What is the time complexity of Invalid Transactions?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Invalid Transactions?

This problem uses Array, Hash Table, String, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Invalid Transactions?

Insight 1: Parsing the transaction string into its components (name, time, amount, city) is crucial. Insight 2: We need to compare each transaction against all other transactions made by the same person to check for proximity in time and difference in city.

Which companies ask LeetCode 1169. Invalid Transactions in interviews?

This problem is frequently asked at PayPal, Wix.

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Invalid Transactions - LeetCode 1169 Solution

Invalid Transactions - Complete Solution Guide

Invalid Transactions is LeetCode problem 1169, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A transaction is possibly invalid if: the amount exceeds $1000 , or; if it occurs within (and including) 60 minutes of another transaction with the same name in a different city . You are given an array of strings transaction where transactions[i] consists of comma-separated values representing the name, time (in minutes), amount, and city of the transaction. Return a list of transactions that are possibly invalid. You may return the answer in any order . Example 1: Input: transactions = ["alice

Detailed Explanation

The problem asks us to identify potentially invalid transactions from a list of transactions. Each transaction contains the name of the person, the time of the transaction, the amount, and the city where the transaction occurred. A transaction is considered invalid if either its amount exceeds $1000, or if it happens within 60 minutes of another transaction by the same person in a different city. The input is a list of strings, each string representing a transaction. The output should be a list of strings representing the invalid transactions, in any order.

Solution Approach

The solution involves iterating through the list of transactions. For each transaction, it first checks if the amount is greater than 1000. If it is, the transaction is marked as invalid. Then, it iterates through the rest of the transactions, comparing each transaction with the current one. If another transaction has the same name and a different city, and occurs within 60 minutes of the current transaction, then both the current transaction and the other transaction are marked as invalid. Finally, a list of invalid transactions is returned.

Step-by-Step Algorithm

Step 1: Parse each transaction string into its individual components: name, time, amount, and city.
Step 2: Iterate through the parsed transactions.
Step 3: For each transaction, check if the amount exceeds $1000. If it does, mark the transaction as invalid.
Step 4: Iterate through the remaining transactions.
Step 5: Compare the current transaction with each of the other transactions. If the name is the same, the city is different, and the time difference is within 60 minutes (inclusive), mark both transactions as invalid.
Step 6: Collect all the transactions marked as invalid into a result list.
Step 7: Return the list of invalid transactions.

Key Insights

Insight 1: Parsing the transaction string into its components (name, time, amount, city) is crucial.
Insight 2: We need to compare each transaction against all other transactions made by the same person to check for proximity in time and difference in city.
Insight 3: Using a hash table or map to group transactions by name can optimize the comparison process.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, String, Sorting.

Companies

Asked at: PayPal, Wix.

Invalid Transactions | LeetCode 1169

Medium

Array Hash Table String Sorting

Time: O(n^2)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<String> invalidTransactions(String[] transactions) {
        if (transactions == null || transactions.length == 0) {
            return new ArrayList<>();
        }

        Set<Integer> invalidIndices = new HashSet<>();
        Map<String, List<Map<String, Object>>> transactionsByName = new HashMap<>();

        for (int i = 0; i < transactions.length; i++) {
            String tStr = transactions[i];
            String[] parts = tStr.split(",");
            String name = parts[0];
            String timeStr = parts[1];
            String amountStr = parts[2];
            String city = parts[3];

            int time = Integer.parseInt(timeStr);
            int amount = Integer.parseInt(amountStr);

            if (amount > 1000) {
                invalidIndices.add(i);
            }

            if (!transactionsByName.containsKey(name)) {
                transactionsByName.put(name, new ArrayList<>());
            }

