LeetCode #3598 — MEDIUM

Longest Common Prefix Between Adjacent Strings After Removals

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

Problem Statement

You are given an array of strings words. For each index i in the range [0, words.length - 1], perform the following steps:

  • Remove the element at index i from the words array.
  • Compute the length of the longest common prefix among all adjacent pairs in the modified array.

Return an array answer, where answer[i] is the length of the longest common prefix between the adjacent pairs after removing the element at index i. If no adjacent pairs remain or if none share a common prefix, then answer[i] should be 0.

Example 1:

Input: words = ["jump","run","run","jump","run"]

Output: [3,0,0,3,3]

Explanation:

  • Removing index 0:
    • words becomes ["run", "run", "jump", "run"]
    • Longest adjacent pair is ["run", "run"] having a common prefix "run" (length 3)
  • Removing index 1:
    • words becomes ["jump", "run", "jump", "run"]
    • No adjacent pairs share a common prefix (length 0)
  • Removing index 2:
    • words becomes ["jump", "run", "jump", "run"]
    • No adjacent pairs share a common prefix (length 0)
  • Removing index 3:
    • words becomes ["jump", "run", "run", "run"]
    • Longest adjacent pair is ["run", "run"] having a common prefix "run" (length 3)
  • Removing index 4:
    • words becomes ["jump", "run", "run", "jump"]
    • Longest adjacent pair is ["run", "run"] having a common prefix "run" (length 3)

Example 2:

Input: words = ["dog","racer","car"]

Output: [0,0,0]

Explanation:

  • Removing any index results in an answer of 0.

Constraints:

  • 1 <= words.length <= 105
  • 1 <= words[i].length <= 104
  • words[i] consists of lowercase English letters.
  • The sum of words[i].length is smaller than or equal 105.

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: You are given an array of strings words. For each index i in the range [0, words.length - 1], perform the following steps: Remove the element at index i from the words array. Compute the length of the longest common prefix among all adjacent pairs in the modified array. Return an array answer, where answer[i] is the length of the longest common prefix between the adjacent pairs after removing the element at index i. If no adjacent pairs remain or if none share a common prefix, then answer[i] should be 0.

Baseline thinking

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

Pattern signal: Array

Example 1

["jump","run","run","jump","run"]

Example 2

["dog","racer","car"]
Step 02

Core Insight

What unlocks the optimal approach

  • Precompute the longest common prefix length for adjacent prefixes and suffixes.
  • After deleting <code>words[i]</code>, compute the longest common prefix for <code>words[i - 1]</code> and <code>words[i + 1]</code> (if they exist).
  • Use the result of the prefix computation up to <code>i - 1</code> and the suffix computation from <code>i + 1</code> onwards.
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
Upper-end input sizes
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 #3598: Longest Common Prefix Between Adjacent Strings After Removals
class Solution {
    private final TreeMap<Integer, Integer> tm = new TreeMap<>();
    private String[] words;
    private int n;

    public int[] longestCommonPrefix(String[] words) {
        n = words.length;
        this.words = words;
        for (int i = 0; i + 1 < n; ++i) {
            tm.merge(calc(words[i], words[i + 1]), 1, Integer::sum);
        }
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            remove(i, i + 1);
            remove(i - 1, i);
            add(i - 1, i + 1);
            ans[i] = !tm.isEmpty() && tm.lastKey() > 0 ? tm.lastKey() : 0;
            remove(i - 1, i + 1);
            add(i - 1, i);
            add(i, i + 1);
        }
        return ans;
    }

    private void add(int i, int j) {
        if (i >= 0 && i < n && j >= 0 && j < n) {
            tm.merge(calc(words[i], words[j]), 1, Integer::sum);
        }
    }

    private void remove(int i, int j) {
        if (i >= 0 && i < n && j >= 0 && j < n) {
            int x = calc(words[i], words[j]);
            if (tm.merge(x, -1, Integer::sum) == 0) {
                tm.remove(x);
            }
        }
    }

    private int calc(String s, String t) {
        int m = Math.min(s.length(), t.length());
        for (int k = 0; k < m; ++k) {
            if (s.charAt(k) != t.charAt(k)) {
                return k;
            }
        }
        return m;
    }
}
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(L + n × log n)
Space
O(n)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
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.

Related Problems
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