Super Ugly Number (Number whose prime factors are in given set)

Super ugly numbers are positive numbers whose all prime factors are in the given prime list. Given a number n, the task is to find n’th Super Ugly number.

It may be assumed that given set of primes is sorted. Also, first Super Ugly number is 1 by convention.

Examples:

Input  : primes[] = [2, 5]
         n = 5
Output : 8
Super Ugly numbers with given prime factors 
are 1, 2, 4, 5, 8, ...
Fifth Super Ugly number is 8

Input  : primes[] = [2, 3, 5]
         n = 50
Output : 243

Input : primes[] = [3, 5, 7, 11, 13]
        n = 9
Output: 21

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

In our previous post we discussed about Ugly Number. This problem is basically extension of Ugly Numbers.

A simple solution for this problem is to one by one pick each number starting from 1 and find its all primes factors, if all prime factors lie in the given set of primes that means number is Super Ugly. Repeat this process until we get n’th Super Ugly Number .

An efficient solution for this problem is similar to Method-2 of Ugly Number. Here is the algorithm :

Let k be size of given array of prime numbers.
Declare a set for super ugly numbers.
Insert first ugly number (which is always 1) into set.
Initialize array multiple_of[k] of size k with 0. Each element of this array is iterator for corresponding prime in primes[k] array.
Initialize nextMultipe[k] array with primes[k]. This array behaves like next multiple variables of each prime in given primes[k] array i.e; nextMultiple[i] = primes[i] * ugly[++multiple_of[i]].
Now loop until there are n elements in set ugly.
a). Find minimum among current multiples of primes in nextMultiple[] array and insert it in the set of ugly numbers.
b). Then find this current minimum is multiple of which prime .
c). Increase iterator by 1 i.e; ++multiple_Of[i], for next multiple of current selected prime and update nextMultiple for it.

Below is implementation of above steps.

// C++ program to find n'th Super Ugly number 
#include<bits/stdc++.h> 
using namespace std; 
  
// Function to get the nth super ugly number 
// primes[]       --> given list of primes f size k 
// ugly           --> set which holds all super ugly 
//                    numbers from 1 to n 
// k              --> Size of prime[] 
int superUgly(int n, int primes[], int k) 
{ 
    // nextMultiple holds multiples of given primes 
    vector<int> nextMultiple(primes, primes+k); 
  
    // To store iterators of all primes 
    int multiple_Of[k]; 
    memset(multiple_Of, 0, sizeof(multiple_Of)); 
  
    // Create a set to store super ugly numbers and 
    // store first Super ugly number 
    set<int> ugly; 
    ugly.insert(1); 
  
    // loop until there are total n Super ugly numbers 
    // in set 
    while (ugly.size() != n) 
    { 
        // Find minimum element among all current 
        // multiples of given prime 
        int next_ugly_no = *min_element(nextMultiple.begin(), 
                                    nextMultiple.end()); 
  
        // insert this super ugly number in set 
        ugly.insert(next_ugly_no); 
  
        // loop to find current minimum is multiple 
        // of which prime 
        for (int j=0; j<k; j++) 
        { 
            if (next_ugly_no == nextMultiple[j]) 
            { 
                // increase iterator by one for next multiple 
                // of current prime 
                multiple_Of[j]++; 
  
                // this loop is similar to find  dp[++index[j]] 
                // it -->  dp[++index[j]] 
                set<int>::iterator it = ugly.begin(); 
                for (int i=1; i<=multiple_Of[j]; i++) 
                    it++; 
  
                nextMultiple[j] = primes[j] * (*it); 
                break; 
            } 
        } 
    } 
  
    // n'th super ugly number 
    set<int>::iterator it = ugly.end(); 
    it--; 
    return *it; 
} 
  
/* Driver program to test above functions */
int main() 
{ 
    int primes[] = {2,  5}; 
    int k = sizeof(primes)/sizeof(primes[0]); 
    int n = 5; 
    cout << superUgly(n, primes, k); 
    return 0; 
} 

Output:

References :
http://stackoverflow.com/questions/34103076/super-ugly-number

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Another method (using priority_queue)
Here we use min heap priority_queue.
The idea is to push the first ugly no. which is 1 into priority_queue and at every iteration take the top of priority_queue and push all the mulitples of that top into priotiy_queue.
Assuming a[] = {2, 3, 5},
so at first iteration 1 is top so 1 is popped and 1 * 2, 1 * 3, 1 * 5 is pushed.
At second iteration min is 2, so it is popped and 2 * 2, 2 * 3, 2 * 5 is pushed and so on.

// C++ program for super ugly number 
#include<bits/stdc++.h> 
using namespace std; 
//function will return the nth super ugly number 
int ugly(int a[], int size, int n){ 
      
    //n cannot be negative hence return -1 if n is 0 or -ve 
    if(n <= 0) 
        return -1; 
   
    if(n == 1) 
        return 1; 
      
    // Declare a min heap priority queue 
    priority_queue<int, vector<int>, greater<int>> pq; 
      
    // Push all the array elements to priority queue 
    for(int i = 0; i < size; i++){ 
        pq.push(a[i]); 
    } 
      
    // once count = n we return no 
    int count = 1, no; 
      
    while(count < n){ 
        // Get the minimum value from priority_queue 
        no = pq.top(); 
        pq.pop(); 
          
        // If top of pq is no then don't increment count. This to avoid duplicate counting of same no. 
        if(no != pq.top()) 
        { 
            count++; 
          
            //Push all the multiples of no. to priority_queue 
            for(int i = 0; i < size; i++){ 
                pq.push(no * a[i]); 
            //    cnt+=1; 
        } 
        } 
    } 
    // Return nth super ugly number 
    return no; 
} 
  
/* Driver program to test above functions */
int main(){ 
    int a[3] = {2, 3,5}; 
    int size = sizeof(a) / sizeof(a[0]); 
    cout << ugly(a, size, 1000)<<endl; 
    return 0; 
} 

Output:

51200000

Time Complexity: O(nlogn)
Auxiliary Space: O(n)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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My Personal Notes

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

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