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

Sparse Arrays

  • h351680058
     + 0 comments

    python 3 solution using hash map, pass all cases

    def matchingStrings(strings, queries):
        d = {}
        for s in queries:
                d[s] = 0
        for ss in strings:
                if ss in d:
                        d[ss]+=1
        ans = [d[num] for num in queries]
        return ans