# Python O(n) Time, O(1) Space

• Hash the number of times each character appears in p. Iterate over s with a sliding window and maintain a similar hash. If these two hashes are ever the same, add that to the result.

Each of the hashes have a finite (a-z, A-Z) number of possible characters, so the space used is O(1)

We iterate over s linearly, comparing constant length hashes at each iteration so each iteration is also O(1), so the runtime is O(n)

``````class Solution(object):
def findAnagrams(self, s, p):
"""
:type s: str
:type p: str
:rtype: List[int]
"""
res = []
n, m = len(s), len(p)
if n < m: return res
phash, shash = [0]*123, [0]*123
for x in p:
phash[ord(x)] += 1
for x in s[:m-1]:
shash[ord(x)] += 1
for i in range(m-1, n):
shash[ord(s[i])] += 1
if i-m >= 0:
shash[ord(s[i-m])] -= 1
if shash == phash:
res.append(i - m + 1)
return res
``````

• is the '[0]*123' come from ASCII code?

• yes indeed! The two lists sHash and pHash contain 123 elements, one for each ASCII value from the strings s and p. This is a really cool way to count the frequency of each letter in the string. I also loved the efficiency of adding a single letter to sHash, and then checking whether the sHash ==pHash.

• @ZhengwuFang Your English is really bad man

• @ngaikw You are really polite and cultivated.

• @ngaikw lol in computer science code judges

• comparing constant length hashes at each iteration so each iteration is also O(1)

Can you explain a little bit why you're comparing two hashes here? What I see is comparison of two lists `phash` and `shash` and the time complexity of comparing two lists should be O(n).

• @lizzy0127 Sure! So the runtime of iterating over a list depends on what the list contains. If we are iterating over a list that's given to us as the input, this would definitely be O(n). If we are iterating over a constant sized list, this is considered an O(1) operation. The two hashes we have here that we iterate over are constant sized.

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