Problem Description

The given problem requires us to find the number of substrings in a string s that contains at least one occurrence of the characters 'a', 'b', and 'c'. The string s consists only of these three characters. A substring is any sequence of consecutive characters from the string. The goal is to count all such possible substrings where each of the three characters appears at least once.

Intuition

To solve this problem, we use a sliding window approach to track the latest positions of 'a', 'b', and 'c' while iterating through the string. The key intuition here is to understand that once we have found a substring containing all three characters, extending this substring to the right (by adding more characters in sequence) will also form valid substrings containing all three characters.

Here’s the step-by-step intuition:

1. Initialize a dictionary to store the latest index of 'a', 'b', and 'c'. By default, they are set to -1, indicating that they haven't been found yet.

2. Iterate through the string character by character, updating the dictionary with the new index of each character encountered.

3. At each step, determine the smallest of the three indices because the smallest index indicates the rightmost position up to which we have seen all three characters together.

4. For the current index i, the count of valid substrings ending at i will be the minimum index among the latest indices of 'a', 'b', and 'c' plus one. This is because any substring starting from index 0 to the minimum index will have all three characters up to the current position i.

5. Sum up all these counts to get the total number of substrings containing all three characters.

This simple yet elegant solution effectively counts all the necessary substrings by considering each valid end position and how many substrings it can generate based on the earlier occurrences of 'a', 'b', and 'c'.

Learn more about Sliding Window patterns.

Solution Approach

The solution approach uses the concept of pointers and a hash map (in Python, a dictionary) to efficiently keep track of the latest occurrence of each character 'a', 'b', and 'c'.

d = {"a": -1, "b": -1, "c": -1}  # Dictionary to store the latest index for 'a', 'b', and 'c'
ans = 0  # This will hold the total count of substrings

The dictionary d serves as our hash map, holding the most recent indices of each character. Initially, all characters are set to -1, indicating they have not been encountered yet.

Then, we start iterating over each character in the string:

for i, c in enumerate(s):
    d[c] = i  # Update the latest index for the character `c`
    ans += min(d["a"], d["b"], d["c"]) + 1  # Count substrings ending at the current index `i`

As we iterate, for each character (c) at index i, we update its latest index in the dictionary. The min(d["a"], d["b"], d["c"]) finds the smallest index of the three, which, as previously mentioned, is the furthest right we can go while still having all three characters. By adding 1 to this minimum value, we get the number of substrings ending at index i that has all three characters.

Finally, the sum of all such substrings is returned as the final answer:

return ans

Algorithm

1. Initialize a dictionary with keys 'a', 'b', 'c', and values -1.

2. Start iterating over each character in the string using a for loop.

3. Each time we find a character, update its latest index in the dictionary.

4. Calculate the number of valid substrings that end at the current index (minimum index of 'a', 'b', 'c' + 1).

5. Accumulate this count to a running total.

6. After the loop finishes, return the total count as the answer.

This algorithm is efficient because it processes each character exactly once and uses constant space for the dictionary, resulting in a time complexity of O(n) and a space complexity of O(1), where n is the length of the string.

Example Walkthrough

Let's use a string s = "abcabc" to illustrate the solution approach.

1. We start by initializing the dictionary d with {"a": -1, "b": -1, "c": -1}.
2. Set our running total of valid substrings ans to 0.

Now we iterate through the string while keeping track of the latest index of each character in d.

• For i = 0 (character = 'a'):

  • Update d to {"a": 0, "b": -1, "c": -1}.
  • The minimum index among 'a', 'b', and 'c' is -1, so ans += -1 + 1, resulting in ans = 0.

• For i = 1 (character = 'b'):

  • Update d to {"a": 0, "b": 1, "c": -1}.
  • The minimum index among 'a', 'b', and 'c' is still -1, so ans += -1 + 1, resulting in ans = 0.

• For i = 2 (character = 'c'):

  • Update d to {"a": 0, "b": 1, "c": 2}.
  • Now we have all three characters, and the minimum index is 0, so ans += 0 + 1, resulting in ans = 1.

• For i = 3 (character = 'a'):

  • Update d to {"a": 3, "b": 1, "c": 2}.
  • The minimum index among 'a', 'b', and 'c' is now 1, so ans += 1 + 1, resulting in ans = 3.

• For i = 4 (character = 'b'):

  • Update d to {"a": 3, "b": 4, "c": 2}.
  • The minimum index among 'a', 'b', and 'c' is 2, so ans += 2 + 1, resulting in ans = 6.

• For i = 5 (character = 'c'):

  • Update d to {"a": 3, "b": 4, "c": 5}.
  • The minimum index among 'a', 'b', and 'c' is 3, so ans += 3 + 1, resulting in ans = 10.

After the loop finishes, we return the total count, which is 10.

Throughout this process, we efficiently tracked the latest positions of 'a', 'b', and 'c', calculated the substrings terminating at each character position, and accumulated this to find the total number of substrings containing all three characters.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n), where n is the length of the string s. This is because the code iterates over each character in the string exactly once. Within the loop, updating the dictionary and calculating the minimum value and the cumulative sum is done in constant time.

The space complexity of the code is O(1). The space is constant because the dictionary d only stores three key-value pairs regardless of the size of the input string, corresponding to the characters 'a', 'b', and 'c'.

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