Fast, Readable Solution - Sliding window
Intuition\nThe intuition behind the solution to count substrings that contain at least one character appearing at least k times revolves around the concept of m
# Intuition\nThe intuition behind the solution to count substrings that contain at least one character appearing at least `k` times revolves around the concept of maintaining a dynamic window (or substring) and efficiently checking conditions as we expand and contract this window. Here\u2019s a breakdown of the key intuitive elements:\n\n### 1. **Sliding Window Technique**:\n   - The sliding window technique allows us to explore substrings without generating all possible combinations explicitly. Instead of checking every possible substring, we maintain a window defined by two pointers (`left` and `right`).\n   - As we move the `right` pointer to expand the window, we include new characters and update our character count. This allows us to dynamically adjust our view of the substring.\n\n### 2. **Dynamic Character Count**:\n   - By using a `HashMap` to keep track of character frequencies, we can quickly determine how many times each character appears in the current substring.\n   - This dynamic counting is crucial because it allows us to check the condition (at least one character appearing `k` times) efficiently without needing to recount characters from scratch for every substring.\n\n### 3. **Condition Checking**:\n   - The inner loop checks if the current substring meets the condition of having at least one character with `k` occurrences. If it does, we can count all valid substrings that can be formed from the current `left` to `right` position.\n   - This is based on the observation that if a substring is valid (i.e., it contains a character with at least `k` occurrences), then all extensions of this substring (by moving `right` further) will also be valid until the condition is violated.\n\n### 4. **Counting Valid Substrings**:\n   - When we find a valid substring, we count all possible substrings that can be formed from the current `left` to `right` position. This is efficient because it leverages the fact that once a valid substring is found, all subsequent substrings that start from the same `left` and end at `right`, `right + 1`, etc., are also valid.\n   - This counting mechanism avoids the need for nested loops, which would be computationally expensive.\n\n### 5. **Adjusting the Window**:\n   - After counting valid substrings, we increment the `left` pointer to explore new potential substrings. This adjustment is crucial because it allows us to maintain the condition while exploring different starting points for substrings.\n   - By decrementing the count of the character at the `left` index, we can check if the condition still holds. If it doesn\u2019t, we continue adjusting the `left` pointer until we find a valid substring again.\n\n### Conclusion\nThe intuition behind this solution is to leverage the sliding window technique combined with dynamic character counting to efficiently explore and count valid substrings. By maintaining a balance between expanding and contracting the window, we can ensure that we consider all possible substrings while adhering to the specified condition, resulting in an optimal and elegant solution.\n\n# Approach\nLet\'s break down how the code ensures that all substrings are considered while counting those that contain at least one character appearing at least `k` times.\n\n### Code Breakdown\n\n1. **Initialization**:\n   - We initialize a `HashMap<Character, Integer> charCount` to keep track of the frequency of each character in the current substring.\n   - We also initialize two pointers: `left` (starting point of the substring) and `right` (ending point of the substring).\n\n2. **Outer Loop**:\n   - The outer loop iterates through the string with the `right` pointer, which expands the current substring by including the character at the `right` index.\n   - For each character added to the substring, we update its count in the `charCount` map.\n\n3. **Inner While Loop**:\n   - The inner `while` loop checks if the current substring (from `left` to `right`) contains at least one character that appears `k` times by calling the `occurs` method.\n   - If the condition is met, it means we have found a valid substring.\n\n4. **Counting Valid Substrings**:\n   - When the condition is satisfied, we count all possible substrings that can be formed from the current `left` to `right` position. This is done using the expression `count += s.length() - right;`.\n   - This expression counts all substrings that start from `left` and end anywhere from `right` to the end of the string. For example, if `right` is at index 3, and the string length is 5, it counts substrings: `s[left..3]`, `s[left..4]`, and `s[left..5]`.