LeetCode #1268 — MEDIUM

Search Suggestions System

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

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The Problem

Problem Statement

You are given an array of strings products and a string searchWord.

Design a system that suggests at most three product names from products after each character of searchWord is typed. Suggested products should have common prefix with searchWord. If there are more than three products with a common prefix return the three lexicographically minimums products.

Return a list of lists of the suggested products after each character of searchWord is typed.

Example 1:

Input: products = ["mobile","mouse","moneypot","monitor","mousepad"], searchWord = "mouse"
Output: [["mobile","moneypot","monitor"],["mobile","moneypot","monitor"],["mouse","mousepad"],["mouse","mousepad"],["mouse","mousepad"]]
Explanation: products sorted lexicographically = ["mobile","moneypot","monitor","mouse","mousepad"].
After typing m and mo all products match and we show user ["mobile","moneypot","monitor"].
After typing mou, mous and mouse the system suggests ["mouse","mousepad"].

Example 2:

Input: products = ["havana"], searchWord = "havana"
Output: [["havana"],["havana"],["havana"],["havana"],["havana"],["havana"]]
Explanation: The only word "havana" will be always suggested while typing the search word.

Constraints:

  • 1 <= products.length <= 1000
  • 1 <= products[i].length <= 3000
  • 1 <= sum(products[i].length) <= 2 * 104
  • All the strings of products are unique.
  • products[i] consists of lowercase English letters.
  • 1 <= searchWord.length <= 1000
  • searchWord consists of lowercase English letters.
Patterns Used
🔍Modified Binary Search→🔤Trie→

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 products and a string searchWord. Design a system that suggests at most three product names from products after each character of searchWord is typed. Suggested products should have common prefix with searchWord. If there are more than three products with a common prefix return the three lexicographically minimums products. Return a list of lists of the suggested products after each character of searchWord is typed.

Baseline thinking

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

Pattern signal: Array · Binary Search · Trie

Example 1

["mobile","mouse","moneypot","monitor","mousepad"]
"mouse"

Example 2

["havana"]
"havana"
Step 02

Core Insight

What unlocks the optimal approach

  • Brute force is a good choice because length of the string is ≤ 1000.
  • Binary search the answer.
  • Use Trie data structure to store the best three matching. Traverse the Trie.
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 #1268: Search Suggestions System
class Trie {
    Trie[] children = new Trie[26];
    List<Integer> v = new ArrayList<>();

    public void insert(String w, int i) {
        Trie node = this;
        for (int j = 0; j < w.length(); ++j) {
            int idx = w.charAt(j) - 'a';
            if (node.children[idx] == null) {
                node.children[idx] = new Trie();
            }
            node = node.children[idx];
            if (node.v.size() < 3) {
                node.v.add(i);
            }
        }
    }

    public List<Integer>[] search(String w) {
        Trie node = this;
        int n = w.length();
        List<Integer>[] ans = new List[n];
        Arrays.setAll(ans, k -> new ArrayList<>());
        for (int i = 0; i < n; ++i) {
            int idx = w.charAt(i) - 'a';
            if (node.children[idx] == null) {
                break;
            }
            node = node.children[idx];
            ans[i] = node.v;
        }
        return ans;
    }
}

class Solution {
    public List<List<String>> suggestedProducts(String[] products, String searchWord) {
        Arrays.sort(products);
        Trie trie = new Trie();
        for (int i = 0; i < products.length; ++i) {
            trie.insert(products[i], i);
        }
        List<List<String>> ans = new ArrayList<>();
        for (var v : trie.search(searchWord)) {
            List<String> t = new ArrayList<>();
            for (int i : v) {
                t.add(products[i]);
            }
            ans.add(t);
        }
        return ans;
    }
}
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 × log n + m)
Space
O(L)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Related Problems
#792Number of Matching Subsequencesmedium→#3291Minimum Number of Valid Strings to Form Target Imedium→#74Search a 2D Matrixmedium→#81Search in Rotated Sorted Array IImedium→#139Word Breakmedium→