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

Sparse Arrays

  • pdog1111 + 2 comments

    Counter makes it even easier

    from collections import Counter, defaultdict

counter = defaultdict(int, Counter([raw_input() for _ in range(input())]))

for _ in range(input()):
    print counter[raw_input()]
    • DanHaggard + 2 comments

      I haven't checked in python2 - but in 3 you don't need defaultdict if you use Counter.

      from collections import Counter, defaultdict

cnt = Counter([input().strip() for _ in range(int(input().strip()))]) 

for _ in range(int(input().strip())):
    print(cnt[input().strip()])
      • PapaThor + 1 comment

        Condensed down to 2 statements (plus the import):

        from collections import Counter

d = Counter([input().strip() for _ in range(int(input()))])
print(*[d[input().strip()] for _ in range(int(input()))], sep='\n')
        • chucklesoclock + 0 comments

          Lovely!

      • tusharkailash29 + 1 comment

        I am new to python and I had difficulty understanding the code. Why are two same two for-loops needed? And how does this statement get us the solution : print(cnt[input().strip()]) ? Thanks.

        • CaptnCrunchCode + 0 comments

          @tusharkailash29 ... or anyone else currently wondering.

          We are comparing two variable input ranges in this question, and the ranges may vary in length. The "for loops" are used to organize the inputs as "S" strings followed by "Q" queries.

          cnt = Counter([input().strip() for _ in range(int(input()))])

          The Counter() function creates a dictionary (key,value) pairs. The "S" string inputs are the keys, and their occurences (a.k.a counts) are the value. For every duplicate key that appears in this loop, it's value (count) increases by one.

          for _ in range(int(input().strip())): print(cnt[input().strip()])

          In the second & third lines, "Q" query inputs are the keys we search in the "cnt" dictionary. Think of "input().strip()" in print(cnt[input().strip()]) as the key in (key,value) pair. Searching by cnt[key] provides the dictionary value associated with it. The second "for loop" repeats the cnt[key] search step for the remaining "Q" query inputs.

    • philippkipp + 3 comments

      Python 3: You don't really need to import anything additionally:

      result = [] for q in queries: result.append(strings.count(q)) return result

      I don't know how to put a codeblock into the comments

      • kakadiyaraj1999 + 0 comments

        Same! :)

      • Tej_Pratap_Singh + 1 comment

        Wow, this is amazing

        • qh5ny8wni40i + 1 comment

          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

      • sanjaybisht8999 + 0 comments

        You can by typing three times ` key

        `