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++
#include <bits/stdc++.h>
using namespace std;
long long int smallestNumber( int n)
{
if (n >= 0 && n <= 9)
return n;
stack< int > digits;
for ( int i=9; i>=2 && n > 1; i--)
{
while (n % i == 0)
{
digits.push(i);
n = n / i;
}
}
if (n != 1)
return -1;
long long int k = 0;
while (!digits.empty())
{
k = k*10 + digits.top();
digits.pop();
}
return k;
}
int main()
{
int n = 100;
cout << smallestNumber(n);
return 0;
}
|
Java
import java.util.Stack;
public class GFG {
static long smallestNumber( int n) {
if (n >= 0 && n <= 9 ) {
return n;
}
Stack<Integer> digits = new Stack<>();
for ( int i = 9 ; i >= 2 && n > 1 ; i--) {
while (n % i == 0 ) {
digits.push(i);
n = n / i;
}
}
if (n != 1 ) {
return - 1 ;
}
long k = 0 ;
while (!digits.empty()) {
k = k * 10 + digits.peek();
digits.pop();
}
return k;
}
static public void main(String[] args) {
int n = 100 ;
System.out.println(smallestNumber(n));
}
}
|
Python3
import math as mt
def smallestNumber(n):
if (n > = 0 and n < = 9 ):
return n
digits = list ()
for i in range ( 9 , 1 , - 1 ):
while (n % i = = 0 ):
digits.append(i)
n = n / / i
if (n ! = 1 ):
return - 1
k = 0
while ( len (digits) ! = 0 ):
k = k * 10 + digits[ - 1 ]
digits.pop()
return k
n = 100
print (smallestNumber(n))
|
C#
using System;
using System.Collections.Generic;
public class GFG {
static long smallestNumber( int n) {
if (n >= 0 && n <= 9) {
return n;
}
Stack< int > digits = new Stack< int >();
for ( int i = 9; i >= 2 && n > 1; i--) {
while (n % i == 0) {
digits.Push(i);
n = n / i;
}
}
if (n != 1) {
return -1;
}
long k = 0;
while (digits.Count!=0) {
k = k * 10 + digits.Peek();
digits.Pop();
}
return k;
}
static public void Main() {
int n = 100;
Console.Write(smallestNumber(n));
}
}
|
PHP
<?php
function smallestNumber( $n )
{
if ( $n >= 0 && $n <= 9)
return $n ;
$digits = array ();
for ( $i = 9; $i >= 2 && $n > 1; $i --)
{
while ( $n % $i == 0)
{
array_push ( $digits , $i );
$n =(int)( $n / $i );
}
}
if ( $n != 1)
return -1;
$k = 0;
while (! empty ( $digits ))
$k = $k * 10 + array_pop ( $digits );
return $k ;
}
$n = 100;
echo smallestNumber( $n );
?>
|
Javascript
<script>
function smallestNumber(n)
{
if (n >= 0 && n <= 9) {
return n;
}
let digits = [];
for (let i = 9; i >= 2 && n > 1; i--) {
while (n % i == 0) {
digits.push(i);
n = Math.floor(n / i);
}
}
if (n != 1) {
return -1;
}
let k = 0;
while (digits.length!=0) {
k = k * 10 + digits[digits.length-1];
digits.pop();
}
return k;
}
let n = 100;
document.write(smallestNumber(n));
</script>
|
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;
}
int main()
{
ll N=100;
cout<<getSmallest(N);
return 0;
}
|
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 ]
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);
}
public static void Main()
{
int N = 100;
Console.WriteLine(GetSmallest(N));
}
}
|
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( "" );
}
const N = 100;
console.log(getSmallest(N));
|
Time Complexity: O(log N)
Auxiliary Space: O(1)
Last Updated :
28 Mar, 2023
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