            Map<String, Object> transaction = new HashMap<>();
            transaction.put("time", time);
            transaction.put("city", city);
            transaction.put("index", i);

            transactionsByName.get(name).add(transaction);
        }

        for (String name : transactionsByName.keySet()) {
            List<Map<String, Object>> userTransactions = transactionsByName.get(name);
            userTransactions.sort(Comparator.comparingInt(t -> (int) t.get("time")));

            int n = userTransactions.size();
            for (int i = 0; i < n; i++) {
                for (int j = i + 1; j < n; j++) {
                    Map<String, Object> t1 = userTransactions.get(i);
                    Map<String, Object> t2 = userTransactions.get(j);

                    if ((int) t2.get("time") - (int) t1.get("time") > 60) {
                        break;
                    }

                    if (!t2.get("city").equals(t1.get("city"))) {
                        invalidIndices.add((int) t1.get("index"));
                        invalidIndices.add((int) t2.get("index"));
                    }
                }
            }
        }

        List<String> invalidTransactionsList = new ArrayList<>();
        for (int i : invalidIndices) {
            invalidTransactionsList.add(transactions[i]);
        }

        return invalidTransactionsList;
    }
}

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Problem Info

Description

Problem 1169: Invalid Transactions - Detailed explanation and solution approaches.

Topics

Array Hash Table String Sorting

Companies

PayPalWix

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1169

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Parse each transaction string into its individual components: name, time, amount, and city.
2
Step 2: Iterate through the parsed transactions.
3
Step 3: For each transaction, check if the amount exceeds $1000. If it does, mark the transaction as invalid.
4
Step 4: Iterate through the remaining transactions.
5
Step 5: Compare the current transaction with each of the other transactions. If the name is the same, the city is different, and the time difference is within 60 minutes (inclusive), mark both transactions as invalid.
6
Step 6: Collect all the transactions marked as invalid into a result list.
7
Step 7: Return the list of invalid transactions.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n^2) due to the nested loops: one outer loop iterating through each transaction and one inner loop iterating through the rest of the transactions to find invalid pairs. The space complexity is O(n) because the `invalid_indices` set or equivalent structure, and the `result` list, can potentially store up to n transaction indices or strings, respectively.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting and Two Pointers

Group transactions by name. Within each group, sort transactions by time. Use two pointers to efficiently find transactions within 60 minutes.

Time: O(n log n), where n is the number of transactions, due to sorting.Space: O(n), where n is the number of transactions, due to grouping.

Trade-offs: This approach can be more efficient than the O(n^2) solution if the sorting can be done in O(n log n) time, and can often improve the actual runtime if the transaction data leads to few comparisons in the nested loop. It is more complex to implement than the straightforward O(n^2) solution.

✨

Key Insights

1Insight 1: Parsing the transaction string into its components (name, time, amount, city) is crucial.
2Insight 2: We need to compare each transaction against all other transactions made by the same person to check for proximity in time and difference in city.
3Insight 3: Using a hash table or map to group transactions by name can optimize the comparison process.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input list of transactions. The solution correctly handles this by returning an empty list.
Edge case 2: All transactions are valid. The solution correctly returns an empty list in this case.
Edge case 3: Transactions with the same name and city, but different times within 60 minutes. These are not invalid unless the amount exceeds $1000. The solution handles this correctly.
Edge case 4: Transactions with very large time values. The solution relies on integer comparisons and subtraction, which might overflow if the time values are too large (though problem constraints state time is between 0 and 1000).
Edge case 5: Duplicate transactions in the input. Solution handles this since each transaction is checked with every other transaction.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge case of an empty input list.

✗Common mistake 2: Incorrectly calculating the time difference (e.g., using absolute difference).

✗Common mistake 3: Forgetting to check for the same name and different city condition when comparing transactions.

✗Common mistake 4: Only marking one of the invalid transactions when both should be marked.

✗Common mistake 5: Not parsing transaction data correctly and failing to handle potential parsing errors (especially in languages like C).

✗Common mistake 6: Not sorting the transactions by time when optimizing with the two-pointer approach and grouping by names.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Fraud detection in financial transactions. This algorithm can be used to identify suspicious transactions that may be fraudulent.

🏦Real-world scenario 2: Anomaly detection in network traffic. Identifying unusual patterns in network traffic can help detect security threats.

🎮Real-world scenario 3: Monitoring user activity in online games. Detecting unusual patterns in player behavior can help identify cheaters or hackers.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 29, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Invalid Transactions"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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