Fixed Size Sliding Window Pattern

Overview

The Fixed Size Sliding Window pattern is a powerful algorithmic technique that involves maintaining a window of fixed size k while traversing through an array or string. This window "slides" through the data structure one element at a time, making it highly efficient for solving various problems that require analyzing subarrays or substrings of a specific length.

When to Use This Pattern

• Subarray Operations: Finding subarrays of fixed length that meet certain criteria
• String Pattern Matching: Locating fixed-length patterns or anagrams
• Array Averages: Computing averages of fixed-size subarrays
• Maximum/Minimum Values: Finding extreme values in fixed-size windows
• Binary Substrings: Analyzing fixed-length binary patterns
• Distinct Elements: Counting unique elements in fixed-size windows
• Sum Calculations: Computing sums of consecutive elements
• Statistical Metrics: Calculating medians or other statistics in windows
• Pattern Recognition: Identifying repeated patterns of fixed length
• Data Streaming: Processing fixed-size chunks of streaming data

Key Characteristics

• Fixed Window Size: Window length remains constant throughout traversal
• Linear Time Complexity: Typically achieves O(n) time complexity
• Constant Space: Usually requires O(1) or O(k) extra space
• Sequential Processing: Elements are processed in order
• Window Sliding: One element added, one removed per slide

Real-World Applications

• Moving Averages: Stock price analysis
• Network Monitoring: Fixed-time traffic analysis
• Image Processing: Pixel window operations
• Signal Processing: Fixed-width filters
• Data Analytics: Time-series analysis

Why Fixed Size Sliding Window Matters

• Efficiency: Eliminates redundant computations
• Optimization: Perfect for fixed-size subarray problems
• Simplicity: Easy to implement and understand
• Performance: Maintains consistent time complexity

Let's explore how the fixed size sliding window technique works in different programming languages and common problem patterns.

Implementation Examples

C++

// Maximum Average Subarray I
class Solution {
public:
    double findMaxAverage(vector<int>& nums, int k) {
        double sum = 0;
        // Calculate sum of first window
        for (int i = 0; i < k; i++) {
            sum += nums[i];
        }
        
        double maxSum = sum;
        // Slide window and update maxSum
        for (int i = k; i < nums.size(); i++) {
            sum = sum + nums[i] - nums[i - k];
            maxSum = max(maxSum, sum);
        }
        
        return maxSum / k;
    }
};

// Find All Anagrams
class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        if (s.length() < p.length()) return {};
        
        vector<int> result;
        vector<int> pCount(26, 0);
        vector<int> window(26, 0);
        
        // Count characters in pattern
        for (char c : p) {
            pCount[c - 'a']++;
        }
        
        // Initialize first window
        for (int i = 0; i < p.length(); i++) {
            window[s[i] - 'a']++;
        }
        
        if (window == pCount) result.push_back(0);
        
        // Slide window
        for (int i = p.length(); i < s.length(); i++) {
            window[s[i - p.length()] - 'a']--;
            window[s[i] - 'a']++;
            
            if (window == pCount) {
                result.push_back(i - p.length() + 1);
            }
        }
        
        return result;
    }
};

Java

// Maximum Average Subarray I
class Solution {
    public double findMaxAverage(int[] nums, int k) {
        double sum = 0;
        // Calculate sum of first window
        for (int i = 0; i < k; i++) {
            sum += nums[i];
        }
        
        double maxSum = sum;
        // Slide window and update maxSum
        for (int i = k; i < nums.length; i++) {
            sum = sum + nums[i] - nums[i - k];
            maxSum = Math.max(maxSum, sum);
        }
        
        return maxSum / k;
    }
}

// Find All Anagrams
class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        if (s.length() < p.length()) return new ArrayList<>();
        
        List<Integer> result = new ArrayList<>();
        int[] pCount = new int[26];
        int[] window = new int[26];
        
        // Count characters in pattern
        for (char c : p.toCharArray()) {
            pCount[c - 'a']++;
        }
        
