从给定的str中查找第一次出现str的索引

Question

Google or Amazone ask the following question in an interview, would my solution be accepted?

problem: find the index of the first occurrence of the given word from the given string

note: Above problem is from a website and following code passed all the test cases. however, I am not sure if this is the most optimum solutions and so would be accepted by big giants.

def strStr(A, B):
    if len(A) == 0 or len(B) == 0:
        return -1
    for i in range(len(A)):
        c = A[i:i+len(B)]
        if c == B:
            return i
    else:
        return -1

Answer 1

Python actually has a built in function for this, which is why this question doesn't seem like a great fit for interviews in python. Something like this would suffice:

def strStr(A, B):
  return A.find(B)

Otherwise, as commenters have mentioned, inputs/outputs and tests are important. You could add some checks that make it slightly more performant (ie check that B is smaller than A), but I think in general, you won't do better than O(n).

Answer 2

If you want to match the entire word to the words in the string, your code would not work.
Eg If my arguments are print(strStr('world hello world', 'wor')) , your code would return 0, but it should return -1.

Answer 3

I checked your function, works well in python3.6

print(strStr('abcdef', 'bcd')) # with your function.    *index start from 0
print("adbcdef".find('bcd')) # python default function. *index start from 1

Answer 4

first occurrence index, use index() or find()

text = 'hello i am homer simpson'

index = text.index('homer')
print(index)

index = text.find('homer')
print(index)

output:
11
11

Answer 5

It is always better to got for the builtin python funtions. But sometimes in the interviews they will ask for you to implemente it yourself. The best thing to do is to start with the simplest version, then think about corner cases and improvements.

Here you have a test with your version, a slightly improved one that avoid to reallocating new strings in each index and the python built-ing:

A = "aaa foo baz fooz bar aaa"
B = "bar"

def strInStr1(A, B):
    if len(A) == 0 or len(B) == 0:
        return -1
    for i in range(len(A)):
        c = A[i:i+len(B)]
        if c == B:
            return i
    else:
        return -1

def strInStr2(A, B):
  size = len(B)
  for i in range(len(A)):
    if A[i] == B[0]:
      if A[i:i+size] == B:
        return i
  return -1


def strInStr3(A, B):
  return A.index(B)


import timeit
setup = '''from __main__ import strInStr1, strInStr2, strInStr3, A, B'''
for f in  ("strInStr1", "strInStr2", "strInStr3"):
  result =  timeit.timeit(f"{f}(A, B)", setup=setup)
  print(f"{f}: ", result)

The results speak for themselves (time in seconds):

strInStr1:  15.809420814999612
strInStr2:  7.687011377005547
strInStr3:  0.8342400040055509

Here you have the live version

Answer 6

There are a few algorithms that you can learn on this topic like

rabin karp algorithm , z algorithm , kmpalgorithm

which all run in run time complexity of O(n+m) where n is the string length and m is the pattern length. Your algorithm runs in O(n*m) runtime complexity . I would suggest starting to learn from rabin karp algorithm, I personally found it the easiest to grasp.

There are also some advanced topics like searching many patterns in one string like the aho-corasick algorithm which is good to read. I think this is what grep uses when searching for multiple patterns. Hope it helps :)