Find shortest subarray containing all elements

Question

Suppose you have an array of numbers, and another set of numbers. You have to find the shortest subarray containing all numbers with minimal complexity.

The array ca

Answer 1

Here's how I solved this problem in linear time using collections.Counter objects

from collections import Counter

def smallest_subsequence(stream, search):
    if not search:
        return []  # the shortest subsequence containing nothing is nothing

    stream_counts = Counter(stream)
    search_counts = Counter(search)

    minimal_subsequence = None

    start = 0
    end = 0
    subsequence_counts = Counter()

    while True:
        # while subsequence_counts doesn't have enough elements to cancel out every
        # element in search_counts, take the next element from search
        while search_counts - subsequence_counts:
            if end == len(stream):  # if we've reached the end of the list, we're done
                return minimal_subsequence
            subsequence_counts[stream[end]] += 1
            end += 1

        # while subsequence_counts has enough elements to cover search_counts, keep
        # removing from the start of the sequence
        while not search_counts - subsequence_counts:
            if minimal_subsequence is None or (end - start) < len(minimal_subsequence):
                minimal_subsequence = stream[start:end]
            subsequence_counts[stream[start]] -= 1
            start += 1

print(smallest_subsequence([1, 2, 5, 8, 7, 6, 2, 6, 5, 3, 8, 5], [5, 7]))
# [5, 8, 7]