>>> vowel_histogram('This restaurant is a long way from our house.')
{'a': 4, 'e': 2, 'i': 2, 'o': 4, 'u': 3}
>>> min_key({'a': 4, 'b': -5, 'c': 2, 'd': -7, 'e': 4})
'd'
If the smallest key happens to occur more than once, a solution which returns any key (associated with that value) will be considered correct.