Leetcode solution 1169: Invalid Transactions

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.

Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.

Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.

Example 1:

Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.

Example 2:

Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]

Example 3:

Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]

Constraints:

transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000.

Problem link

Video Tutorial

You can find the detailed video tutorial here

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Thought Process

It's an implementation problem and it's up to you how fast you can type to determine whether you want to flex your muscles on object oriented design. It's actually quite a small moev, whether you want to convert the comma separated line into a Transaction object or not.

To meet the two conditions, we can simply loop over all the transactions while at the same time, keep a map to store all the transactions under the same name. Then the problem is compare the city for all the transactions under the same name that time is less or equal than 60 mins. At first I was trying to sort the collection based on the time in ascending order. However, that does not really help reduce the O(N^2) complexity since worst case is still all transactions are within the 60 mins bound, and one "bad" transaction can essentially cause all the previous "good" transactions to become bad, so we have to go back and revisit the previous ones. (as shown in below graph)

Example on one transaction make all previous ones invalid

Caveats

It's better to use a Set for de-duplication than a list since we might add duplicate transactions to the collection (e.g., a transaction > 1000 amount could also be invalid due to same name with different city under time < 60min)

Solutions

 1 public class Transaction implements Comparable<Transaction> {
 2         public String name;
 3         public int timeInMin;
 4         public int amount;
 5         public String city;
 6         public String txnString;
 7 
 8         /**
 9          * Constructor
10          * @param txn a string "{name},{time},{amount},{city}"
11          */
12         public Transaction(String txn) {
13             this.txnString = txn;
14 
15             String[] txns = txn.split(",");
16             this.name = txns[0];
17             this.timeInMin = Integer.parseInt(txns[1]);
18             this.amount = Integer.parseInt(txns[2]);
19             this.city = txns[3];
20         }
21 
22         // sort transactions order in timeInMin ascending order
23         @Override
24         public int compareTo(Transaction o) {
25             return this.timeInMin - o.timeInMin;
26         }
27     }
28 
29     public List<String> invalidTransactions(String[] transactions) {
30         // need to use a set, since an invalid one could be a one > 1000, and that one could also be invalid due to
31         // same person different locations
32         Set<String> invalidTxns = new HashSet<>();
33 
34         Map<String, List<Transaction>> lookup = new HashMap<>();
35 
36         for (String t : transactions) {
37             Transaction txn = new Transaction(t);
38             if (!lookup.containsKey(txn.name)) {
39                 lookup.put(txn.name, new ArrayList<Transaction>());
40             }
41 
42             lookup.get(txn.name).add(txn);
43 
44             if (txn.amount > 1000) {
45                 invalidTxns.add(txn.txnString);
46                 continue;
47             }
48         }
49 
50         for (Map.Entry<String, List<Transaction>> entry : lookup.entrySet()) {
51             List<Transaction> txns = entry.getValue();
52             if (txns.size() <= 1) {
53                 continue;
54             }
55 
56             // Sorting in this case didn't really help on time complexity
57             Collections.sort(txns);
58 
59             // have to perform two loops, essentially it's O(N^2) if they are all in range
60             for (int i = 1; i < txns.size(); i++) {
61                 Transaction cur = txns.get(i);
62 
63                 for (int j = i - 1; j >= 0; j--) {
64                     if (cur.timeInMin - txns.get(j).timeInMin > 60) {
65                         break;
66                     }
67 
68                     if (cur.city.equals(txns.get(j).city)) {
69                         continue;
70                     }
71 
72                     invalidTxns.add(cur.txnString);
73                     invalidTxns.add(txns.get(j).txnString);
74                 }
75             }
76         }
77         List<String> res = new ArrayList<>(invalidTxns);
78         return res;
79     }

Time Complexity: O(N^2),we have to loop through the transactions in a nested way for worst case all transactions are under the same name and all within 60min.

Space Complexity: O(N),the result list and set and the extra hashmap we used

References

None

来源：oschina

链接：https://my.oschina.net/u/4375228/blog/3398765

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