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 the nth Super Ugly number.
It may be assumed that a given set of primes is sorted. Also, the 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
In our previous post, we discussed Ugly Number. This problem is basically an extension of Ugly Numbers.
A simple solution for this problem is to one by one pick each number starting from 1 and find it’s all primes factors, if all prime factors lie in the given set of primes that means the number is Super Ugly. Repeat this process until we get nth Super Ugly Number.
An efficient solution for this problem is similar to Method-2 of Ugly Number. Here is the algorithm:
- Let k be the size of a given array of prime numbers.
- Declare a set for super ugly numbers.
- Insert the first ugly number (which is always 1) into the set.
- Initialize array multiple_of[k] of size k with 0. Each element of this array is an iterator for the 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 the implementation of the above steps.
CPP
// 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; } |
Java
import java.util.*; class Main { public static void main(String[] args) { int[] primes = { 2, 5 }; int k = primes.length; int n = 5; System.out.println(superUgly(n, primes, k)); } // 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[] public static int superUgly(int n, int[] primes, int k) { // nextMultiple holds multiples of given primes int[] nextMultiple = Arrays.copyOf(primes, k); // To store iterators of all primes int[] multiple_Of = new int[k]; Arrays.fill(multiple_Of, 0); // Create a set to store super ugly numbers and // store first Super ugly number Set<Integer> ugly = new HashSet<>(); ugly.add(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 = Integer.MAX_VALUE; for (int i = 0; i < k; i++) { next_ugly_no = Math.min(next_ugly_no, nextMultiple[i]); } // insert this super ugly number in set ugly.add(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]] List<Integer> uglyList = new ArrayList<>(ugly); int it = uglyList.get(multiple_Of[j] - 1); nextMultiple[j] = primes[j] * it; break; } } } // n'th super ugly number List<Integer> uglyList = new ArrayList<>(ugly); return uglyList.get(uglyList.size() - 1); } } // This code is contributed by divyansh2212 |
Python3
# Python3 program to find n'th Super Ugly number from typing import List # 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[] def superUgly(n: int, primes: List[int], k: int) -> int: # nextMultiple holds multiples of given primes nextMultiple = primes[:] # To store iterators of all primes multiple_Of = [0] * k # Create a set to store super ugly numbers and # store first Super ugly number ugly = set([1]) # loop until there are total n Super ugly numbers # in set while len(ugly) != n: # Find minimum element among all current # multiples of given prime next_ugly_no = min(nextMultiple) # insert this super ugly number in set ugly.add(next_ugly_no) # loop to find current minimum is multiple # of which prime for j in range(k): if next_ugly_no == nextMultiple[j]: # increase iterator by one for next multiple # of current prime multiple_Of[j] += 1 # this loop is similar to find dp[++index[j]] # it --> dp[++index[j]] it = sorted(ugly)[multiple_Of[j]] nextMultiple[j] = primes[j] * it break # n'th super ugly number return max(ugly) # Driver program to test above functions if __name__ == '__main__': primes = [2, 5] k = len(primes) n = 5 print(superUgly(n, primes, k)) # This code is contributed by lokeshpotta20. |
C#
using System; using System.Collections.Generic; class MainClass { public static void Main(string[] args) { int[] primes = { 2, 5 }; int k = primes.Length; int n = 5; Console.WriteLine(superUgly(n, primes, k)); } public static int superUgly(int n, int[] primes, int k) { // nextMultiple holds multiples of given primes int[] nextMultiple = new int[k]; Array.Copy(primes, nextMultiple, k); // To store iterators of all primes int[] multiple_Of = new int[k]; Array.Fill(multiple_Of, 0); // Create a set to store super ugly numbers and // store first Super ugly number HashSet<int> ugly = new HashSet<int>(); ugly.Add(1); // loop until there are total n Super ugly numbers // in set while (ugly.Count != n) { // Find minimum element among all current // multiples of given prime int next_ugly_no = Int32.MaxValue; for (int i = 0; i < k; i++) { next_ugly_no = Math.Min(next_ugly_no, nextMultiple[i]); } // insert this super ugly number in set ugly.Add(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]++; List<int> uglyList = new List<int>(ugly); int it = uglyList[multiple_Of[j] - 1]; nextMultiple[j] = primes[j] * it; break; } } } List<int> finalUglyList = new List<int>(ugly); return finalUglyList[finalUglyList.Count - 1]; } } |
Javascript
function superUgly(n, primes, k) { // nextMultiple holds multiples of given primes let nextMultiple = primes.slice(); // To store iterators of all primes let multipleOf = new Array(k).fill(0); // Create a set to store super ugly numbers and // store first Super ugly number let ugly = new Set(); ugly.add(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 let nextUglyNo = Math.min(...nextMultiple); // insert this super ugly number in set ugly.add(nextUglyNo); // loop to find current minimum is multiple // of which prime for (let j = 0; j < k; j++) { if (nextUglyNo === nextMultiple[j]) { // increase iterator by one for next multiple // of current prime multipleOf[j]++; // this loop is similar to find dp[++index[j]] // it --> dp[++index[j]] let it = ugly.values(); for (let i = 1; i <= multipleOf[j]; i++) it.next(); nextMultiple[j] = primes[j] * it.next().value; break; } } } // n'th super ugly number let result = Array.from(ugly).pop(); return result; } // Driver program let primes = [2, 5]; let k = primes.length; let n = 5; console.log(superUgly(n, primes, k)); // Output: 8 |
Output
8
This article is contributed by Shashank Mishra .
