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  1. Practice
  2. Data Structures
  3. Arrays
  4. Sparse Arrays
  5. Discussions

Sparse Arrays

  • qh5ny8wni40i + 2 comments

    Or you could just condense it to:

      return [strings.count(q) for q in queries]

    • syedasjadsyed + 0 comments

      Nice! That list comprehension is much more readable than the map I ended up with:

      return map(lambda query: strings.count(query), queries)

      Edit: but a lot of these one-liners could really benefit from using better data structures

    • jinyi_lie + 0 comments

      woah, I'm amazed by this shortest solution

      this is my code, before i read the discussions The algorithm is: 1. i mapped the list to dict with the value is count, because dict won't have duplicate value. each of these strings key at least show 1 times. 2. Then, i try to find the the query by query_key in the mapped dict

      string_dict = {}
results = []
for string in strings:
    try:
        if string_dict[string] is not None:
            string_dict[string] += 1
    except: 
        string_dict[string] = 1



for query in queries:
    try:
        results.append(string_dict[query])
    except: 
        results.append(0)
        pass

return results