Ugly Numbers

3.4

Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, … shows the first 11 ugly numbers. By convention, 1 is included.

Given a number n, the task is to find n’th Ugly number.

Input  : n = 7
Output : 8

Input  : n = 10
Output : 12

Input  : n = 15
Output : 24

Input  : n = 150
Output : 5832

Method 1 (Simple)

Loop for all positive integers until ugly number count is smaller than n, if an integer is ugly than increment ugly number count.

To check if a number is ugly, divide the number by greatest divisible powers of 2, 3 and 5, if the number becomes 1 then it is an ugly number otherwise not.

For example, let us see how to check for 300 is ugly or not. Greatest divisible power of 2 is 4, after dividing 300 by 4 we get 75. Greatest divisible power of 3 is 3, after dividing 75 by 3 we get 25. Greatest divisible power of 5 is 25, after dividing 25 by 25 we get 1. Since we get 1 finally, 300 is ugly number.

Implementation:

C/C++

// CPP program to find nth ugly number
# include<stdio.h>
# include<stdlib.h>

/*This function divides a by greatest divisible 
  power of b*/
int maxDivide(int a, int b)
{
  while (a%b == 0)
   a = a/b; 
  return a;
}    

/* Function to check if a number is ugly or not */
int isUgly(int no)
{
  no = maxDivide(no, 2);
  no = maxDivide(no, 3);
  no = maxDivide(no, 5);
  
  return (no == 1)? 1 : 0;
}    

/* Function to get the nth ugly number*/
int getNthUglyNo(int n)
{
  int i = 1; 
  int count = 1;   /* ugly number count */ 

  /*Check for all integers untill ugly count 
    becomes n*/ 
  while (n > count)
  {
    i++;      
    if (isUgly(i))
      count++; 
  }
  return i;
}

/* Driver program to test above functions */
int main()
{
    unsigned no = getNthUglyNo(150);
    printf("150th ugly no. is %d ",  no);
    getchar();
    return 0;
}

Java

// Java program to find nth ugly number
class UglyNumber
{
    /*This function divides a by greatest divisible 
    power of b*/
    int maxDivide(int a, int b)
    {
        while(a % b == 0)
            a = a/b;
        return a;
    }
    
    /* Function to check if a number is ugly or not */
    int isUgly(int no)
    {
        no = maxDivide(no, 2);
        no = maxDivide(no, 3);
        no = maxDivide(no, 5);
        
        return (no == 1)? 1 : 0;
    }
    
    /* Function to get the nth ugly number*/
    int getNthUglyNo(int n)
    {
        int i = 1;
        int count = 1; // ugly number count 
        
        // check for all integers until count becomes n */
        while(n > count)
        {
            i++;
            if(isUgly(i) == 1)
                count++;
        }
        return i;
    }
    
    /* Driver program to test above functions */
    public static void main(String args[])
    {
        UglyNumber obj = new UglyNumber(); 
        int no = obj.getNthUglyNo(150);
        System.out.println("150th ugly no. is "+ no);
    }
}

// This code has been contributed by Amit Khandelwal (Amit Khandelwal 1)


Output:

150th ugly no. is 5832 

This method is not time efficient as it checks for all integers until ugly number count becomes n, but space complexity of this method is O(1)

 

Method 2 (Use Dynamic Programming)

Here is a time efficient solution with O(n) extra space. The ugly-number sequence is 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, …
     because every number can only be divided by 2, 3, 5, one way to look at the sequence is to split the sequence to three groups as below:
     (1) 1×2, 2×2, 3×2, 4×2, 5×2, …
     (2) 1×3, 2×3, 3×3, 4×3, 5×3, …
     (3) 1×5, 2×5, 3×5, 4×5, 5×5, …

     We can find that every subsequence is the ugly-sequence itself (1, 2, 3, 4, 5, …) multiply 2, 3, 5. Then we use similar merge method as merge sort, to get every ugly number from the three subsequence. Every step we choose the smallest one, and move one step after.

1 Declare an array for ugly numbers:  ugly[n]
2 Initialize first ugly no:  ugly[0] = 1
3 Initialize three array index variables i2, i3, i5 to point to 
   1st element of the ugly array: 
        i2 = i3 = i5 =0; 
4 Initialize 3 choices for the next ugly no:
         next_mulitple_of_2 = ugly[i2]*2;
         next_mulitple_of_3 = ugly[i3]*3
         next_mulitple_of_5 = ugly[i5]*5;
5 Now go in a loop to fill all ugly numbers till 150:
For (i = 1; i < 150; i++ ) 
{
    /* These small steps are not optimized for good 
      readability. Will optimize them in C program */
    next_ugly_no  = Min(next_mulitple_of_2,
                        next_mulitple_of_3,
                        next_mulitple_of_5); 

    ugly[i] =  next_ugly_no       

    if (next_ugly_no  == next_mulitple_of_2) 
    {             
        i2 = i2 + 1;        
        next_mulitple_of_2 = ugly[i2]*2;
    } 
    if (next_ugly_no  == next_mulitple_of_3) 
    {             
        i3 = i3 + 1;        
        next_mulitple_of_3 = ugly[i3]*3;
     }            
     if (next_ugly_no  == next_mulitple_of_5)
     {    
        i5 = i5 + 1;        
        next_mulitple_of_5 = ugly[i5]*5;
     } 
     
}/* end of for loop */ 
6.return next_ugly_no

Example:
Let us see how it works

initialize
   ugly[] =  | 1 |
   i2 =  i3 = i5 = 0;

First iteration
   ugly[1] = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
            = Min(2, 3, 5)
            = 2
   ugly[] =  | 1 | 2 |
   i2 = 1,  i3 = i5 = 0  (i2 got incremented ) 

