1169. Invalid Transactions

1169. Invalid Transactions (Medium)

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.

Related Topics:
Array, String

Solution 1. Brute force

// OJ: https://leetcode.com/problems/invalid-transactions/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
struct Transaction {
    string name;
    int time;
    int amount;
    string city;
    Transaction(string t) {
        istringstream ss(t);
        getline(ss, name, ',');
        string token;
        getline(ss, token, ',');
        time = stoi(token);
        getline(ss, token, ',');
        amount = stoi(token);
        ss >> city;
    }
};
class Solution {
public:
    vector<string> invalidTransactions(vector<string>& A) {
        vector<Transaction> B;
        unordered_set<int> invalid;
        for (int i = 0; i < A.size(); ++i) {
            auto t = Transaction(A[i]);
            if (t.amount > 1000) invalid.insert(i);
            B.push_back(t);
        }
        for (int i = 0; i < A.size(); ++i) {
            for (int j = i + 1; j < A.size(); ++j) {
                if (B[i].name == B[j].name && abs(B[i].time - B[j].time) <= 60 && B[i].city != B[j].city) {
                    invalid.insert(i);
                    invalid.insert(j);
                }
            }
        }
        vector<string> ans;
        for (int i : invalid) ans.push_back(A[i]);
        return ans;
    }
};

Solution 1. Sliding Window

// OJ: https://leetcode.com/problems/invalid-transactions/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
struct Transaction {
    string name;
    int time;
    int amount;
    string city;
    Transaction(string t) {
        istringstream ss(t);
        getline(ss, name, ',');
        string token;
        getline(ss, token, ',');
        time = stoi(token);
        getline(ss, token, ',');
        amount = stoi(token);
        ss >> city;
    }
};
class Solution {
public:
    vector<string> invalidTransactions(vector<string>& A) {
        unordered_map<string, vector<int>> m;
        vector<Transaction> B;
        unordered_set<int> invalid;
        for (int i = 0; i < A.size(); ++i) {
            auto t = Transaction(A[i]);
            if (t.amount > 1000) invalid.insert(i);
            m[t.name].push_back(i);
            B.push_back(t);
        }
        for (auto &[p, ids] : m) {
            sort(begin(ids), end(ids), [&](int a, int b) { return B[a].time < B[b].time; });
            int i = 0, j = 0, k = 0, N = ids.size();
            unordered_map<string, int> cities;
            while (j < N) {
                while (B[ids[j]].time - B[ids[i]].time > 60) {
                    auto &c = B[ids[i++]];
                    if (--cities[c.city] == 0) cities.erase(c.city);
                }
                if (B[ids[j]].time - B[ids[i]].time <= 60) {
                    auto &c = B[ids[j++]];
                    cities[c.city]++;
                    if (cities.size() > 1) {
                        for (k = max(k, i); k < j; ++k) invalid.insert(ids[k]);
                    }
                }
            }
        }
        vector<string> ans;
        for (int i : invalid) ans.push_back(A[i]);
        return ans;
    }
};