Do daily

Author: Zanger67

my-submissions/m3306.md

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+Beats 90% runtime

my-submissions/m3306.py

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+class Solution:
+    def countOfSubstrings(self, word: str, k: int) -> int:
+        vows   = set('aeiou')
+        conses = set(ascii_letters) - vows
+
+        output               = 0        # count of the num of substrings
+        consonant_cnt        = 0        # num of consants in current str
+        extra_vowels_on_left = 0        # how many vowels on left side of 
+                                        # string that aren't required (i.e. 
+                                        # we can choose to include them 
+                                        # or exclude them)
+
+        indices = deque()   # Deque to keep track of leftmost required char
+        vowels = {}         # Tracks right-most occurance of vowel
+
+        for i, c in enumerate(word) :
+            if c in vows :
+                vowels[c] = i
+            else :
+                consonant_cnt += 1
+            indices.appendleft((i, c))
+
+            # Consonant count must match exactly
+            while indices and consonant_cnt > k :
+                indx_popped, popped = indices.pop()
+                if popped in conses :
+                    consonant_cnt -= 1
+                elif popped in vows and vowels[popped] == indx_popped :
+                    vowels.pop(popped)
+                extra_vowels_on_left = 0
+
+            # Remove unnecessary vowels
+            while indices and indices[-1][1] in vows and indices[-1][0] != vowels[indices[-1][1]] :
+                extra_vowels_on_left += 1
+                indices.pop()
+
+            # If extra vowels, we can choose to add x number of vowels to create a new string
+            if indices and consonant_cnt == k and len(vowels) == 5 :
+                output += 1 + extra_vowels_on_left
+
+        return output