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Day-17_Finding_All_Anagrams_in_a_String
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Day-17_Finding_All_Anagrams_in_a_String
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'''
Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
The order of output does not matter.
Example 1:
Input:
s: "cbaebabacd" p: "abc"
Output:
[0, 6]
Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".
Example 2:
Input:
s: "abab" p: "ab"
Output:
[0, 1, 2]
Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".
'''
class Solution:
def findAnagrams(self, s: str, p: str) -> List[int]:
ans = []
magic = 1 # will store a hash for p
for i in p:
magic *= ord(i) # calculating hash as product of ascci representation of each alphabet
tmp = 1 # will store the hash for the sliding window
for i in range(len(s)):
tmp *= ord(s[i])
if i >= len(p) - 1:
if magic == tmp:
ans.append(i+1-len(p))
tmp //= ord(s[i+1-len(p)])
return ans