2573. Find the String with LCP

Problem Description

The task is to construct the alphabetically smallest string word for a given lcp matrix which represents the longest common prefixes of the substrings of word. The lcp matrix is an n x n grid where lcp[i][j] equals the length of the longest common prefix between word[i,n-1] and word[j,n-1]. The challenge is to determine the string word if it exists or return an empty string otherwise. It tests one's ability to reason about string prefixes and lexicographical order.

Intuition

The key insight to solve this problem involves underpinning the relationship between the matrix entries and the characters of the strings at different indices. Since the alphabetically smallest string is required, a natural approach is to attempt to fill the string with the lexicographically earliest character ('a') where possible, and then move on to the next character only if necessary.

We start by initializing a string s with an empty character for each position and iterate over the lowercase alphabet. For each character c, we look for indices in s that have not yet been set and that we can set with c without violating the given lcp conditions. For every such index, we also set the character c at all positions j where lcp[i][j] is non-zero, since they must share that prefix.

To ensure that the final string does not violate any lcp constraints, an additional validation step verifies that the lcp matrix entries are consistent with what we would expect given the assigned characters. If an inconsistency is found—a missing character or an incorrect prefix length—the solution concludes that no valid string can be created, thus returning an empty string.

By cautiously filling out the available spots with the smallest possible letter and checking for the lcp matrix consistency along the process, we should be able to either arrive at the desired smallest string or determine that such a string doesn't exist.

Learn more about Greedy, Union Find and Dynamic Programming patterns.

Solution Approach

The solution uses a straightforward yet clever approach by iterating over the list of ASCII lowercase letters, which guarantees that we are always selecting the alphabetically smallest possible character for every index.

Here's a breakdown of how the code implements the solution step by step:

1. Initialize the string s as a list of empty strings of length n, which matches the size of the lcp matrix. Each element of s represents a character in the answer string.

2. Start iterating over each character c in the ASCII lowercase alphabet. The idea is to try and fill s with the smallest possible characters first, maintaining lexicographic ordering.

3. Using a while loop, increment the index i until finding an index where s[i] is still an empty string or until the end of s is reached. This finds the first unfilled position in s where we can potentially place a new character.

4. If index i reaches the end of the string, all characters have been assigned, and the loop breaks.

5. For each index j from i to n-1, if lcp[i][j] is not zero, which means there is a common prefix of non-zero length, assign the current character c to s[j]. This step assigns the character c to all positions that share a prefix with i.

6. After trying to fill s with all characters of the alphabet, check for any unfilled positions left in s. If there are any, return an empty string since it means we weren't able to construct a string fitting the lcp matrix.

7. Perform validation on the built string from back to front to ensure all lcp conditions are met. This step verifies that:

   • For any two indices i and j that have the same character, the corresponding lcp[i][j] must be 1 if either i or j is n-1. Otherwise, lcp[i][j] must equal the value of lcp[i+1][j+1] + 1.
   • For any two indices i and j with different characters, lcp[i][j] must be 0. If any of these conditions are not true, it means that the lcp matrix is not consistent with the string we have built, and thus the function returns an empty string.

8. If the string passes all the validation checks, join the list s to form the final answer string, and return it.

This method utilizes a greedy approach to select the lexicographically smallest letters, combined with a validation step to confirm that the selected characters do not violate the lcp conditions. The intelligent ordering of operations ensures an efficient and correct solution.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Suppose we have an lcp matrix:

1lcp = [
2   [2, 1, 0],
3   [1, 2, 0],
4   [0, 0, 1]
5]

This lcp matrix is of size 3x3, so our word will have three characters. We'll try to build the smallest possible word.

1. Initialize the string s: s starts as ['', '', ''].

2. Iterate over ASCII lowercase: We start from 'a', the smallest character.

3. Find the first empty string position: The first index i with an empty string is 0.

4. End check: Our string s is not full.

5. Assign 'a' to positions with non-zero lcp: Looking at lcp[0], there is a 1 at index 1. We can set s[0] and s[1] to 'a'.

   After this, our s looks like ['a', 'a', ''].

6. Move to next index: Since s[2] is still empty, we need to fill it with the next character. Since 'a' can't be placed due to lcp[0][2] and lcp[1][2] being 0, we try with the next character, which is 'b'.

   Now s looks like ['a', 'a', 'b'].

7. Check for unfilled positions: There are no unfilled positions left.

8. Validate the built string: Validate s using the lcp conditions. For s[0] and s[1], the lcp value is 1, which is correct since they are the same and this matches the precondition set out in the problem description. For s[0] and s[2], as well as s[1] and s[2], the lcp is 0, which is consistent with them having different characters.

9. String passes validation: No inconsistencies found.

10. Return the result: Join s to form the final string. The result for this example is "aab".

Thus, by carefully placing the smallest possible letters and validating against the lcp matrix, we correctly identified that the alphabetically smallest string representation that could produce the given lcp matrix is "aab".

Solution Implementation

Time and Space Complexity

Time Complexity

The given code has the following steps which contribute to its time complexity:

1. Initializing a list s of length n - O(n).
2. A nested loop structure where it iterates over the ascii_lowercase characters and then over the range from i to n to populate the string array s. Since ASCII lowercase has a fixed number of characters (26), and we assume that each character will be the start of a new string only once, we can approximate the time complexity for this loop as O(n). Even though there are nested loops, the outer loop only increments i and does not start from the beginning each time.
3. Checking if there is any empty string "" in s. This is done by scanning the list once, resulting in a complexity of O(n).
4. Finally, there's a nested loop that compares elements in the lcp array to verify the longest common prefix properties. This loop has a worst-case time complexity of O(n^2) as it iterates over each element twice in the worst case.

Therefore, the total time complexity can be approximately expressed as O(n) + O(n) + O(n^2) which simplifies to O(n^2) since the quadratic term dominates for large n.

Space Complexity

The space complexity is determined by:

1. The list s of length n - O(n).
2. No additional significant space is utilized within the loops.

Thus, the space complexity of the code is O(n).

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