);
        }
    }

    // if true, then no number 'k' can be formed
    if ($n != 1)
        return -1;

    // pop digits from the stack 'digits'
    // and add them to 'k'
    $k = 0;
    while (!empty($digits))
        $k = $k * 10 + array_pop($digits);

    // required smallest number
    return $k;
}

    // Driver code
    $n = 100;
    echo smallestNumber($n);

// This code is contributed by mits
?>

Javascript

<script>

// Javascript implementation to find
// smallest number k such that
// the product of digits of k is equal to n

    // function to find smallest number k such that
// the product of digits of k is equal to n
    function smallestNumber(n)
    {
        // if 'n' is a single digit number, then
        // it is the required number
        if (n >= 0 && n <= 9) {
            return n;
        }

        // stack the store the digits
        let digits = [];

        // repeatedly divide 'n' by the numbers
        // from 9 to 2 until all the numbers are
        // used or 'n' > 1
        for (let i = 9; i >= 2 && n > 1; i--) {
            while (n % i == 0) {
                // save the digit 'i' that divides 'n'
                // onto the stack
                digits.push(i);
                n = Math.floor(n / i);
            }
        }

        // if true, then no number 'k' can be formed
        if (n != 1) {
            return -1;
        }

        // pop digits from the stack 'digits'
        // and add them to 'k'
        let k = 0;
        while (digits.length!=0) {
            k = k * 10 + digits[digits.length-1];
            digits.pop();
        }

        // required smallest number
        return k;
    }

    // Driver program to test above
    let n = 100;
    document.write(smallestNumber(n));


// This code is contributed by patel2127

</script>

Output

455

Time Complexity: O(log N)
Space Complexity: O(log N)

We can store the required number k in string for large numbers as shown below.

Also, the above approach can be space optimized if we store our answer directly in a string and return the reverse of it as the final answer.

C++

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

string getSmallest(ll N) {

    string ans;

    for(int i=9;i>=2 && N>1;i--)
    {
        while(N%i==0)
        {
            ans.push_back(i+48);
            N/=i;
        }
    }

    if(N!=1)
    return "-1";
    else if(ans.length()==0)
    return "1";

    reverse(ans.begin(),ans.end());

    return ans;
    }

// driver's code
int main()
{
    ll N=100;
    cout<<getSmallest(N);
    return 0;
}
// this code is contributed by prophet1999

Java

import java.util.*;

public class Main {
    public static String getSmallest(long N) {
        String ans = "";
        for (int i = 9; i >= 2 && N > 1; i--) {
            while (N % i == 0) {
                ans += (char)(i + '0');
                N /= i;
            }
        }
        if (N != 1) {
            return "-1";
        } else if (ans.length() == 0) {
            return "1";
        }
        return new StringBuilder(ans).reverse().toString();
    }

    public static void main(String[] args) {
        long N = 100;
        System.out.println(getSmallest(N));
    }
}

Python3

def getSmallest(N):
    ans = ""

    for i in range(9, 1, -1):
        while N > 1 and N % i == 0:
            ans += str(i)
            N //= i

    if N != 1:
        return "-1"
    elif len(ans) == 0:
        return "1"

    return ans[::-1]

# driver's code
if __name__ == '__main__':
    N = 100
    print(getSmallest(N))

C#

using System;

public class Program
{
  static string GetSmallest(int N)
  {
    string ans = "";

    for (int i = 9; i > 1; i--)
    {
      while (N > 1 && N % i == 0)
      {
        ans += i.ToString();
        N /= i;
      }
    }

    if (N != 1)
    {
      return "-1";
    }
    else if (ans.Length == 0)
    {
      return "1";
    }

    char[] charArray = ans.ToCharArray();
    Array.Reverse(charArray);
    return new string(charArray);
  }

  // driver's code
  public static void Main()
  {
    int N = 100;
    Console.WriteLine(GetSmallest(N));
  }
}

// this code is contributed by shivhack999

Javascript

function getSmallest(N) {
  let ans = "";

  for (let i = 9; i > 1; i--) {
    while (N > 1 && N % i === 0) {
      ans += i.toString();
      N /= i;
    }
  }

  if (N !== 1) {
    return "-1";
  } else if (ans.length === 0) {
    return "1";
  }

  return ans.split("").reverse().join("");
}

// driver's code
const N = 100;
console.log(getSmallest(N));

Output

455

Time Complexity: O(log N)

Auxiliary Space: O(1)

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