Seeds (Or Seed Roots) of a number

04 May, 2021
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++
Java
Python3
C#
PHP
Javascript
// C++ program to find Seed of a number#include <bits/stdc++.h>using namespace std;const int MAX = 10000; int prodDig[MAX]; // Stores product of digits of x in prodDig[x]int getDigitProduct(int x){    // If x has single digit    if (x < 10)      return x;     // If digit product is already computed    if (prodDig[x] != 0)       return prodDig[x];     // If digit product is not computed before.    int prod = (x % 10) * getDigitProduct(x/10);     return (prodDig[x] = prod);} // Prints all seeds of nvoid findSeed(int n){    // Find all seeds using prodDig[]    vector<int> res;    for (int i=1; i<=n/2; i++)        if (i*getDigitProduct(i) == n)            res.push_back(i);     // If there was no seed    if (res.size() == 0)    {        cout << "NO seed exists\n";        return;    }     // Print seeds    for (int i=0; i<res.size(); i++)        cout << res[i] << " ";} // Driver codeint main(){    long long int n = 138;    findSeed(n);    return 0;}
// Java program to find Seed of a numberimport java.util.*; class GFg{static int MAX = 10000;static int[] prodDig=new int[MAX]; // Stores product of digits of x in prodDig[x]static int getDigitProduct(int x){    // If x has single digit    if (x < 10)    return x;     // If digit product is already computed    if (prodDig[x] != 0)    return prodDig[x];     // If digit product is not computed before.    int prod = (x % 10) * getDigitProduct(x/10);     return (prodDig[x] = prod);} // Prints all seeds of nstatic void findSeed(int n){    // Find all seeds using prodDig[]    List<Integer> res = new ArrayList<Integer>();    for (int i=1; i<=n/2; i++)        if (i*getDigitProduct(i) == n)            res.add(i);     // If there was no seed    if (res.size() == 0)    {        System.out.println("NO seed exists");        return;    }     // Print seeds    for (int i=0; i<res.size(); i++)        System.out.print(res.get(i)+" ");} // Driver codepublic static void main(String[] args){    int n = 138;    findSeed(n); }}// this code is contributed by mits
# Python3 program to find Seed of a number MAX = 10000; prodDig = [0] * MAX; # Stores product of digits of# x in prodDig[x]def getDigitProduct(x):         # If x has single digit    if (x < 10):        return x;     # If digit product is already computed    if (prodDig[x] != 0):        return prodDig[x];     # If digit product is not computed before.    prod = (int(x % 10) *            getDigitProduct(int(x / 10)));     prodDig[x] = prod;    return prod; # Prints all seeds of ndef findSeed(n):     # Find all seeds using prodDig[]    res = [];    for i in range(1, int(n / 2 + 2)):        if (i * getDigitProduct(i) == n):            res.append(i);     # If there was no seed    if (len(res) == 0):        print("NO seed exists");        return;     # Print seeds    for i in range(len(res)):        print(res[i], end = " "); # Driver coden = 138;findSeed(n); # This code is contributed by mits
// C# program to find Seed of a numberusing System;using System.Collections; class GFG{     static int MAX = 10000;static int[] prodDig=new int[MAX]; // Stores product of digits of x in prodDig[x]static int getDigitProduct(int x){    // If x has single digit    if (x < 10)    return x;     // If digit product is already computed    if (prodDig[x] != 0)    return prodDig[x];     // If digit product is not computed before.    int prod = (x % 10) * getDigitProduct(x/10);     return (prodDig[x] = prod);} // Prints all seeds of nstatic void findSeed(int n){    // Find all seeds using prodDig[]    ArrayList res = new ArrayList();    for (int i=1; i<=n/2; i++)        if (i*getDigitProduct(i) == n)            res.Add(i);     // If there was no seed    if (res.Count == 0)    {        Console.WriteLine("NO seed exists");        return;    }     // Print seeds    for (int i=0; i<res.Count; i++)        Console.WriteLine(res[i]+" ");} // Driver codestatic void Main(){    int n = 138;    findSeed(n); }}// this code is contributed by mits
<?php// PHP program to find Seed of a number $MAX = 10000; $prodDig = array_fill(0, $MAX, 0); // Stores product of digits of x in prodDig[x]function getDigitProduct($x){    global $prodDig;         // If x has single digit    if ($x < 10)    return $x;     // If digit product is already computed    if ($prodDig[$x] != 0)    return $prodDig[$x];     // If digit product is not computed before.    $prod = (int)($x % 10) *                  getDigitProduct((int)($x / 10));     $prodDig[$x] = $prod;    return $prod;} // Prints all seeds of nfunction findSeed($n){    // Find all seeds using prodDig[]    $res = array();    for ($i = 1; $i <= (int)($n / 2 + 1); $i++)        if ($i * getDigitProduct($i) == $n)            array_push($res, $i);     // If there was no seed    if (count($res) == 0)    {        echo "NO seed exists\n";        return;    }     // Print seeds    for ($i = 0; $i < count($res); $i++)        echo $res[$i] . " ";} // Driver code$n = 138;findSeed($n); // This code is contributed by mits?>
<script> // Javascript program to find Seed of a number   var MAX = 10000;var prodDig=Array.from({length: MAX}, (_, i) => 0); // Stores product of digits of x in prodDig[x]function getDigitProduct(x){    // If x has single digit    if (x < 10)    return x;     // If digit product is already computed    if (prodDig[x] != 0)    return prodDig[x];     // If digit product is not computed before.    var prod = (x % 10) *    getDigitProduct(parseInt(x/10));     return (prodDig[x] = prod);} // Prints all seeds of nfunction findSeed(n){    // Find all seeds using prodDig    var res = [];         for (var i=1; i<=parseInt(n/2); i++)        if (i*getDigitProduct(i) == n)            res.push(i);     // If there was no seed    if (res.length == 0)    {        document.write("NO seed exists");        return;    }     // Print seeds    for (i=0; i<res.length; i++)        document.write(res[i]+" ");} // Driver codevar n = 138;findSeed(n);  // This code is contributed by 29AjayKumar </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.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Mithun Kumar
29AjayKumar
number-digits
Mathematical
Mathematical
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Algorithm to solve Rubik's Cube
Merge two sorted arrays with O(1) extra space
Program to print prime numbers from 1 to N.
Find next greater number with same set of digits
Segment Tree | Set 1 (Sum of given range)
Check if a number is Palindrome
Count ways to reach the n'th stair
Count all possible paths from top left to bottom right of a mXn matrix
Fizz Buzz Implementation
Product of Array except itself