Here are a few suggestions.\n\nfrom collections import defaultdict\n\ndef contains_duplicates(arr): # PEP recommends not using camel case\n arr_count = defaultdict(lambda: 0) # you should use defaultdict to specify a default value\n for i in arr:\n arr_count[i] += 1\n if arr_count[i] >= 2:\n return True # Python has a boolean type; you shouldn't use 0s and 1s\n return False\n",

Regarding your style, you appear to not know that sets are implemented similarly to dicts in Python, and you were unnecessarily complicated in your code. I would have implemented it like Gavin H did.\n\nAlso, other elements of style that you should work on (see PEP8): \n\n\nsingle spaces around assignment operators\nuse underscores to separate words in names as opposed to camel-case\nlearn dict methods (you could have used the dict.get method instead of try/except)\nreturn True or False, not 1 or 0\n",

There are a number of coding style issues with your code, but the one I like least is that it will always process the entire array even if the first two elements are the same.\n\nMy take on this would be the following it may not be the shortest but going back later would be better:\n\ndef has_dups(array):\n \"\"\" Return true if there are any duplicates in the array \"\"\"\n result = False # Return value\n seen = {} # Dictionary of values seen\n for val in array:\n if seen.get(val, False):\n result = True # We have a duplicate\n break # Return NOW\n else:\n seen[val] = True # Add to seen\n return result\n\n\nIn a commercial organisation you are usually looking for code that:\n\n\nMeets the specification - yours did not as it returns 0/1 rather than False/True\nMeets the standards - yours has several PEP8 issues\nIs clear and maintainable - yours lacks comments and relies on exceptions to save time in an interview it is a good idea to at least mark where you would comment the code.\nIs testable - this means one entry and exit and a low complexity\n\n\nEdit\n\nThe OP has pointed out that their code returns on the first non-exception which I had missed - this emphasises points 3 and 4 above.",

A more concise solution to the problem is:\n\ndef has_dups(arr):\n return len(set(arr)) != len(arr)\n\n\nOn the set creation any dups would be removed so the count would tell us the result. I think the set creation from an iterator is O(n) so it would fulfill that requirement.\n\nCan't promise that is what they are looking for though!",

Check PEP-8\n\nfrom collections import defaultdict\ndef is_duplicate(string):\n freq = defaultdict(lambda: 0)\n for c in string:\n freq[c] += 1\n return all(v == 1 for v in freq.values())\n",

you would have used \"in\" instead of try and except.\n\ndef containsDuplicate(arr):\n hashi={}\n res = 0 \n for i in arr:\n if i not in hashi:\n hashi[i]=1\n else:\n res = 1\n break\n return res\n",

It could be that they didn't like the fact that you return 0 or 1 instead of False or True. ",

Trying to reverse engineer the interview rejection reason is typically a difficult and often meaningless exercise, so let's reform the question to something more fun and continue to improve on your answer, as you would have if you had been given more time.\n\nWrite a function that is given an array of integers. It should return true if any value appears at least twice in the array, and it should return false if every element is distinct.\n\ndef containsDuplicate(arr):\n hashi={}\n for i in arr:\n try:\n hashi[i]+=1\n return 1\n except KeyError:\n hashi[i]=1\n return 0\n\n\nWell, python has a style guide that suggests some spacing, so let's do that. (No one would ever reject someone for this, btw)\n\nPython also uses True and False more then the C style 0 and 1s for booleans.\n\ndef containsDuplicate(arr):\n hashi = {}\n for i in arr:\n try:\n return True\n except KeyError:\n hashi[i] = 1\n return False\n\n\nNow, Python officially subscribes to EAFP where you use exceptions rather then checking, but some people disagree. If you used checking, we have:\n\ndef containsDuplicate(arr):\n hashi = {}\n for i in arr:\n if i in hashi:\n return True\n else:\n hashi[i] = True\n return False\n\n\nHowever, time to get clever. Using sets, we can do:\n\ndef containsDuplicate(a):\n return len(set(a)) != len(a)\n",