Master Fixed-Length Sliding Window: A Universal Approach

Core Concept

We aim to calculate the maximum number of vowels in any substring with a length of exactly k. While brute-forcing all substrings results in a time complexity of O(nk), this approach is too slow. Can we achieve O(1) substring property calculations? Yes! For instance:

In the string “abci”, if we already know the vowel count in substring “abc”, then to compute it for “bci”:

1. Check if the leaving character (‘a’) is a vowel.
2. Check if the entering character (‘i’) is a vowel.

This works because the middle characters (‘b’ and ‘c’) remain unchanged in both substrings.

Example Walkthrough

Input: s = “abciiidef”, k = 3

Step-by-step:

1. Traverse s from left to right.
2. Count vowels in the first k − 1 = 2 characters. Initially, there is 1 vowel.
3. Start processing the sliding window:
   • s[2] = ‘c’ enters window, forming “abc” (1 vowel). Update max count. Then s[0] = ‘a’ exits window, reducing the count to 0.
   • s[3] = ‘i’ enters window, forming “bci” (1 vowel). Update max count. Then s[1] = ‘b’ exits, keeping the count at 1.
   • s[4] = ‘i’ enters window, forming “cii” (2 vowels). Update max count. Then s[2] = ‘c’ exits, keeping the count at 2.
   • s[5] = ‘i’ enters window, forming “iii” (3 vowels). Update max count. Then s[3] = ‘i’ exits, reducing the count to 2.
   • s[6] = ‘d’ enters window, forming “iid” (2 vowels). Update max count. Then s[4] = ‘i’ exits, reducing the count to 1.
   • s[7] = ‘e’ enters window, forming “ide” (2 vowels). Update max count. Then s[5] = ‘i’ exits, reducing the count to 1.
   • s[8] = ‘f’ enters window, forming “def” (1 vowel). Update max count. Traversal complete.

Fixed-Length Sliding Window Pattern

This pattern follows three simple steps: Enter-Update-Exit

1. Enter: Element at index i enters the window; update statistics. Repeat if i < k − 1.
2. Update: Update the answer (usually max/min value).
3. Exit: Element at index i − k + 1 exits the window; update statistics.

This pattern is universally applicable to all fixed-length sliding window problems.

Implementation

class Solution {
    public int maxVowels(String S, int k) {
        char[] s = S.toCharArray();
        int ans = 0;
        int vowel = 0;
        for (int i = 0; i < s.length; i++) {
            // 1. Enter window
            if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' ||
                s[i] == 'o' || s[i] == 'u') {
                vowel++;
            }
            if (i < k - 1) { // Window size less than k
                continue;
            }
            // 2. Update answer
            ans = Math.max(ans, vowel);
            // 3. Exit window
            char out = s[i - k + 1];
            if (out == 'a' || out == 'e' || out == 'i' ||
                out == 'o' || out == 'u') {
                vowel--;
            }
        }
        return ans;
    }
}

Complexity Analysis

• Time Complexity: O(n), where n is the length of s.
• Space Complexity: O(1), using only a few extra variables.

Key Benefits

1. Universal applicability to fixed-length sliding window problems.
2. Simple and easy-to-remember three-step process: Enter-Update-Exit.
3. Efficient O(n) time complexity.
4. Minimal space usage (O(1)).

Best Practices

1. Initialize your window with the first k − 1 elements.
2. Update the answer after forming the complete window.
3. Update statistics for both entering and exiting elements.
4. Handle boundary conditions carefully, especially during window formation.

Applications

This pattern can be adapted for various fixed-length sliding window problems, such as:

• Finding the maximum/minimum sum of k consecutive elements.
• Finding the maximum/minimum average of k consecutive elements.
• Counting occurrences of specific patterns in k-length windows.
• Calculating statistics over k-length sliding windows.

Mastering this pattern empowers you to solve a wide range of problems efficiently and effectively!