Ugly Numbers Dynamic Programming c++

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 into 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 a similar merge method as merge sort, to get every ugly number from the three subsequences. 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

Code using DP:





reference