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