LeetCode #2573 — HARD

Find the String with LCP

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

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. If there is no such string, return an empty string.

A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b. For example, "aabd" is lexicographically smaller than "aaca" because the first position they differ is at the third letter, and 'b' comes before 'c'.

Example 1:

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".

Example 2:

Input: lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,1]]
Output: "aaaa"
Explanation: 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: ""
Explanation: 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
Patterns Used
📊Dynamic Programming→🎯Greedy→🔗Union-Find→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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. If there is no such string, return an empty string. A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b. For example, "aabd" is lexicographically smaller than "aaca" because the first position they differ is at the third letter, and 'b' comes before 'c'.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Greedy · Union-Find

Example 1

[[4,0,2,0],[0,3,0,1],[2,0,2,0],[0,1,0,1]]

Example 2

[[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,1]]

Example 3

[[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,3]]
Step 02

Core Insight

What unlocks the optimal approach

  • Use the LCP array to determine which groups of elements must be equal.
  • Match the smallest letter to the group that contains the smallest unassigned index.
  • Build the LCP matrix of the resulting string then check if it is equal to the target LCP.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2573: Find the String with LCP
class Solution {
    public String findTheString(int[][] lcp) {
        int n = lcp.length;
        char[] s = new char[n];
        int i = 0;
        for (char c = 'a'; c <= 'z'; ++c) {
            while (i < n && s[i] != '\0') {
                ++i;
            }
            if (i == n) {
                break;
            }
            for (int j = i; j < n; ++j) {
                if (lcp[i][j] > 0) {
                    s[j] = c;
                }
            }
        }
        for (i = 0; i < n; ++i) {
            if (s[i] == '\0') {
                return "";
            }
        }
        for (i = n - 1; i >= 0; --i) {
            for (int j = n - 1; j >= 0; --j) {
                if (s[i] == s[j]) {
                    if (i == n - 1 || j == n - 1) {
                        if (lcp[i][j] != 1) {
                            return "";
                        }
                    } else if (lcp[i][j] != lcp[i + 1][j + 1] + 1) {
                        return "";
                    }
                } else if (lcp[i][j] > 0) {
                    return "";
                }
            }
        }
        return String.valueOf(s);
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n^2)
Space
O(n)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.

Related Problems
#45Jump Game IImedium→#55Jump Gamemedium→#122Best Time to Buy and Sell Stock IImedium→#376Wiggle Subsequencemedium→#410Split Array Largest Sumhard→