LeetCode #3606 — EASY

Coupon Code Validator

Build confidence with an intuition-first walkthrough focused on array fundamentals.

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The Problem

Problem Statement

You are given three arrays of length n that describe the properties of n coupons: code, businessLine, and isActive. The ith coupon has:

  • code[i]: a string representing the coupon identifier.
  • businessLine[i]: a string denoting the business category of the coupon.
  • isActive[i]: a boolean indicating whether the coupon is currently active.

A coupon is considered valid if all of the following conditions hold:

  1. code[i] is non-empty and consists only of alphanumeric characters (a-z, A-Z, 0-9) and underscores (_).
  2. businessLine[i] is one of the following four categories: "electronics", "grocery", "pharmacy", "restaurant".
  3. isActive[i] is true.

Return an array of the codes of all valid coupons, sorted first by their businessLine in the order: "electronics", "grocery", "pharmacy", "restaurant", and then by code in lexicographical (ascending) order within each category.

Example 1:

Input: code = ["SAVE20","","PHARMA5","SAVE@20"], businessLine = ["restaurant","grocery","pharmacy","restaurant"], isActive = [true,true,true,true]

Output: ["PHARMA5","SAVE20"]

Explanation:

  • First coupon is valid.
  • Second coupon has empty code (invalid).
  • Third coupon is valid.
  • Fourth coupon has special character @ (invalid).

Example 2:

Input: code = ["GROCERY15","ELECTRONICS_50","DISCOUNT10"], businessLine = ["grocery","electronics","invalid"], isActive = [false,true,true]

Output: ["ELECTRONICS_50"]

Explanation:

  • First coupon is inactive (invalid).
  • Second coupon is valid.
  • Third coupon has invalid business line (invalid).

Constraints:

  • n == code.length == businessLine.length == isActive.length
  • 1 <= n <= 100
  • 0 <= code[i].length, businessLine[i].length <= 100
  • code[i] and businessLine[i] consist of printable ASCII characters.
  • isActive[i] is either true or false.

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 three arrays of length n that describe the properties of n coupons: code, businessLine, and isActive. The ith coupon has: code[i]: a string representing the coupon identifier. businessLine[i]: a string denoting the business category of the coupon. isActive[i]: a boolean indicating whether the coupon is currently active. A coupon is considered valid if all of the following conditions hold: code[i] is non-empty and consists only of alphanumeric characters (a-z, A-Z, 0-9) and underscores (_). businessLine[i] is one of the following four categories: "electronics", "grocery", "pharmacy", "restaurant". isActive[i] is true. Return an array of the codes of all valid coupons, sorted first by their businessLine in the order: "electronics", "grocery", "pharmacy", "restaurant", and then by code in lexicographical (ascending) order within each category.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

["SAVE20","","PHARMA5","SAVE@20"]
["restaurant","grocery","pharmacy","restaurant"]
[true,true,true,true]

Example 2

["GROCERY15","ELECTRONICS_50","DISCOUNT10"]
["grocery","electronics","invalid"]
[false,true,true]
Step 02

Core Insight

What unlocks the optimal approach

  • Filter out any coupon where <code>isActive[i]</code> is false, <code>code[i]</code> is empty or contains non‑alphanumeric/underscore chars, or <code>businessLine[i]</code> is not in the allowed set
  • Store each remaining coupon as a pair <code>(businessLine[i], code[i])</code>
  • Define a priority map, e.g. <code>{"electronics":0, "grocery":1, "pharmacy":2, "restaurant":3}</code>
  • Sort the list of pairs by <code>(priority[businessLine], code)</code> and return the <code>code</code> values in order
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 #3606: Coupon Code Validator
class Solution {
    public List<String> validateCoupons(String[] code, String[] businessLine, boolean[] isActive) {
        List<Integer> idx = new ArrayList<>();
        Set<String> bs
            = new HashSet<>(Arrays.asList("electronics", "grocery", "pharmacy", "restaurant"));

        for (int i = 0; i < code.length; i++) {
            if (isActive[i] && bs.contains(businessLine[i]) && check(code[i])) {
                idx.add(i);
            }
        }

        idx.sort((i, j) -> {
            int cmp = businessLine[i].compareTo(businessLine[j]);
            if (cmp != 0) {
                return cmp;
            }
            return code[i].compareTo(code[j]);
        });

        List<String> ans = new ArrayList<>();
        for (int i : idx) {
            ans.add(code[i]);
        }
        return ans;
    }

    private boolean check(String s) {
        if (s.isEmpty()) {
            return false;
        }
        for (char c : s.toCharArray()) {
            if (!Character.isLetterOrDigit(c) && c != '_') {
                return false;
            }
        }
        return true;
    }
}
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 × 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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.