How do you solve Coupon Code Validator on LeetCode?

Coupon Code Validator (LeetCode #3606) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Validating input is crucial.

What is the time complexity of Coupon Code Validator?

The optimal solution for Coupon Code Validator runs in O(n log n) time and uses O(n) space. Filtering is O(n) and sorting is O(n log n), leading to overall O(n log n).

What is the brute force solution for Coupon Code Validator?

The brute force approach for Coupon Code Validator is Brute Force, with O(n²) time complexity and O(n) space complexity. This approach checks each coupon individually for validity. It’s straightforward but inefficient due to repeated checks. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Coupon Code Validator?

Coupon Code Validator uses the following concepts: Array, Hash Table, String, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Coupon Code Validator?

Explain your thought process clearly. Discuss edge cases. Optimize as you code.

What are common mistakes when solving Coupon Code Validator?

Ignoring empty codes. Not checking business line validity.

#3606

Coupon Code Validator

Easy

ArrayHash TableStringSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(n)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

This approach efficiently filters and sorts valid coupons in one pass, leveraging a dictionary for business line priority.

⚙️

Algorithm

3 steps

1Step 1: Create a priority mapping for business lines.
2Step 2: Filter valid coupons and store them in a list with their business line.
3Step 3: Sort the list based on business line priority and code.

solution.py4 lines

1def validateCoupons(code, businessLine, isActive):
2    priority = {'electronics': 0, 'grocery': 1, 'pharmacy': 2, 'restaurant': 3}
3    validCoupons = [(businessLine[i], code[i]) for i in range(len(code)) if isActive[i] and code[i] and all(c.isalnum() or c == '_' for c in code[i]) and businessLine[i] in priority]
4    return [c for b, c in sorted(validCoupons, key=lambda x: (priority[x[0]], x[1]))]

ℹ

Complexity note: Filtering is O(n) and sorting is O(n log n), leading to overall O(n log n).

1Validating input is crucial.
2Sorting can be optimized with priority mapping.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.