Minimum Window Subsequence 🎯

Welcome to this comprehensive guide on the Minimum Window Subsequence problem! This is a fascinating topic that will help you understand and solve real-world coding problems. Let's dive in!

Understanding the Problem 📝

The Minimum Window Subsequence problem requires finding the smallest possible substring within a given string S that contains all the unique characters in another given string T, in the same order as they appear in T.

For example, let's consider S = "ABCDEFE" and T = "DEF". Here, the minimum window subsequence is "DEF" itself, since it appears in S and contains all the characters in T.

Algorithm Overview 💡

To solve the Minimum Window Subsequence problem, we'll use a Sliding Window approach. This method involves moving a window of characters across the string and keeping track of the characters that need to be included in the solution.

Here's an outline of the algorithm:

Initialize an empty dictionary freq_t to store the frequency of characters in T.
Initialize left and right pointers to 0 and 1, respectively, to define the initial window in S.
While right is less than the length of S, perform the following steps:
- Increment the frequency of the character at index right in a new dictionary freq_s.
- If the frequency of a character in freq_s equals or exceeds the frequency of that character in freq_t, remove that character from freq_t.
- While freq_t still has characters and the frequency of the last character in freq_s is greater than the frequency of the last character in freq_t, remove the last character from freq_s.
- Calculate the length of the current window by subtracting left from right. If the window length is 0 or greater than the current minimum window length, update the minimum window length.
- Move the left pointer one position to the right.
Once the loop is finished, the current window will be the minimum window subsequence.

Code Example ✅

Here's a Python code example that demonstrates the Minimum Window Subsequence problem using the Sliding Window approach:

python

def min_window_subsequence(s, t):
    freq_t = {}
    for char in t:
        if char in freq_t:
            freq_t[char] += 1
        else:
            freq_t[char] = 1

    freq_s = {}
    left, right, min_len = 0, 0, float('inf')
    while right < len(s):
        freq_s[s[right]] = freq_s.get(s[right], 0) + 1

        while len(freq_t) > 0 and all([freq_t[char] <= freq_s[char] for char in freq_t]):
            char = list(freq_t.keys())[0]
            del freq_t[char]

        while len(freq_t) > 0 and len(freq_s) > len(freq_t) and freq_s[list(freq_s.keys())[-1]] > freq_t[list(freq_t.keys())[-1]]:
            char = list(freq_s.keys())[-1]
            del freq_s[char]

        if len(freq_t) == 0 or right - left + 1 < min_len:
            min_len = right - left + 1
            min_window = s[left:right + 1]

        left += 1
        right += 1

    return min_window

Practice Time 💡

Now that you've learned about the Minimum Window Subsequence problem and the Sliding Window approach, it's time to put your knowledge into practice! Try solving the following problem and see if you can come up with the correct solution: