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++
// 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; } |
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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<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 code public static void main(String[] args) { int n = 138; findSeed(n); } } // this code is contributed by mits |
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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 |
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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<res.Count; i++) Console.WriteLine(res[i]+" "); } // Driver code static void Main() { int n = 138; findSeed(n); } } // this code is contributed by mits |
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PHP
<?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 $prod; } // Prints all seeds of n function findSeed($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 ?> |
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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
