71 lines
2.3 KiB
Python
71 lines
2.3 KiB
Python
class Solution:
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def checkInclusion(self, s1: str, s2: str) -> bool:
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'''
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Sliding window, better space utilization
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'''
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if len(s1) > len(s2): return False
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def get_letter_index(letter):
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'''
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Problem only concerns with lowercase letters
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'''
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return ord(letter) - ord('a')
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s1_count, s2_count = [0] * 26, [0] * 26
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for i in range(len(s1)):
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s1_count[get_letter_index(s1[i])] += 1
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s2_count[get_letter_index(s2[i])] += 1
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# Keep a `count_matches` variable that tells us how many
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# counts of a-z from s1 match a-z in the window of s2
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count_matches = 0
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for i in range(26):
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if s1_count[i] == s2_count[i]:
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count_matches += 1
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# It's possible that we get a perfect match right after the
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# above initial computation. If yes, yay!
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if count_matches == 26:
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return True
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# Start from the very next character after figuring out the
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# initial `count_matches` value which would be at len(s1)
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l = 0
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for r in range(len(s1), len(s2)):
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#
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# Adding rightmost letter
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#
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li = get_letter_index(s2[r])
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s2_count[li] += 1
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if s1_count[li] == s2_count[li]:
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count_matches += 1
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# If after adding new right letter, count increased by 1 for
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# the letter index, then that means total matches also reduced
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# by 1
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elif s1_count[li] + 1 == s2_count[li]:
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count_matches -= 1
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#
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# Removing leftmost letter
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#
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li = get_letter_index(s2[l])
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s2_count[li] -= 1
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if s1_count[li] == s2_count[li]:
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count_matches += 1
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# If after removing leftmost letter, count decreased by 1, then
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# our total matches also reduced by 1
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elif s1_count[li] - 1 == s2_count[li]:
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count_matches -= 1
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if count_matches == 26: return True
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l += 1
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return count_matches == 26
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