Minimum Window Subsequence LeetCode Dynamic Programming Solution

by ADMIN 65 views

Introduction

In this article, we will discuss the Minimum Window Subsequence problem on LeetCode, which involves finding the minimum contiguous substring of a given string S such that a target string T is a subsequence of it. We will explore a dynamic programming solution to this problem, which is an efficient approach to solve this challenge.

Problem Statement

Given two strings S and T, find the minimum contiguous substring W of S such 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 "".

Example Use Cases

  • Input: S = "abcda", T = "ad" Output: "ad"
  • Input: S = "abcda", T = "a" Output: "a"
  • Input: S = "abcda", T = "ab" Output: "ab"

Dynamic Programming Solution

To solve this problem, we can use dynamic programming to build 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 initialize the table with zeros and then fill it up row by row.

Code

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

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
            if dp[i][j] > max_length:
                max_length = dp[i][j]
                start = i - max_length
                end = i

if max_length == 0:
    return ""

return S[start:end]

Explanation

The dynamic programming solution works as follows:

  1. We initialize a 2D table dp of size (m + 1) x (n + 1), where m and n are the lengths of S and T, respectively.
  2. We iterate over the characters of S and T simultaneously, and for each pair of characters, we check if they match.
  3. If the characters match, we update the value of dp[i][j] to be the length of the minimum window in S that covers the first i characters of T, which is dp[i - 1][j - 1] + 1.
  4. We keep track of the maximum length of the minimum window found so far and the corresponding start and end indices.
  5. Finally, we return the minimum window substring S[start:end].

Time Complexity

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

Space Complexity

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

Conclusion

In this article, we discussed the Minimum Window Subsequence problem on LeetCode and presented a dynamic programming solution to it. The solution works by building a 2D table where each cell represents the length of the minimum window in S that covers the first i characters of T. We then iterate over the characters of S and T simultaneously and update the table accordingly. Finally, we return the minimum window substring S[start:end]. The time and space complexities of the solution are O(m * n) and O(m * n), respectively.

Example Use Cases

  • Input: S = "abcda", T = "ad" Output: "ad"
  • Input: S = "abcda", T = "a" Output: "a"
  • Input: S = "abcda", T = "ab" Output: "ab"

Advice

  • When solving the Minimum Window Subsequence problem, make sure to use a dynamic programming approach to build a 2D table where each cell represents the length of the minimum window in S that covers the first i characters of T.
  • Iterate over the characters of S and T simultaneously and update the table accordingly.
  • Keep track of the maximum length of the minimum window found so far and the corresponding start and end indices.
  • Finally, return the minimum window substring S[start:end].

Related Problems

  • Minimum Window Substring: Find the minimum window substring of S that contains all characters of T.
  • Longest Common Subsequence: Find the longest common subsequence of S and T.
  • Edit Distance: Find the minimum number of operations (insertions, deletions, and substitutions) required to transform S into T.
    Minimum Window Subsequence LeetCode Dynamic Programming Solution: Q&A ====================================================================

Introduction

In our previous article, we discussed the Minimum Window Subsequence problem on LeetCode and presented a dynamic programming solution to it. In this article, we will answer some frequently asked questions (FAQs) related to the problem and its solution.

Q: What is the Minimum Window Subsequence problem?

A: The Minimum Window Subsequence problem is a string matching problem where we need to find the minimum contiguous substring of a given string S such that a target string T is a subsequence of it.

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 does the dynamic programming solution work?

A: The dynamic programming solution works by building 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 then iterate over the characters of S and T simultaneously and update the table accordingly.

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 use a different approach to solve the Minimum Window Subsequence problem?

A: Yes, we can use a different approach to solve the Minimum Window Subsequence problem. For example, we can use a sliding window approach or a greedy algorithm.

Q: What are some common pitfalls to avoid when solving the Minimum Window Subsequence problem?

A: Some common pitfalls to avoid when solving the Minimum Window Subsequence problem include:

  • Not initializing the 2D table correctly
  • Not updating the table correctly
  • Not keeping track of the maximum length of the minimum window found so far
  • Not returning the correct minimum window substring

Q: How can we optimize the dynamic programming solution?

A: We can optimize the dynamic programming solution by:

  • Using a more efficient data structure to store the 2D table
  • Using a more efficient algorithm to update the table
  • Using a more efficient approach to find the maximum length of the minimum window found so far

Q: Can we use the dynamic programming solution to solve other string matching problems?

A: Yes, we can use the dynamic programming solution to solve other string matching problems. For example, we can use it to solve the Longest Common Subsequence problem or the Edit Distance problem.

Conclusion

In this article, we answered some frequently asked questions (FAQs) related to the Minimum Window Subsequence problem and its dynamic programming. We hope that this article has been helpful in clarifying any doubts you may have had about the problem and its solution.

Related Problems

  • Minimum Window Substring: Find the minimum window substring of S that contains all characters of T.
  • Longest Common Subsequence: Find the longest common subsequence of S and T.
  • Edit Distance: Find the minimum number of operations (insertions, deletions, and substitutions) required to transform S into T.

Advice

  • When solving the Minimum Window Subsequence problem, make sure to use a dynamic programming approach to build a 2D table where each cell represents the length of the minimum window in S that covers the first i characters of T.
  • Iterate over the characters of S and T simultaneously and update the table accordingly.
  • Keep track of the maximum length of the minimum window found so far and the corresponding start and end indices.
  • Finally, return the minimum window substring S[start:end].

Example Use Cases

  • Input: S = "abcda", T = "ad" Output: "ad"
  • Input: S = "abcda", T = "a" Output: "a"
  • Input: S = "abcda", T = "ab" Output: "ab"

Code

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

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
            if dp[i][j] > max_length:
                max_length = dp[i][j]
                start = i - max_length
                end = i

if max_length == 0:
    return ""

return S[start:end]