Problem Description

The goal of this problem is to encode a given string s in such a way that the encoded version has the shortest possible length. The encoding rule must follow the format k[encoded_string], where encoded_string represents a sequence of the string that is repeated k times, and k is a positive integer that represents the number of times encoded_string occurs consecutively within s.

For example, if the string is "aaaabaaaab", it can be encoded as 2[a4[ab]], which means that a4[ab] (which represents "aaaab") is repeated twice.

The string should only be encoded if doing so reduces its length. For strings that would not be shortened by encoding, the original string should be returned. In cases where multiple encoding methods result in the shortest length, any of those methods are considered a valid solution.

Intuition

The intuition behind the solution is to apply dynamic programming, which is a method often used in optimization problems. Dynamic programming involves breaking down a complex problem into simpler subproblems and solving each of those subproblems just once, storing their solutions—often in a two-dimensional array.

The main idea of the solution is to:

1. Iterate over all possible substrings of s and determine the shortest encoding for each substring.
2. For each substring, check if it has a repetitive pattern that can be encoded. Patterns are identified by checking if the substring t can be found within the concatenation of t with itself, but not at the very beginning (this implies a repetition).
3. If such a pattern exists and it helps reduce the length of the current substring, encode it using the format k[encoded_substring]. If not, keep the substring unencoded.
4. For longer substrings that do not have a repetitive pattern that can be encoded, or such encoding does not reduce the length, explore breaking the substring into two smaller encoded subproblems (substrings) whose total encoded length is minimal.

By solving the subproblems from shortest to longest substrings and building up solutions, we ensure that when we determine whether to encode a longer substring, we are making the decision based on the optimal (shortest) encodings of all possible subparts of that substring.

Essentially, the dynamic programming array f[i][j] holds the shortest encoded version of the substring starting from index i to index j in the original string s. By the end of the iterations, f[0][n-1] will give us the shortest encoded version of the entire string.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution approach implements dynamic programming to solve the problem, where a two-dimensional array f is used to store the shortest encoded strings for all the substrates from i to j. The implementation goes as follows:

1. Function g(i, j) takes the start index i and the end index j as inputs and returns the optimal encoded string for the substring s[i:j+1]. If the substring is shorter than 5 characters, it's not worth encoding because the encoded format would not be shorter than the original substring.

2. In function g(i, j), t is the current substring. In order to find repetitive patterns within t, (t + t).index(t, 1) is used. This expression checks whether t repeats in a concatenation of itself after the first character. If k is the index where t is found, it means t is repeating every k characters within t. If k is less than the length of t, we have found a repeating pattern, and we encode it as "{cnt}[{f[i][i + k - 1]}]", where cnt is the count of repetition.

3. For every substring, we store our finding in the f array. The elements f[i][j] will represent the shortest encoded form of the substring s[i:j+1].

4. A nested loop is used to fill the table f in a bottom-up manner. The outer loop decrements i from n-1 to 0 to ensure we solve for all smaller substrates first. The inner loop increments j from i to n-1 to cover all substrings starting from i.

5. For every pair of i and j, we first check whether it's worth encoding the current substring using function g(i, j). If the substring length from i to j is more than 4 (since we don't want to encode shorter substrings), we then check for potential splits within this substring that could lead to an overall shorter length.

6. To check for splits, we iterate through all possible split positions k from i to j and calculate the combined length of encoded substrings f[i][k] and f[k+1][j]. If this combined length is shorter than the currently stored encoded string for s[i:j+1], we update f[i][j] with this shorter version.

7. After filling the array f, the shortest encoded version of the entire string s will be stored in f[0][n-1], which is the solution to the original problem.

The algorithm capitalizes on the property that to get the shortest encoding of a longer substring, you need to know the shortest encodings of all its subparts. This leads to an optimization problem where dynamic programming excels. By storing solutions of small problems and using them to solve larger problems, the algorithm works efficiently without recalculating the encoded strings for the same substrates multiple times.

Example Walkthrough

Let's take the string "abbbabbb" to illustrate the solution approach:

1. Initialize a table f where f[i][j] will eventually contain the shortest encoded version of the substring s[i:j+1].

2. Consider substrings s[i:j+1] of length less than 5, which are not worth encoding. So f[i][j] would just be s[i:j+1]. For instance, f[0][3] for substring "abbb" remains "abbb" because its length is less than 5.

3. For substrings of length 5 or more, check if they have a repeatable pattern that can be encoded. For example, the substring "abbbabbb" has a repetitive substring which is "abbb" that repeats twice.

4. Use (t + t).index(t, 1) to find the repeating pattern within t. For our example, t = "abbb", and by searching in (t + t) = "abbbabbb", t starts repeating at index 4, which is equal to the length of t, indicating that it repeats every 4 characters.

5. Since the pattern t repeats 2 times, encode the substring as "2[abbb]" and store it in table f at f[0][7], replacing "abbbabbb".

6. If we were to consider a longer string and f is partially filled with shorter substrings' encodings, check for optimal splits by trying all possible split points. For every possible split at position k for the substring s[i:j+1], calculate the total length of f[i][k] + f[k+1][j]. Choose the split that offers the shortest encoded length.

7. After checking for all possible splits and patterns, the final encoded strings are ready in f. For our example, since the substring "abbbabbb" has already been optimally encoded as "2[abbb]", this would be the value of f[0][7], and thus the output for the entire string.

By systematically applying these steps to increasingly longer substrates of the original string and using the dynamic programming array to store intermediate solutions, the algorithm ensures efficiency and avoids redundant calculations, leading to an optimal solution.

Solution Implementation

Time and Space Complexity

The provided Python code is a dynamic programming solution for the problem of encoding the minimum length of a string where the string can be encoded by the number of repetitions and a pattern inside the brackets. To analyze the time and space complexity, we need to consider the operations performed inside the nested loops and the recursive calls.

Time Complexity

The time complexity of the code is determined by the three nested loops and the string operations inside the helper function g.

• The outer loop runs from n-1 down to 0, thus running n times.
• The second loop, for variable j, will run at most n times for each i.
• The third loop is used to find the optimal partition of the string for encoding and runs at most n times for each pair of i and j.

Additionally, the helper function g includes a string pattern check using (t + t).index(t, 1) which can be considered O(n) in the worst case because it could potentially scan the entire doubled string to find the pattern start. Thus, this operation could contribute significantly to the execution time.

Considering all these factors, the worst-case time complexity is O(n^3) for the loops multiplied by O(n) for the string pattern checking, leading to an overall time complexity of O(n^4).

Space Complexity

The space complexity includes the space required for the dynamic programming table f and the stack space used by the helper function g.

• The dynamic programming table f is a 2D array of size n x n, contributing O(n^2) space complexity.
• The helper function g uses a temporary string t, but this does not increase the asymptotic space complexity since it requires space proportional to the input string s.

So the overall space complexity of the code is O(n^2) due to the 2D array f.

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