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

• Difficulty Level : Medium
• Last Updated : 09 Aug, 2022

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

Recommended Practice

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 ` `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 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

 `= 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

 ``

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++14

 `#include ``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<

Output

`455`

Time Complexity: O(log N)

Auxiliary Space: O(1)

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