263. Ugly Number && 264. Ugly Number II && 313. Super Ugly Number

263. Ugly Number

Write a program to check whether a given number is an ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while 14 is not ugly since it includes another prime factor 7.

Note that 1 is typically treated as an ugly number.

 

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 (E) Happy Number (E) Count Primes (M) Ugly Number II

 

 

public class Solution {
    public boolean isUgly(int num) {
        if(num == 0)
            return false;
        num = removeFactor(num, 2);
        num = removeFactor(num, 3);
        num = removeFactor(num, 5);
        return num == 1;
        
    }
    
    private int removeFactor(int num, int factor)
    {
        while(num%factor==0)
        {
            num = num/factor;
        }
        return num;
    }
}

 

 

264. Ugly Number II

Write a program to find the n-th ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.

Note that 1 is typically treated as an ugly number.

Hint:

1. The naive approach is to call isUgly for every number until you reach the nth one. Most numbers are not ugly. Try to focus your effort on generating only the ugly ones.
2. An ugly number must be multiplied by either 2, 3, or 5 from a smaller ugly number.
3. The key is how to maintain the order of the ugly numbers. Try a similar approach of merging from three sorted lists: L1, L2, and L3.
4. Assume you have Uk, the kth ugly number. Then Uk+1 must be Min(L1 * 2, L2 * 3, L3 * 5).

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 (H) Merge k Sorted Lists (E) Count Primes (E) Ugly Number (M) Perfect Squares (M) Super Ugly Number

 

Solution using priority queue. 

public class Solution {
    public int nthUglyNumber(int n) {
        if(n==1) return 1;
        PriorityQueue<Long> q = new PriorityQueue();
        q.add(1l);
        
        for(long i=1; i<n; i++) {
            long tmp = q.poll();
            while(!q.isEmpty() && q.peek()==tmp) tmp = q.poll();
            
            q.add(tmp*2);
            q.add(tmp*3);
            q.add(tmp*5);
        }
        return q.poll().intValue();
    }
}

 

Solution by merging 3 sorted array.

multiplier array (which is the same as ugly number array)

　　 1,　　2,　　3,　　4,　　5,　　6,　　8,　　9, ....

2:  2*1,  2*2,  2*3,  2*4,  2*5,  2*6,  2*8,  2*9,....

3:  3*1,  3*2,  3*3,  3*4,  3*5,  3*6,  3*8,  3*9,....

5:  5*1,  5*2,  5*3,  5*4,  5*5,  5*6,  5*8,  5*9,....

 

public class Solution {
  public int nthUglyNumber(int n) {
    if (n == 1)
      return 1;

    int[] firstNUglyNumbers = new int[n];
    firstNUglyNumbers[0] = 1;
    int[] factors = new int[]{2, 3, 5};
    int[] nextMultiplierIndices = new int[]{0, 0, 0};

    int i = 1;
    while (i < n) {
      int factorIndex = 0;
      int nextUgly = factors[0] * firstNUglyNumbers[nextMultiplierIndices[0]];
      for (int f = 0; f < 3; ++f) {
        int temp = factors[f] * firstNUglyNumbers[nextMultiplierIndices[f]];
        if (temp < nextUgly) {
          nextUgly = temp;
          factorIndex = f;
        }
      }

      if (nextUgly != firstNUglyNumbers[i - 1]) {
        firstNUglyNumbers[i] = nextUgly;
        ++i;
      }
      nextMultiplierIndices[factorIndex] += 1;
    }

    return firstNUglyNumbers[n - 1];
  }
}

 

 

313. Super Ugly Number

Write a program to find the nth super ugly number.

Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k. For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32] is the sequence of the first 12 super ugly numbers given primes = [2, 7, 13, 19] of size 4.

Note:
(1) 1 is a super ugly number for any given primes.
(2) The given numbers in primes are in ascending order.
(3) 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.

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(M) Ugly Number II

 

Slight modify the first solution above will get a TimeLimitExceeded solutions

public class Solution {
    public int nthSuperUglyNumber(int n, int[] primes) {
        Arrays.sort(primes);
        if(n==1) 
            return 1;
        PriorityQueue<Long> q = new PriorityQueue();
        q.add(1l);
        
        for(long i=1; i<n; i++) {
            long tmp = q.poll();
            while(!q.isEmpty() && q.peek()==tmp) 
                tmp = q.poll();
            
            for(Integer p: primes)
            {
                q.add(tmp*p);
            }
        }
        return q.poll().intValue();
    }
}

 

 Slight modify the second solution above will get a solutions

public class Solution {
    public int nthSuperUglyNumber(int n, int[] primes) {
        if (n == 1)
          return 1;
    
        int[] firstNUglyNumbers = new int[n];
        firstNUglyNumbers[0] = 1;
        int[] factors = primes;
        int[] nextMultiplierIndices = new int[primes.length];
    
        int i = 1;
        while (i < n) {
          int factorIndex = 0;
          int nextUgly = factors[0] * firstNUglyNumbers[nextMultiplierIndices[0]];
          for (int f = 0; f < primes.length; ++f) {
            int temp = factors[f] * firstNUglyNumbers[nextMultiplierIndices[f]];
            if (temp < nextUgly) {
              nextUgly = temp;
              factorIndex = f;
            }
          }
    
          if (nextUgly != firstNUglyNumbers[i - 1]) {
            firstNUglyNumbers[i] = nextUgly;
            ++i;
          }
          nextMultiplierIndices[factorIndex] += 1;
        }
    
        return firstNUglyNumbers[n - 1];
      
    }
}