Question

I want to generate 2n-1 random integers in the range [1,n] with each element appearing twice except for random value, which only appears once.
for example:

n = 3  
seq = [1, 2, 3, 1, 3]  

in this example 2 appears only once.

My algorithm is to use dictionary, like this:

-------------
| num |times|
|  1  |  2  |
|  2  |  1  |
|  3  |  2  |

where the keys are from 1 to n and the value represents the number of occurrences of the key. I fill the dictionary with values of two and reduce the value to 1 for one random key.

1. Is there any other algorithm?
2. how to do it if n is very large lead to can not be stored in memory?

Answer 1

I am not sure that I am 100 sure what you are after, but here is a try:

import random as rn

x = range(3)*2 #generate a list where each number appears twice

rn.shuffle(x) #shuffle it
x.pop()       #remove one number

Result:

>>> x
[2, 0, 2, 1, 0] #the result is a list where every number appears twice, except for
                #one number which was removed at random, also the numbers are 
                #randomly arranged

EDIT:

Here is a try to do this for very large n (an n that a list of that size cannot be stored in your ram). I cannot see how to shuffle the integers. However, I can remove one at random. Let’s say you want to write the list to txt file.

drop = rn.range(0,n) #choose a random integer to drop

with open('my_file.txt','w') as f:
    for ind,ele in enumerate(xrange(n)):  
        if ind == drop: #do not write the element to txt file
            pass
        else:
            f.write(str(ele) + '\n') #write every except for one element to txt file

with open('my_file.txt','a') as f:
    for ele in xrange(n):
        f.write(str(ele) + '\n') # write every element to txt file

In the end we were wrote n-1 element to txt file two times, and 1 element once, that element was chosen at random.

For n = 5 the txt file looks like this:

0
2
3
4
0
1
2
3
4

In the above case 1 is only showing up once, every other number is showing up twice.