        // Initialize first window
        for (int i = 0; i < p.length(); i++) {
            window[s.charAt(i) - 'a']++;
        }
        
        if (Arrays.equals(window, pCount)) result.add(0);
        
        // Slide window
        for (int i = p.length(); i < s.length(); i++) {
            window[s.charAt(i - p.length()) - 'a']--;
            window[s.charAt(i) - 'a']++;
            
            if (Arrays.equals(window, pCount)) {
                result.add(i - p.length() + 1);
            }
        }
        
        return result;
    }
}

Python

# Maximum Average Subarray I
class Solution:
    def findMaxAverage(self, nums: List[int], k: int) -> float:
        # Calculate sum of first window
        curr_sum = sum(nums[:k])
        max_sum = curr_sum
        
        # Slide window and update max_sum
        for i in range(k, len(nums)):
            curr_sum = curr_sum + nums[i] - nums[i - k]
            max_sum = max(max_sum, curr_sum)
        
        return max_sum / k

# Find All Anagrams
class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        if len(s) < len(p):
            return []
        
        result = []
        p_count = [0] * 26
        window = [0] * 26
        
        # Count characters in pattern
        for c in p:
            p_count[ord(c) - ord('a')] += 1
        
        # Initialize first window
        for i in range(len(p)):
            window[ord(s[i]) - ord('a')] += 1
        
        if window == p_count:
            result.append(0)
        
        # Slide window
        for i in range(len(p), len(s)):
            window[ord(s[i - len(p)]) - ord('a')] -= 1
            window[ord(s[i]) - ord('a')] += 1
            
            if window == p_count:
                result.append(i - len(p) + 1)
        
        return result

Visual Demonstration

Fixed-size window (k=3) sliding through the array

This animation demonstrates how a fixed-size sliding window operates on an array. Notice how the window (highlighted in blue) maintains its constant size of k=3 elements while sliding through the array. This visualization helps understand how the window processes consecutive subarrays of fixed length.

Common Patterns and Their Complexities

Pattern	Description	Time Complexity	Space Complexity
Sum/Average	Calculate sum/average of k elements	O(n)	O(1)
Pattern Match	Find fixed-length patterns	O(n)	O(k)
Distinct Elements	Count unique elements in k-window	O(n)	O(k)
Max/Min Value	Find extreme values in k-window	O(n)	O(1)
Median Finding	Find median in k-window	O(n log k)	O(k)

Practice Problems

Easy Level

1. Maximum Average Subarray I

   • Find maximum average value in subarray of size k
   • Time: O(n), Space: O(1)

2. Substrings of Size Three with Distinct Characters

   • Count substrings of size 3 with unique characters
   • Time: O(n), Space: O(1)

Medium Level

1. Find All Anagrams in a String

   • Find all anagrams of pattern in string
   • Time: O(n), Space: O(1)

2. Permutation in String

   • Check if string contains permutation of pattern
   • Time: O(n), Space: O(1)

3. Maximum Sum of Distinct Subarrays With Length K

   • Find maximum sum of k-length subarray with distinct elements
   • Time: O(n), Space: O(k)

4. Maximum Number of Vowels in a Substring of Given Length

   • Find maximum number of vowels in k-length substring
   • Time: O(n), Space: O(1)

Hard Level

1. Sliding Window Maximum

   • Find maximum element in each k-sized window
   • Time: O(n), Space: O(k)

2. Sliding Window Median

   • Find median in each k-sized window
   • Time: O(n log k), Space: O(k)

3. Maximum Sum of 3 Non-Overlapping Subarrays

   • Find three non-overlapping subarrays with maximum sum
   • Time: O(n), Space: O(n)

Tips for Fixed Size Sliding Window Problems

1. Initialize the first window correctly
2. Maintain window state efficiently while sliding
3. Handle edge cases: empty input, k > array length
4. Consider using auxiliary data structures for complex conditions
5. Optimize space usage by reusing variables when possible

Happy Coding! 🚀