Seeds (Or Seed Roots) of a number

1.5

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 C++ implementation of the idea.

// 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 n
void 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 code
int main()
{
    long long int n = 138;
    findSeed(n);
    return 0;
}

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 http://code.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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