Second iteration
    ugly[2] = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
             = Min(4, 3, 5)
             = 3
    ugly[] =  | 1 | 2 | 3 |
    i2 = 1,  i3 =  1, i5 = 0  (i3 got incremented ) 

Third iteration
    ugly[3] = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
             = Min(4, 6, 5)
             = 4
    ugly[] =  | 1 | 2 | 3 |  4 |
    i2 = 2,  i3 =  1, i5 = 0  (i2 got incremented )

Fourth iteration
    ugly[4] = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
              = Min(6, 6, 5)
              = 5
    ugly[] =  | 1 | 2 | 3 |  4 | 5 |
    i2 = 2,  i3 =  1, i5 = 1  (i5 got incremented )

Fifth iteration
    ugly[4] = Min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5)
              = Min(6, 6, 10)
              = 6
    ugly[] =  | 1 | 2 | 3 |  4 | 5 | 6 |
    i2 = 3,  i3 =  2, i5 = 1  (i2 and i3 got incremented )

Will continue same way till I < 150

C/C++

// C++ program to find n'th Ugly number
# include<bits/stdc++.h>
using namespace std;

/* Function to get the nth ugly number*/
unsigned getNthUglyNo(unsigned n)
{
    unsigned ugly[n]; // To store ugly numbers
    unsigned i2 = 0, i3 = 0, i5 = 0;
    unsigned next_multiple_of_2 = 2;
    unsigned next_multiple_of_3 = 3;
    unsigned next_multiple_of_5 = 5;
    unsigned next_ugly_no = 1;

    ugly[0] = 1;
    for (int i=1; i<n; i++)
    {
       next_ugly_no = min(next_multiple_of_2,
                           min(next_multiple_of_3,
                               next_multiple_of_5));
       ugly[i] = next_ugly_no;
       if (next_ugly_no == next_multiple_of_2)
       {
           i2 = i2+1;
           next_multiple_of_2 = ugly[i2]*2;
       }
       if (next_ugly_no == next_multiple_of_3)
       {
           i3 = i3+1;
           next_multiple_of_3 = ugly[i3]*3;
       }
       if (next_ugly_no == next_multiple_of_5)
       {
           i5 = i5+1;
           next_multiple_of_5 = ugly[i5]*5;
       }
    } /*End of for loop (i=1; i<n; i++) */
    return next_ugly_no;
}

/* Driver program to test above functions */
int main()
{
    int n = 150;
    cout << getNthUglyNo(n);
    return 0;
}

Java

// Java program to find nth ugly number
import java.lang.Math;

class UglyNumber
{
    /* Function to get the nth ugly number*/
    int getNthUglyNo(int n)
    {
        int ugly[] = new int[n];  // To store ugly numbers
        int i2 = 0, i3 = 0, i5 = 0;
        int next_multiple_of_2 = 2;
        int next_multiple_of_3 = 3;
        int next_multiple_of_5 = 5;
        int next_ugly_no = 1;
        
        ugly[0] = 1;
        
        for(int i = 1; i < n; i++)
        {
            next_ugly_no = Math.min(next_multiple_of_2,
                                  Math.min(next_multiple_of_3,
                                        next_multiple_of_5));
            
            ugly[i] = next_ugly_no;
            if (next_ugly_no == next_multiple_of_2)
            {
               i2 = i2+1;
               next_multiple_of_2 = ugly[i2]*2;
            }
            if (next_ugly_no == next_multiple_of_3)
            {
               i3 = i3+1;
               next_multiple_of_3 = ugly[i3]*3;
            }
            if (next_ugly_no == next_multiple_of_5)
            {
               i5 = i5+1;
               next_multiple_of_5 = ugly[i5]*5;
            }
        } /*End of for loop (i=1; i<n; i++) */
        return next_ugly_no;
    }

    /* Driver program to test above functions */
    public static void main(String args[])
    {
        int n = 150;
        UglyNumber obj = new UglyNumber();
        System.out.println(obj.getNthUglyNo(n));
    }
}

// This code has been contributed by Amit Khandelwal (Amit Khandelwal 1)

Python

# Python program to find n'th Ugly number

# Function to get the nth ugly number
def getNthUglyNo(n):

    ugly = [0] * n # To store ugly numbers

    # 1 is the first ugly number
    ugly[0] = 1

    # i2, i3, i5 will indicate indices for 2,3,5 respectively
    i2 = i3 =i5 = 0

    # set initial multiple value
    next_multiple_of_2 = 2
    next_multiple_of_3 = 3
    next_multiple_of_5 = 5

    # start loop to find value from ugly[1] to ugly[n]
    for l in range(1, n):

        # choose the min value of all available multiples
        ugly[l] = min(next_multiple_of_2, next_multiple_of_3, next_multiple_of_5)

        # increment the value of index accordingly
        if ugly[l] == next_multiple_of_2:
            i2 += 1
            next_multiple_of_2 = ugly[i2] * 2

        if ugly[l] == next_multiple_of_3:
            i3 += 1
            next_multiple_of_3 = ugly[i3] * 3

        if ugly[l] == next_multiple_of_5: 
            i5 += 1
            next_multiple_of_5 = ugly[i5] * 5

    # return ugly[n] value
    return ugly[-1]

def main():

    n = 150

    print getNthUglyNo(n)


if __name__ == '__main__':
    main()

#This code is contributed by Neelam Yadav


Output :
5832

Algorithmic Paradigm: Dynamic Programming
Time Complexity: O(n)
Auxiliary Space: O(n)

Asked in: Goldman-Sachs

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

Please write comments if you find any bug in the above program or other ways to solve the same problem.

GATE CS Corner    Company Wise Coding Practice

Recommended Posts:



3.4 Average Difficulty : 3.4/5.0
Based on 188 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.