How to Solve Find the String with LCP

This problem requires reconstructing a string from a given LCP matrix, ensuring that each substring prefix matches the matrix values. By modeling the solution with state transition dynamic programming, you can assign characters while maintaining lexicographical order. Conflicts in LCP values indicate impossible configurations, which must be detected to return an empty string.

Problem Statement

Given an n x n integer matrix lcp representing the longest common prefix lengths of a string word with itself, reconstruct the lexicographically smallest string word of length n consisting of lowercase English letters. Each lcp[i][j] equals the length of the longest common prefix of word[i...] and word[j...].

Return the alphabetically smallest string that matches the provided LCP matrix. If no such string exists due to inconsistent LCP values, return an empty string. Lexicographical order is determined by the first differing character; for example, "aabd" is smaller than "aaca".

Examples

Example 1

Input: lcp = [[4,0,2,0],[0,3,0,1],[2,0,2,0],[0,1,0,1]]

Output: "abab"

lcp corresponds to any 4 letter string with two alternating letters. The lexicographically smallest of them is "abab".

Example 2

Input: lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,1]]

Output: "aaaa"

lcp corresponds to any 4 letter string with a single distinct letter. The lexicographically smallest of them is "aaaa".

Example 3

Input: lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,3]]

Output: ""

lcp[3][3] cannot be equal to 3 since word[3,...,3] consists of only a single letter; Thus, no answer exists.

Constraints

• 1 <= n == lcp.length == lcp[i].length <= 1000
• 0 <= lcp[i][j] <= n

Solution Approach

Initialize characters and validate LCP constraints

Start by assigning each position a character placeholder and immediately check that the LCP matrix satisfies basic consistency rules, such as lcp[i][i] equal to the remaining substring length and symmetry for lcp[i][j] and lcp[j][i]. Any violation at this stage indicates no valid string exists.

Use state transition dynamic programming to propagate character equivalences

Iterate over the matrix and for each lcp[i][j] > 0, enforce that word[i] == word[j] and continue this equality for the next lcp[i][j]-1 characters. Track equivalence classes to handle multiple overlapping constraints efficiently. Resolve conflicts by marking impossible states to detect invalid configurations early.

Assign lexicographically smallest letters and construct the string

Once equivalence classes are established, assign the smallest available letter to each class in order. This guarantees lexicographical minimality. Finally, assemble the string from these assignments. If any LCP rule is violated during assignment, return an empty string.

Complexity Analysis

Time complexity is O(n^2) due to checking all pairs in the LCP matrix and propagating equivalences. Space complexity is O(n) to store equivalence classes and the resulting string, though the LCP matrix itself requires O(n^2) storage.

What Interviewers Usually Probe

• Candidate identifies matrix symmetry and self-prefix conditions.
• Candidate uses dynamic programming to propagate equality constraints efficiently.
• Candidate correctly detects impossible LCP configurations and returns empty string.

Common Pitfalls or Variants

Common pitfalls

• Ignoring symmetry and failing to check lcp[i][j] == lcp[j][i].
• Assigning letters before resolving all equivalence classes, which may violate LCP rules.
• Misinterpreting lexicographical ordering when multiple choices of letters exist.

Follow-up variants

• Find the largest lexicographical string matching an LCP matrix.
• Handle LCP matrices with wildcard entries represented by -1.
• Extend the problem to uppercase letters or a custom alphabet.

How GhostInterview Helps

• GhostInterview provides step-by-step state transition guidance for propagating LCP constraints.
• It highlights conflicts in LCP values early, preventing wasted computation on impossible configurations.
• GhostInterview suggests optimal letter assignment strategies to achieve the smallest lexicographical string efficiently.

Topic Pages

• Array
• String
• Dynamic Programming
• Greedy
• Union Find
• Matrix

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FAQ

What is the best approach for solving Find the String with LCP?

Using state transition dynamic programming to propagate character equivalences ensures all LCP constraints are satisfied.

How do I detect if no string can satisfy a given LCP matrix?

Check for violations such as lcp[i][j] != lcp[j][i] or inconsistent prefix lengths during propagation; any conflict indicates impossibility.

Can this problem be solved greedily instead of dynamic programming?

A purely greedy approach may fail due to overlapping equality constraints; DP ensures correct propagation across the matrix.

Why is lexicographical order important in this problem?

After equivalence classes are determined, assigning the smallest available letters guarantees the string is alphabetically minimal.

What patterns does Find the String with LCP primarily test?

It tests state transition dynamic programming with array and string constraints, focusing on prefix equality propagation and conflict detection.