Intuition - Optimized Sliding Window
Naive Approach\nIn the naive approach, we aim to find the most frequent substring by generating all substrings of sizes between minSize and maxSize, checking co
# Naive Approach\nIn the naive approach, we aim to find the most frequent substring by generating **all substrings** of sizes between `minSize` and `maxSize`, checking constraints, and tracking their frequencies. However, this results in significant overhead due to:\n1. **Overlapping Substrings**:\n   - For example, substrings `"abc"`, `"abcd"`, and `"abcde"` share the same prefix `"abc"`, but are treated as separate entities, leading to redundant work.\n2. **Redundant Size Checks**:\n   - Tracking all possible sizes introduces unnecessary complexity since substrings of smaller sizes inherently "cover" their larger extensions.\n3. **High Computational Cost**:\n   - Generating substrings and validating constraints involves nested loops, resulting in **O(n^3)** time complexity.\n\n### Complexity Analysis\n- **Time Complexity**:\n  - Generating all substrings takes **O(n^2)** in the worst case.\n  - Checking constraints for each substring adds another **O(k)**, where `k` is the substring length.\n  - Total complexity: **O(n^3)**.\n- **Space Complexity**:\n  - Storing all substrings in a hash map consumes **O(n^2)** space.\n\nThis approach is impractical for large inputs and redundant due to overlapping substrings.\n\n---\n\n# Optimized Approach: First Character as the Anchor\n\n### Intuition\nInstead of generating all substrings, we leverage the **first character** of each substring as a natural **anchor** or **lower bound** for substring analysis. This eliminates redundancy and focuses only on valid substrings.\n\n#### Why the First Character Acts as the Lower Bound:\n1. **Foundation for Substrings**:\n   - All substrings starting with the same character inherently depend on that character\'s occurrence.\n   - For example, in `"abcdeabcde"`, substrings starting with `"a"` include `"abc"`, `"abcd"`, and `"abcde"`.\n   - Tracking substrings anchored at `"a"` ensures we capture all substrings starting with `"a"` without repeatedly analyzing overlapping substrings.\n   \n2. **Efficient Grouping**:\n   - Instead of checking every substring size, we track substrings of **exactly `minSize`**, which act as representatives for their larger extensions.\n   - For example:\n     - `"abcde"` can be represented by its prefix `"abc"` (of size `minSize`), as any occurrence of `"abcde"` inherently implies an occurrence of `"abc"`.\n   - This avoids redundant counting and reduces unnecessary size checks.\n\n3. **Frequency Maximization**:\n   - By focusing on the most frequent occurrence of substrings starting with a given first character, we inherently maximize substring frequency.\n\n---\n\n### Why Focus on Substrings of Size `minSize`?\n1. **Constraint Satisfaction**:\n   - Substrings smaller than `minSize` are invalid, while larger substrings are inherently counted through their `minSize` prefixes.\n   - For example:\n     - If `"abcde"` appears twice, its prefix `"abc"` (of size `minSize`) will also appear twice. Tracking `"abc"` suffices.\n2. **Eliminating Redundancy**:\n   - Larger substrings are automatically included in the counts of their `minSize` prefixes, so there\u2019s no need to explicitly track them.\n3. **Efficiency**:\n   - Limiting analysis to substrings of size `minSize` reduces both computational overhead and space usage.\n\n---\n\n### Complexity Analysis\n- **Time Complexity**:\n  - The sliding window iterates through the string in **O(n)**.\n  - Extracting and validating substrings of size `minSize` takes **O(minSize)** per position.\n  - Overall complexity: **O(n * minSize)**.\n- **Space Complexity**:\n  - Storing substring frequencies in a hash map results in **O(n)** space.\n\n---\n\n# Code\n```java []\nclass Solution {\n    public int maxFreq(String s, int maxLetters, int minSize, int maxSize) {\n        int maxFreq = 0;\n        Map<Character, Integer> window = new HashMap<>();\n        Map<String, Integer> substringToCount = new HashMap<>();\n\n        int left = 0;\n        for (int right = 0; right < s.length(); right++) {\n            // Expand window\n            char cr = s.charAt(right);\n            window.put(cr, window.getOrDefault(cr, 0) + 1);\n\n            // Fixed size sliding window\n            if (right - left + 1 == minSize) {\n                // Check if the window has at most \'maxLetters\' distinct characters\n                if (window.size() <= maxLetters) {\n                    String sub = s.substring(left, right + 1);\n                    substringToCount.put(sub, substringToCount.getOrDefault(sub, 0) + 1);\n                    maxFreq = Math.max(maxFreq, substringToCount.get(sub));\n                }\n\n                // Shrink window\n                char cl = s.charAt(left);\n                window.put(cl, window.get(cl) - 1);\n                if (window.get(cl) == 0)\n                    window.remove(cl);\n                left++;\n            }\n        }\n\n        return maxFreq;\n    }\n}\n\n```
['Java']