Leetcode #2573: Find the String with LCP
In this guide, we solve Leetcode #2573 Find the String with LCP in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
We define the lcp matrix of any 0-indexed string word of n lowercase English letters as an n x n grid such that: lcp[i][j] is equal to the length of the longest common prefix between the substrings word[i,n-1] and word[j,n-1]. Given an n x n matrix lcp, return the alphabetically smallest string word that corresponds to lcp.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Greedy, Union Find, Array, String, Dynamic Programming, Matrix
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: lcp = [[4,0,2,0],[0,3,0,1],[2,0,2,0],[0,1,0,1]]
Output: "abab"
Explanation: lcp corresponds to any 4 letter string with two alternating letters. The lexicographically smallest of them is "abab".
Python Solution
class Solution:
def findTheString(self, lcp: List[List[int]]) -> str:
n = len(lcp)
s = [""] * n
i = 0
for c in ascii_lowercase:
while i < n and s[i]:
i += 1
if i == n:
break
for j in range(i, n):
if lcp[i][j]:
s[j] = c
if "" in s:
return ""
for i in range(n - 1, -1, -1):
for j in range(n - 1, -1, -1):
if s[i] == s[j]:
if i == n - 1 or j == n - 1:
if lcp[i][j] != 1:
return ""
elif lcp[i][j] != lcp[i + 1][j + 1] + 1:
return ""
elif lcp[i][j]:
return ""
return "".join(s)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.