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Problem solution · Python

ABC278 C — FF

ABC278 C — FF: a Python solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to KATO-Hiro AtCoder Solutions.

Technique
Hash-based lookup
Source
KATO-Hiro AtCoder Solutions
Length
29 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Hash-based lookup

For ABC278 C — FF, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 29 lines of Python from the credited upstream file abc278_c.py.
  • The implementation visibly relies on hash lookup, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from KATO-Hiro AtCoder Solutions by KATO-Hiro and is used under the CC0-1.0 licence.

Open problem ↗Original source ↗
Full codeABC278 C — FF · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*-  def main():    from collections import defaultdict    import sys     input = sys.stdin.readline     n, q = map(int, input().split())    d = defaultdict(set)     for _ in range(q):        ti, ai, bi = map(int, input().split())         if ti == 1:            d[ai].add(bi)        elif ti == 2:            d[ai].discard(bi)        else:            if bi in d[ai] and ai in d[bi]:                print("Yes")            else:                print("No")     if __name__ == "__main__":    main()

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