A Seed of a number n is a number x such that multiplication of x with its digits is equal to n. The task is to find all seeds of a given number n. If no seed exists, then print the same.
Examples:
Input : n = 138
Output : 23
23 is a seed of 138 because
23*2*3 is equal to 138
Input : n = 4977
Output : 79 711
79 is a seed of 4977 because
79 * 7 * 9 = 4977.
711 is also a seed of 4977 because
711 * 1 * 1 * 7 = 4977
Input : n = 9
Output : No seed exists
Input : n = 738
Output : 123
Asked in Epic
The idea is to traverse all numbers from 1 to n/2. For every number being traversed, find product of digits with the number. An important optimization done in below program is to avoid re-computations of digit products. We store the products in an array. If a product has already been computed, we return it, else we compute it.
Below is the implementation of the idea.
C++
#include <bits/stdc++.h>
using namespace std;
const int MAX = 10000;
int prodDig[MAX];
int getDigitProduct( int x)
{
if (x < 10)
return x;
if (prodDig[x] != 0)
return prodDig[x];
int prod = (x % 10) * getDigitProduct(x/10);
return (prodDig[x] = prod);
}
void findSeed( int n)
{
vector< int > res;
for ( int i=1; i<=n/2; i++)
if (i*getDigitProduct(i) == n)
res.push_back(i);
if (res.size() == 0)
{
cout << "NO seed exists\n" ;
return ;
}
for ( int i=0; i<res.size(); i++)
cout << res[i] << " " ;
}
int main()
{
long long int n = 138;
findSeed(n);
return 0;
}
|
Java
import java.util.*;
class GFg{
static int MAX = 10000 ;
static int [] prodDig= new int [MAX];
static int getDigitProduct( int x)
{
if (x < 10 )
return x;
if (prodDig[x] != 0 )
return prodDig[x];
int prod = (x % 10 ) * getDigitProduct(x/ 10 );
return (prodDig[x] = prod);
}
static void findSeed( int n)
{
List<Integer> res = new ArrayList<Integer>();
for ( int i= 1 ; i<=n/ 2 ; i++)
if (i*getDigitProduct(i) == n)
res.add(i);
if (res.size() == 0 )
{
System.out.println( "NO seed exists" );
return ;
}
for ( int i= 0 ; i<res.size(); i++)
System.out.print(res.get(i)+ " " );
}
public static void main(String[] args)
{
int n = 138 ;
findSeed(n);
}
}
|
Python3
MAX = 10000 ;
prodDig = [ 0 ] * MAX ;
def getDigitProduct(x):
if (x < 10 ):
return x;
if (prodDig[x] ! = 0 ):
return prodDig[x];
prod = ( int (x % 10 ) *
getDigitProduct( int (x / 10 )));
prodDig[x] = prod;
return prod;
def findSeed(n):
res = [];
for i in range ( 1 , int (n / 2 + 2 )):
if (i * getDigitProduct(i) = = n):
res.append(i);
if ( len (res) = = 0 ):
print ( "NO seed exists" );
return ;
for i in range ( len (res)):
print (res[i], end = " " );
n = 138 ;
findSeed(n);
|
C#
using System;
using System.Collections;
class GFG{
static int MAX = 10000;
static int [] prodDig= new int [MAX];
static int getDigitProduct( int x)
{
if (x < 10)
return x;
if (prodDig[x] != 0)
return prodDig[x];
int prod = (x % 10) * getDigitProduct(x/10);
return (prodDig[x] = prod);
}
static void findSeed( int n)
{
ArrayList res = new ArrayList();
for ( int i=1; i<=n/2; i++)
if (i*getDigitProduct(i) == n)
res.Add(i);
if (res.Count == 0)
{
Console.WriteLine( "NO seed exists" );
return ;
}
for ( int i=0; i<res.Count; i++)
Console.WriteLine(res[i]+ " " );
}
static void Main()
{
int n = 138;
findSeed(n);
}
}
|
PHP
<?php
$MAX = 10000;
$prodDig = array_fill (0, $MAX , 0);
function getDigitProduct( $x )
{
global $prodDig ;
if ( $x < 10)
return $x ;
if ( $prodDig [ $x ] != 0)
return $prodDig [ $x ];
$prod = (int)( $x % 10) *
getDigitProduct((int)( $x / 10));
$prodDig [ $x ] = $prod ;
return $prod ;
}
function findSeed( $n )
{
$res = array ();
for ( $i = 1; $i <= (int)( $n / 2 + 1); $i ++)
if ( $i * getDigitProduct( $i ) == $n )
array_push ( $res , $i );
if ( count ( $res ) == 0)
{
echo "NO seed exists\n" ;
return ;
}
for ( $i = 0; $i < count ( $res ); $i ++)
echo $res [ $i ] . " " ;
}
$n = 138;
findSeed( $n );
?>
|
Javascript
<script>
var MAX = 10000;
var prodDig=Array.from({length: MAX}, (_, i) => 0);
function getDigitProduct(x)
{
if (x < 10)
return x;
if (prodDig[x] != 0)
return prodDig[x];
var prod = (x % 10) *
getDigitProduct(parseInt(x/10));
return (prodDig[x] = prod);
}
function findSeed(n)
{
var res = [];
for ( var i=1; i<=parseInt(n/2); i++)
if (i*getDigitProduct(i) == n)
res.push(i);
if (res.length == 0)
{
document.write( "NO seed exists" );
return ;
}
for (i=0; i<res.length; i++)
document.write(res[i]+ " " );
}
var n = 138;
findSeed(n);
</script>
|
Output :
23
Further Optimization :
We can further optimize above code. The idea is to make a call to getDigitProduct(i) only if i is divisible by n. Please refer https://ide.geeksforgeeks.org/oLYduu for implementation.
This article is contributed by Rakesh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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