\n\n5. **Adjusting the Left Pointer**:\n   - After counting, we need to check if we can still maintain the condition of having at least one character appearing `k` times while moving the `left` pointer to the right.\n   - We decrement the count of the character at the `left` index in the `charCount` map. If the count of that character drops to zero, we remove it from the map.\n   - The `left` pointer is then incremented to explore the next potential substring.\n\n6. **Continuing the Process**:\n   - The outer loop continues to expand the substring by moving the `right` pointer until all characters in the string have been processed.\n   - The inner loop ensures that we only count valid substrings while maintaining the condition of at least one character appearing `k` times.\n\n### Ensuring All Substrings Are Considered\n\n- **Comprehensive Coverage**: By using the two-pointer technique, we systematically explore all possible substrings. The `right` pointer expands the substring, while the `left` pointer contracts it when necessary. This guarantees that every possible starting point (`left`) and ending point (`right`) combination is considered.\n  \n- **Counting Mechanism**: The counting mechanism (`count += s.length() - right;`) ensures that for every valid substring found, all possible extensions of that substring are counted. This means that if a valid substring is found at a certain `right` position, all substrings starting from `left` to `right`, `left` to `right + 1`, and so on, are included in the count.\n\n- **Dynamic Adjustment**: The dynamic adjustment of the `left` pointer allows the algorithm to explore new substrings while ensuring that the condition of having at least one character appearing `k` times is maintained. This means that as we move through the string, we are continuously checking and counting valid substrings without missing any.\n\n### Conclusion\n\nOverall, the combination of the sliding window technique, the frequency count of characters, and the systematic counting of valid substrings ensures that all possible substrings are considered while adhering to the specified condition. This results in an efficient and comprehensive solution to the problem.\n# Complexity\n- Time complexity:\n  - The overall approach is efficient because it processes each character in the string a limited number of times (once when added and once when removed), leading to a linear time complexity of O(n), where n is the length of the string.\n   - This efficiency is achieved by avoiding the need to generate all substrings explicitly and instead focusing on maintaining a valid window of characters.\n\n- Space complexity:\n- O(m), where m is the number of unique characters in the string.\nIn practical scenarios with a limited character set, this can be treated as O(1). However, in the general case, it is O(m).\n\nLet\'s analyze it step by step:\n1. Character Count Storage:\nThe solution uses a HashMap<Character, Integer> to keep track of the frequency of characters in the current substring.\nIn the worst case, if all characters in the string are unique, the HashMap will store an entry for each character. Therefore, the space required for the HashMap is O(m), where m is the number of unique characters in the string.\n2. Input String:\nThe input string s itself takes O(n) space, where n is the length of the string. However, this space is typically not counted in the space complexity analysis of the algorithm itself, as it is part of the input.\n3. Other Variables:\nThe algorithm uses a few integer variables (count, left, right) and a constant amount of additional space for the loop control and function calls. This contributes O(1) to the space complexity.\n\n# Code\n```java []\n\nclass Solution {\n\n    public int numberOfSubstrings(String s, int k) {\n        return countSubstrings(s, k);\n    }\n\n    private int countSubstrings(String s, int k) {\n        int count = 0;\n        HashMap<Character, Integer> charCount = new HashMap<>();\n        int left = 0;\n\n        for (int right = 0; right < s.length(); right++) {\n            charCount.put(s.charAt(right), charCount.getOrDefault(s.charAt(right), 0) + 1);\n\n            while (occurs(charCount, k)) {\n                count += s.length() - right; \n                charCount.put(s.charAt(left), charCount.get(s.charAt(left)) - 1);\n                if (charCount.get(s.charAt(left)) == 0) {\n                    charCount.remove(s.charAt(left));\n                }\n                left++;\n            }\n        }\n\n        return count;\n    }\n\n    private boolean occurs(HashMap<Character, Integer> charCount, int k) {\n        for (int count : charCount.values()) {\n            if (count >= k) {\n                return true;\n            }\n        }\n        return false;\n    }\n}\n```
['Sliding Window', 'Java']