Minimum Window Subsequence LeetCode Dynamic Programming Solution

Problem Description

Given strings S and T, find the minimum (contiguous) substring W of S, so that T is a subsequence of W. If there is no such window in S that covers all characters in T, return the empty string "&...".

Approach Overview

To solve this problem, we will use dynamic programming. The idea is to create a 2D table where each cell [i][j] represents the length of the minimum window in S that covers the first i characters of T. We will then use this table to find the minimum window in S that covers all characters in T.

Dynamic Programming Solution

Step 1: Initialize the Table

We will create a 2D table dp of size (len(S) + 1) x (len(T) + 1). The extra row and column are for handling edge cases.

def minWindow(S, T):
    m, n = len(S), len(T)
    dp = [[0] * (n + 1) for _ in range(m + 1)]

Step 2: Fill the Table

We will fill the table in a bottom-up manner. For each cell [i][j], we will consider two possibilities:

• If the current character in S is equal to the current character in T, we can either include this character in the window or not. In the first case, the length of the window will be dp[i-1][j-1] + 1. In the second case, the length of the window will be dp[i-1][j].
• If the current character in S is not equal to the current character in T, we cannot include this character in the window. In this case, the length of the window will be dp[i-1][j].

    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if S[i-1] == T[j-1]:
                dp[i][j] = dp[i-1][j-1] + 1
            else:
                dp[i][j] = dp[i-1][j]

Step 3: Find the Minimum Window

We will find the minimum window in S that covers all characters in T by finding the maximum value in the last row of the table.

    max_len = 0
    min_window = ""
    for i in range(m + 1):
        if dp[i][n] > max_len:
            max_len = dp[i][n]
            min_window = S[i-max_len+1:i+1]
    return min_window

Example Use Cases

print(minWindow("bba", "ab"))  # Output: "ba"
print(minWindow("ab", "bba"))  # Output: "ab"
print(minWindow("c", "c"))  # Output: "c"
print(minWindow("c", "cbbd"))  # Output: "c"
print(minWindow("c", "cbbdab"))  # Output: "c"

Time Complexity

The time complexity of this solution is O(m * n), where m and n are the lengths of S and T respectively.

Space Complexity

The space complexity of this solution is O(m * n), where m and n are the lengths of S and T respectively.

Conclusion

Q: What is the Minimum Window Subsequence problem?

A: The Minimum Window Subsequence problem is a classic problem in computer science where we are given two strings S and T, and we need to find the minimum (contiguous) substring W of S, so that T is a subsequence of W.

Q: What is a subsequence?

A: A subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

Q: How do we solve the Minimum Window Subsequence problem?

A: We can solve the Minimum Window Subsequence problem using dynamic programming. The idea is to create a 2D table where each cell [i][j] represents the length of the minimum window in S that covers the first i characters of T.

Q: What is the time complexity of the dynamic programming solution?

A: The time complexity of the dynamic programming solution is O(m * n), where m and n are the lengths of S and T respectively.

Q: What is the space complexity of the dynamic programming solution?

A: The space complexity of the dynamic programming solution is O(m * n), where m and n are the lengths of S and T respectively.

Q: Can we solve the Minimum Window Subsequence problem using other methods?

A: Yes, we can solve the Minimum Window Subsequence problem using other methods such as greedy algorithm or two-pointer technique. However, the dynamic programming solution is more efficient and easier to implement.

Q: What are some common use cases of the Minimum Window Subsequence problem?

A: The Minimum Window Subsequence problem has many use cases in computer science and real-world applications such as:

• Text search and retrieval
• Data compression
• Bioinformatics
• Natural language processing

Q: How do we handle edge cases in the Minimum Window Subsequence problem?

A: We can handle edge cases in the Minimum Window Subsequence problem by:

• Checking if the input strings are empty
• Handling cases where T is a subsequence of S but not a contiguous substring
• Handling cases where S is a subsequence of T but not a contiguous substring

Q: Can we optimize the dynamic programming solution further?

A: Yes, we can optimize the dynamic programming solution further by:

• Using a more efficient data structure such as a hash table or a trie
• Implementing a more efficient algorithm such as the two-pointer technique
• Using parallel processing or distributed computing to speed up the solution

Q: What are some common pitfalls in solving the Minimum Window Subsequence problem?

A: Some common pitfalls in solving the Minimum Window Subsequence problem include:

• Not handling edge cases properly
• Not using an efficient algorithm or data structure
• Not testing the solution thoroughly

Q: How do we test the dynamic programming solution?

A: We can test the dynamic programming solution by:

• Using a test framework such as Pytest or Unittest
• Writing unit tests to cover different edge cases
• Using a debugger to step through the code and identify any issues

Q: Can we use the dynamic programming solution in a real-world application?

A: Yes, we can use the dynamic programming solution in a real-world application such as:

• A text search engine
• A data compression algorithm
• A bioinformatics tool

Q: What are some future directions for the Minimum Window Subsequence problem?

A: Some future directions for the Minimum Window Subsequence problem include:

• Developing more efficient algorithms and data structures
• Applying the problem to new domains and applications
• Investigating the theoretical limits of the problem