Problem solution · Python

ABC344 E — Insert or Erase

ABC344 E — Insert or Erase: 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
82 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 ABC344 E — Insert or Erase, 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

  • 82 lines of Python from the credited upstream file abc344_e.py.
  • The implementation visibly relies on sequence storage, 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.

Full codeABC344 E — Insert or Erase · PythonPython
Use this to learn the idea, then write your own version.
# -*- coding: utf-8 -*- from collections import defaultdictfrom itertools import pairwisefrom typing import List HEAD, TAIL = 0, -1  class DoublyLinkedList:     def __init__(self, array: List) -> None:        self.next = defaultdict(int)        self.prev = defaultdict(int)         for first, second in pairwise([HEAD] + array + [TAIL]):            self.next[first] = second            self.prev[second] = first     def insert(self, value_left: int, value_mid: int) -> None:        assert value_left in self.next.keys() and value_left in self.prev.keys()        assert not value_mid in self.next.keys() and not value_mid in self.prev.keys()         value_right = self.next[value_left]         self.next[value_left] = value_mid        self.next[value_mid] = value_right         self.prev[value_right] = value_mid        self.prev[value_mid] = value_left     def remove(self, value_mid: int) -> None:        assert value_mid in self.next.keys() and value_mid in self.prev.keys()         value_left, value_right = self.prev[value_mid], self.next[value_mid]         self.next[value_left] = value_right        self.prev[value_right] = value_left         del self.next[value_mid]        del self.prev[value_mid]     def fetch_all_values(self) -> List:        results = list()        pos = HEAD         while self.next[pos] != TAIL:            results.append(self.next[pos])            pos = self.next[pos]         return results  def main():    import sys     input = sys.stdin.readline     n = int(input())    a = list(map(int, input().split()))    d = DoublyLinkedList(array=a)     q = int(input())     for _ in range(q):        qi = list(map(int, input().split()))         if qi[0] == 1:            _, x, y = qi            d.insert(x, y)        else:            _, x = qi            d.remove(x)     ans = d.fetch_all_values()     print(*ans)  if __name__ == "__main__":    main() 

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