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 https://ide.geeksforgeeks.org/oLYduu for implementation.

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