Problem Description

In this problem, you are given two strings s1 and s2. Your task is to find the shortest contiguous substring in s1 such that the entire string s2 is a subsequence of that substring. A subsequence is a sequence that can be derived from the other sequence by deleting some or no elements without changing the order of the remaining elements.

For example, if s1 is "abcdebdde" and s2 is "bde", you have to find the shortest part of s1 that contains all the characters in s2 in the same order.

Here are the conditions you need to satisfy:

• If no such window exists, return an empty string "".
• If there are multiple substrings of s1 that satisfy the condition, return the one with the left-most starting index.

In the example given, the minimum window in s1 where s2 is a subsequence is "bcde".

Intuition

The key to solving this problem is understanding dynamic programming and the process of subsequence matching. We must scan through s1 and s2 to find all possible match positions in a way that allows us to efficiently calculate the minimum window length where s2 is a subsequence.

The intuition behind the solution is to first find all the matchings between characters of s2 with s1, and keep track of the starting index of the sequence in s1 that matches up to a certain point in s2.

To solve this, we create a 2D array f where f[i][j] represents the starting index in s1 from which we have a matching sequence for s2 up to its j-th character at i-th index of s1.

Once we have this information stored in the f array, we iterate through s1 to find the character that matches the last character of s2. Each time we find such a match, we look into our f array for the starting index and calculate the window size.

The minimum window size is kept updated during the scan, and finally, we have the starting index and the size of our minimum window which we use to return our result.

The entire algorithm takes into account:

• Dynamic Programming to solve the subsequence matching.
• Iterating s1 and s2 smartly to minimize unnecessary comparisons.
• Keeping track of the minimum window during the iteration.
• Handling edge cases effectively, like when no window exists.

Learn more about Dynamic Programming and Sliding Window patterns.

Solution Approach

The solution provided uses a two-dimensional dynamic programming approach:

1. Initialization:

   • The 2D array f is created with a size of (m+1) x (n+1) where m is the length of s1 and n is the length of s2. This array will help track matches between s1 and s2.

2. Filling the DP array:

   • We iterate over both strings starting from index 1 (because we've initialized from 0 as part of the dynamic programming table setup), comparing each character of s1 with each character of s2.
   • When a match (a == b) is found:
     • If it is the first character of s2 (j == 1), we record this position i as a starting point of a matching sequence because it could potentially start a new subsequence.
     • For other characters, we copy the starting index from f[i - 1][j - 1], which means we extend the subsequence found until the previous character of s2.
   • When there is no match, we get the starting index from the previous value f[i - 1][j] because we want to retain the starting position of the best match found so far for s2 up to j.

3. Identifying the minimum window:

   • Now, we look for the last character of s2 in s1 to try and close a potential window.
   • For each matching position i in s1 where s1[i] == s2[n - 1] and a subsequence match has been found (f[i][n] is not zero), we calculate the window size by subtracting the starting index j (which is f[i][n] - 1) from i. We keep track of the smallest window found in variables k for size and p for the starting index.

4. Returning the result:

   • If we have not found any window (k > m), we return an empty string.
   • Otherwise, we return the substring of s1 starting from p with the length k.

The choice of dynamic programming in this solution is crucial as it eliminates redundant comparisons and stores intermediate results, allowing efficient computation of the final minimum length substring. This algorithm has an O(m * n) time complexity due to the nested loops required to fill the DP table, where m and n are the lengths of s1 and s2, respectively.

The solution is concise and the use of dynamic programming provides optimal substructure and overlapping subproblems, two key characteristics exploited by this paradigm to achieve efficiency. Each entry in the DP table only depends on previously computed values, making the implementation straightforward and logical once the table relationships are understood.

Example Walkthrough

Let's illustrate the solution approach using a small example.

Suppose s1 is "axbxcxdx" and s2 is "bcd".

Following the solution approach:

1. Initialization:

   • We set up a 2D array f with a size of (8+1) x (3+1) since s1 is of length 8 and s2 is of length 3.

2. Filling the DP array:

   • We iterate over s1 and s2, filling up the f array.
   • Let's iterate over s1: "axbxcxdx" and s2: "bcd".
     • When i = 1 and j = 1, we find that s1[i] != s2[j], so we don't update f[1][1].
     • When i = 2 and j = 1, we find s1[i] == s2[j] ('b' == 'b'), so we update f[2][1] to 2. This marks the start of a possible subsequence.
     • Continuing this process, we find the 'c' of s2 in s1 at position 4, so f[4][2] is updated to 2, indicating that from position 2 in s1, we have 'bc' of s2.
     • We find the 'd' of s2 in s1 at position 6, so f[6][3] is updated to 2, meaning from position 2 in s1, we have the full 'bcd' of s2.

3. Identifying the minimum window:

   • We scan f looking for the smallest window where s1[i] == s2[3] ('d' in this case) and f[i][3] is not zero.
   • We find this at i = 6 for s2[3] ('d'), where f[i][3] is 2. The window size is 6 - 2 + 1 = 5.

4. Returning the result:

   • The smallest window has a size of 5 starting at index 2 in s1. So, the result is the substring "bxcxd".

Following this process, we've identified the subsequence 'bcd' within s1 and found the minimum window. The algorithm efficiently computes this by tracking possible starting points for subsequences in s1 and keeping track of these starting points as we match characters of s2. This allows us to quickly compute the length of potential windows without redundant comparisons.

Solution Implementation

Time and Space Complexity

The given Python code snippet is designed to find the smallest window in string s1 which contains all the characters of string s2 in the same order. It utilizes dynamic programming to achieve this.

Time Complexity

The time complexity of the code can be analyzed by looking at the nested loops:

1. The first pair of nested loops, where the outer loop runs for m iterations (m being the length of s1) and the inner loop runs for n iterations (n being the length of s2), establishes a time complexity of O(m * n) for the dynamic programming table population.

2. The second pair of nested loops also runs up to m times,and the inner operations are constant time since they're only comparing and updating values based on previously computed results. Therefore, the second nested loop does not exceed O(m) in complexity.

When combined, the time complexity is dictated by the larger of these loops, which is the first one. Therefore, the overall time complexity of the algorithm is O(m * n).

Space Complexity

The space complexity of the code is determined by the size of the dynamic programming table f. Since the table is of size (m + 1) * (n + 1), where m is the length of s1 and n is the length of s2, the space complexity is O(m * n). No additional significant space is consumed since only integer variables for bookkeeping purposes are used outside the table.

In summary, the time complexity is O(m * n) and the space complexity is O(m * n).

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