Problem Description

Given a string s and an integer k, the task is to count how many substrings of length k exist within s that contain no repeated characters. A substring is a contiguous sequence of characters within a string. For example, in the string abcde, abc, bcd, and cde are substrings of length 3. We are specifically interested in substrings where every character is unique, meaning no character appears more than once in that substring.

Intuition

The intuition behind the solution is to use the "sliding window" technique. This approach involves maintaining a window of characters that can expand or shrink as needed while scanning the string from left to right. Here's how the intuition develops:

1. If k is greater than the length of the string or if k is greater than 26 (the number of letters in the English alphabet), there can't be any valid substrings of length k with all unique characters, so we return 0 immediately.

2. Starting at the beginning of the string, we expand our window to include characters until we hit a size of k or we find a duplicate character.

3. To efficiently keep track of the characters inside our current window and their counts, we use a Counter (a type of dictionary or hashmap in Python), incrementing the count for each new character we add to the window.

4. If at any point a character's count exceeds 1 (meaning we've found a duplicate), or the window's size exceeds k, we shrink the window from the left by removing characters and decrementing their respective counts.

5. Every time our window size becomes exactly k and all characters are unique (i.e., no count is greater than 1), we have found a valid substring. We then increment the counter for the final answer (ans).

Notice that for each position i in the string, we adjust the left side of our window (j) to ensure the window size does not exceed k and there are no repeating characters. The solution iteratively keeps account of all such substrings and returns the count as the final result.

Learn more about Sliding Window patterns.

Solution Approach

The solution employs the sliding window approach along with a Counter from the collections module to keep track of the number of occurrences of each character within the current window. Here's a step-by-step breakdown:

1. Initialize Variables: A counter named cnt is used to count occurrences of characters, ans for storing the total count of valid substrings found, and j for marking the start position of the sliding window.

2. Early Termination Checks: Check if k is larger than the string length n or greater than 26. If so, we return 0 because it's not possible to have substrings of length k with unique characters.

3. Iterate Over the String: Use a for-loop with enumerate(s) to get both index i and character c.

4. Expand the Window: Increase the count of the current character c in cnt.

5. Shrink the Window if Necessary: a. While the count of character c in cnt is more than 1 (meaning c has appeared before in the current window), decrease the count of the leftmost character s[j] and increase j to move the window's left edge to the right. b. Also, shrink the window if the current window length (i - j + 1) exceeds k.

6. Check Window Validity: If the window length is exactly k, and all characters in the window are unique (the while loop has ensured that the counts are 1), increment ans which signifies a valid substring.

7. Return Result: After finishing the loop, the ans variable contains the number of substrings with length k and no repeating characters, which is returned as the solution.

This solution uses a dynamic slide of the window that only moves the left bound forward (j), and never backward, ensuring that the algorithm's time complexity is O(n), as each character is processed at most twice (once when added to the window, once when removed). This makes the algorithm efficient for this problem.

Example Walkthrough

Let's illustrate the solution approach with a concrete example. Consider the string s = "abcba" and k = 3. We want to find the total number of valid substrings of length 3 with no repeating characters.

1. Initialize Variables: Initialize cnt = Counter(), ans = 0, j = 0.

2. Early Termination Checks: Length of s is 5, and k is 3, which is less than 26, so we move on without terminating early.

3. Iterate Over the String: We start a for-loop that iterates over each character.

   • On the first iteration i = 0, c = 'a'. We add 'a' to the counter.
   • Then, the window size is 1, which is less than 3. Since there are no duplicates, we continue.

4. On the second iteration i = 1, c = 'b'. We add 'b' to the counter.

   • Now the window size is 2, which is still less than 3. No duplicates so, we continue.

5. In the third iteration i = 2, c = 'c'. We add 'c' to the counter.

   • Our window is now abc (i - j + 1 = 3), which is equal to k and contains no repeating characters.
   • We found a valid substring, so ans = ans + 1.

6. Fourth iteration i = 3, c = 'b'. Counter for 'b' becomes 2, indicating a duplicate.

   • We proceed to shrink the window. We reduce the count of 'a' since it's the leftmost character in the current window and move j from 0 to 1.
   • The window length is again 3 after adjusting j, but now it's bcb which is not valid because it contains repeating characters 'b'.

7. Fifth iteration i = 4, c = 'a'. The counter for 'a' is now 2, another duplicate.

   • Continuing to shrink the window, we decrease the count for 'b' and increment j to 2.
   • The window size (i - j + 1 = 3 - 2 + 1) now becomes 2. Since it's less than k, we continue to expand the window in the next iterations.

8. At this point, the substring cba from the previous steps is no longer in our current window. We have a window of size 2 with unique characters cb. Unfortunately, this is the last iteration, and there are no more characters in s to consider.

9. Return Result: No other valid substrings of length 3 are found, so the final ans remains 1.

In conclusion, there is only one valid substring of length 3 with all unique characters in the string abcba, which is abc. Hence the function would return 1.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the string s. This is because the code uses a sliding window approach, traversing the string once. Each character is added to the sliding window, and if a condition is met (more than one occurrence of the same character, or the window size exceeds k), the window adjusts by removing characters from the start. The number of adjustments made is proportional to n as well.

The space complexity of the code is O(k) if k < 26, or O(26) (which simplifies to O(1)) if k >= 26, due to the length of the Counter dictionary never having more characters than the alphabet case (k characters when k < 26, and 26 characters of the alphabet when k >= 26 because no more than 26 unique characters can be in the string at a time given it consists of just the alphabet).

Learn more about how to find time and space complexity quickly using problem constraints.