Isomorphic Strings in Python

Question

Here's my code:

def isIso(x,y):
  if len(x) != len(y):
    return False
  for i in range(len(x)):
    count = 0
    if x.count(x[i]) != y.count(y[i]):
      return False
  return True

Why do all solutions for this question online involve mapping or dictionaries? I'm wondering why everyone seems to be overcomplicating the solution to this problem. Is it a time complexity thing? The time complexity of this is n which is not ideal - are people using maps/dictionaries because of a better time complexity?

Answer 1

The time complexity of this is n which is not ideal

No! Your time complexity is not on the order of n. It is on the order of n ² .

str.count must loop through the entire string every time, n operations. And you call it n times. So the result is n*n = n ² complexity, much much worse than if you store the counts in a dictionary and look them up.

The simplest implementation in Python that would be on the order of n time is:

from collections import Counter
def is_isomoprhic(x, y):
    xc, yc = Counter(x), Counter(y)
    return all(xc[a] == xc[b] for a, b in zip(x, y))

Answer 2

您将找到的最简单的解决方案---谢谢！！！

len(set(zip(list(s), list(s1)))) == len(set(s))

Answer 3

Using maketrans and translate .

def is_isometric(x: str, y: str) -> bool:
    trans = str.maketrans(x, y)
    return x.translate(trans) == y