# Seeds (Or Seed Roots) of a number

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // C++ program to find Seed of a number #include 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 n void findSeed(int n) {     // Find all seeds using prodDig[]     vector 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

## Java

 // Java program to find Seed of a number import 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 n static void findSeed(int n) {     // Find all seeds using prodDig[]     List 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.size() == 0)     {         System.out.println("NO seed exists");         return;     }        // Print seeds     for (int i=0; i

## Python3

 # 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 n def 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 code n = 138; findSeed(n);    # This code is contributed by mits

## C#

 // C# program to find Seed of a number using 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 n static 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

## PHP



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 contribute.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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Improved By : Mithun Kumar

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