How do you solve Invalid Transactions on LeetCode?

Invalid Transactions (LeetCode #1169) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Transactions can be invalid due to amount or timing/location conflicts.

What is the brute force solution for Invalid Transactions?

The brute force approach for Invalid Transactions is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each transaction against every other transaction to determine if it is invalid. This is simple but inefficient for larger input sizes. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Invalid Transactions?

Invalid Transactions uses the following concepts: Array, Hash Table, String, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Invalid Transactions?

Always clarify the problem statement and constraints before jumping into coding. Think about how to optimize your solution after implementing the brute force version. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Invalid Transactions?

Not considering edge cases like multiple transactions from the same user in the same city. Overlooking the need to check both conditions for invalid transactions.

#1169

Invalid Transactions

Medium

ArrayHash TableStringSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a single pass through the transactions while utilizing a hashmap to store transactions by name and city. This allows us to efficiently check for invalid transactions without needing nested loops.

⚙️

Algorithm

3 steps

1Step 1: Create a list to store invalid transactions and a hashmap to store transactions by name and city.
2Step 2: Iterate through the transactions and check if the amount exceeds $1000; if so, add it to the invalid list.
3Step 3: For each transaction, check the hashmap for any transactions with the same name in different cities within 60 minutes and add them to the invalid list.

solution.py24 lines

1# Full working Python code
2
3def invalidTransactions(transactions):
4    from collections import defaultdict
5    invalid = set()
6    transaction_map = defaultdict(list)
7    n = len(transactions)
8    for i in range(n):
9        name, time, amount, city = transactions[i].split(',')
10        time = int(time)
11        amount = int(amount)
12        if amount > 1000:
13            invalid.add(transactions[i])
14        transaction_map[name].append((time, city, transactions[i]))
15    for name, trans in transaction_map.items():
16        for i in range(len(trans)):
17            time1, city1, trans1 = trans[i]
18            for j in range(len(trans)):
19                if i != j:
20                    time2, city2, trans2 = trans[j]
21                    if city1 != city2 and abs(time1 - time2) <= 60:
22                        invalid.add(trans1)
23                        invalid.add(trans2)
24    return list(invalid)

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the transactions and use a hashmap for quick lookups. The space complexity is O(n) due to the storage of transactions in the hashmap.

1Transactions can be invalid due to amount or timing/location conflicts.
2Using a hashmap allows for faster lookups and reduces the need for nested loops.

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