Solution #3306 - Mridul/Edited - 11/03/2025

Daily Questions/#3306 - Count of Substrings Containing Every Vowel and K Consonants II/Explanation.md

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+Initialization:
+
+n: Length of the word.
+
+freq: Array to store the frequency of each character.
+
+currConsonant: Counter for the current number of consonants in the substring.
+
+cnt: Counter for the valid substrings.
+
+left: Left pointer for the sliding window.
+
+Iterating through the word:
+
+For each character ch in the word:
+
+If ch is a consonant, increment currConsonant.
+
+Increment the frequency of ch in freq.
+
+Checking the conditions:
+
+While the current substring contains at least k consonants and all vowels:
+
+Increment cnt by the number of valid substrings ending at the current position.
+
+Move the left pointer to reduce the window size.
+
+Decrement the frequency of the character at the left pointer.
+
+If the character is a consonant, decrement currConsonant.
+
+This logic ensures that we count all substrings that meet the conditions specified in the problem.

Daily Questions/#3306 - Count of Substrings Containing Every Vowel and K Consonants II/Solution.java

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+class Solution {
+    public long countOfSubstrings(String word, int k) {
+        return atleast(k, word)-atleast(k+1, word);
+    }
+    private boolean isAllVowelPresent(int[] freq)
+    {
+        return (freq['a'-'a']>0 && freq['e'-'a']>0 && freq['i'-'a']>0 && 
+            freq['o'-'a']>0 && freq['u'-'a']>0);
+    }
+    private boolean isConsonant(char ch)
+    {
+        return (ch!='a' && ch!='e' && ch!='i' && ch!='o' && ch!='u');
+    }
+    private long atleast(int k, String word)
+    {
+        int n=word.length();
+        int[] freq=new int[26];
+        int currConsonant=0;
+        long cnt=0;
+        int left=0;
+        for(int right=0;right<n;right++)
+        {
+            char ch=word.charAt(right);
+            if(isConsonant(ch)) currConsonant++;
+            freq[ch-'a']++;
+            while(currConsonant>=k && isAllVowelPresent(freq))
+            {
+                cnt+=(n-right);
+                char c=word.charAt(left);
+                if(isConsonant(c)) currConsonant--;
+                freq[c-'a']--;
+                left++;
+            }
+        }
+        return cnt;
+    }
+}