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 multiples of that top into priority_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.
Implementation:
CPP
// 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, 10)<<endl; return 0; } |
Java
import java.util.PriorityQueue; public class SuperUglyNumber { // function will return the nth super ugly number public static 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 PriorityQueue<Integer> pq = new PriorityQueue<Integer>(); // Push all the array elements to priority queue for(int i = 0; i < size; i++){ pq.add(a[i]); } // once count = n we return no int count = 1, no = 0; while(count < n){ // Get the minimum value from priority_queue no = pq.poll(); // If top of pq is no then don't increment count. // This to avoid duplicate counting of same no. if(no != pq.peek()) { count++; // Push all the multiples of no. to priority_queue for(int i = 0; i < size; i++){ pq.add(no * a[i]); } } } // Return nth super ugly number return no; } /* Driver program to test above functions */ public static void main(String[] args) { int[] a = {2, 3, 5}; int size = a.length; System.out.println(ugly(a, size, 10)); } } |
Python3
# Python3 program for super ugly number # function will return the nth super ugly number def ugly(a, size, 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 pq = [] # Push all the array elements to priority queue for i in range(size): pq.append(a[i]) # once count = n we return no count = 1 no = 0 pq = sorted(pq) while(count < n): # sorted(pq) # Get the minimum value from priority_queue no = pq[0] del pq[0] # If top of pq is no then don't increment count. # This to avoid duplicate counting of same no. if(no != pq[0]): count += 1 # Push all the multiples of no. to priority_queue for i in range(size): pq.append(no * a[i]) # cnt+=1 pq = sorted(pq) # Return nth super ugly number return no # /* Driver program to test above functions */ if __name__ == '__main__': a = [2, 3,5] size = len(a) print(ugly(a, size, 1000)) # This code is contributed by mohit kumar 29. |
C#
using System; using System.Collections.Generic; class Program { static 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 List<int> pq = new List<int>(); // Push all the array elements to priority queue for(int i = 0; i < size; i++) { pq.Add(a[i]); } // once count = n we return no int count = 1; int no = 0; pq.Sort(); while(count < n) { // Get the minimum value from priority_queue no = pq[0]; pq.RemoveAt(0); // If top of pq is no then don't increment count. // This to avoid duplicate counting of same no. if(no != pq[0]) { count++; // Push all the multiples of no. to priority_queue for(int i = 0; i < size; i++) { pq.Add(no * a[i]); } } pq.Sort(); } // Return nth super ugly number return no; } // Driver program to test above functions static void Main() { int[] a = {2, 3, 5}; int size = a.Length; Console.WriteLine(Ugly(a, size, 1000)); } } |
Javascript
<script> // Javascript program for super ugly number //function will return the nth super ugly number function ugly(a,size,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 let pq=[];; // Push all the array elements to priority queue for(let i = 0; i < size; i++){ pq.push(a[i]); } // once count = n we return no let count = 1, no; pq.sort(function(a,b){return a-b;}) ; while(count < n){ // Get the minimum value from priority_queue no = pq.shift(); // If top of pq is no then don't increment count. This to avoid duplicate counting of same no. if(no != pq[0]) { count++; //Push all the multiples of no. to priority_queue for(let i = 0; i < size; i++){ pq.push(no * a[i]); // cnt+=1; } } pq.sort(function(a,b){return a-b;}) ; } // Return nth super ugly number return no; } /* Driver program to test above functions */ let a=[2, 3,5]; let size = a.length; document.write(ugly(a, size, 1000)); // This code is contributed by unknown2108 </script> |
Output
12
Time Complexity: O(n*size*logn)
Auxiliary Space: O(n)
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