Finding the odd number out in an array

An easier (and more efficient) way of doing this than your initial approach is with a Counter object:

 from collections import Counter

 singlet = Counter(nums).most_common()[-1][0]

The Counter object will create a dictionary-like object with the keys being the values in your list and the values being the number of times they appear. The most_common method will return a list of tuples of (value, count) sorted by count in decreasing order.

If you don't know how many singlets there will be, you can get a list of them with:

[k for k, v in Counter(nums).items() if v == 1]

---

Complexity:

I said my top solution was more efficient because your original implementation iterates through your list and for each item calls both remove and in which is going to get you to something like O(n²) complexity. In the Counter implementation the construction of the Counter object only does a single pass through the entire list. There is probably a sort going on when most_common is called so I'm guessing the complexity is about O(n log n). @Stefan Pochman has corrected me on this: Python uses the Timsort algorithm which will be very efficient in a case like this (if all but one of the numbers appear twice, the list is effectively almost completely sorted already) so its complexity will about about O(n).

You already nums_copy.remove(i) so you can't nums_copy.remove(i) again

You could do:

a = [0, 0, 1, 1, 2, 2, 6, 6, 9, 10, 10]

def get_single_instance(array):
  d = {}

  for item in a:
    if item not in d:
      d[item] = 1
    else:
      d[item] += 1

  print d

  for k, v in d.iteritems():
    if v == 1:
      return k

print get_single_instance(a)

Result: 9

The best algorithm is to use XOR to find the odd number.

def find_number(nums):
    s = 0 
    for n in nums:
        s ^= n 
    return s 


a = [0, 0, 1, 1, 2, 2, 6, 6, 9, 10, 10] 
print(find_number(a))