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Invalid Transactions LeetCode Solution
Problem – Invalid Transactions
A transaction is possibly invalid if:
- the amount exceeds
$1000, or; - if it occurs within (and including)
60minutes 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,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 between1and10. - Each
{time}consist of digits, and represent an integer between0and1000. - Each
{amount}consist of digits, and represent an integer between0and2000.
Invalid Transactions LeetCode Solution in Java
class Solution {
public List<String> invalidTransactions(final String[] transactions) {
final List<String> invalid = new ArrayList<>();
final Map<String, List<Transaction>> map = new HashMap<>();
/*
* build a map with name as key and value as list of transactions for that name
*/
for (final String transaction : transactions) {
final Transaction tran = new Transaction(transaction);
if (map.containsKey(tran.name)) {
map.get(tran.name).add(tran);
} else {
final List<Transaction> list = new ArrayList<>();
list.add(tran);
map.put(tran.name, list);
}
}
for (final String transaction : transactions) {
final Transaction tran = new Transaction(transaction);
if (!isValid(map.get(tran.name), tran)) {
invalid.add(transaction);
}
}
return invalid;
}
public boolean isValid(final List<Transaction> transactions, final Transaction transaction) {
/* if there is only one transaction and the amount is less than 1000 */
if (transactions.size() <= 1 && transaction.amount < 1000)
return true;
/* check against all other transaction to check it this one is valid */
for (final Transaction tran : transactions) {
if (transaction.invalidTransaction(tran.city, tran.time)) {
return false;
}
}
return true;
}
class Transaction {
String name;
int time;
int amount;
String city;
Transaction(final String transaction) {
final String[] t = transaction.split(",");
this.name = t[0];
this.time = Integer.parseInt(t[1]);
this.amount = Integer.parseInt(t[2]);
this.city = t[3];
}
/*
* 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.
*/
public boolean invalidTransaction(final String city, final int time) {
return invalidAmount() || differentCity(city, time);
}
private boolean differentCity(final String city, final int time) {
return !this.city.equals(city) && Math.abs(this.time - time) <= 60;
}
private boolean invalidAmount() {
return this.amount > 1000;
}
}
}Invalid Transactions LeetCode Solution in Python
class Solution(object):
def invalidTransactions(self, transactions):
"""
:type transactions: List[str]
:rtype: List[str]
"""
r = {}
inv = []
for i in transactions:
split = i.decode("utf-8").split(",")
name = str(split[0])
time = int(split[1])
amount = int(split[2])
city = str(split[3])
if time not in r:
r[time] = {
name: [city]
}
else:
if name not in r[time]:
r[time][name]=[city]
else:
r[time][name].append(city)
for i in transactions:
split = i.decode("utf-8").split(",")
name = str(split[0])
time = int(split[1])
amount = int(split[2])
city = str(split[3])
if amount > 1000:
inv.append(i)
continue
for j in range(time-60, time+61):
if j not in r:
continue
if name not in r[j]:
continue
if len(r[j][name]) > 1 or (r[j][name][0] != city):
inv.append(i)
break
return inv Invalid Transactions LeetCode Solution in C++
class transaction {
public:
int time;
int amount;
string city;
string tran;
bool marked;
transaction (int t, int a, string c, string tr) {
tran = tr;
time = t;
amount = a;
city = c;
marked = false;
}
};
class Solution {
public:
vector<string> invalidTransactions(vector<string>& transactions) {
if(transactions.empty())
return {};
unordered_map<string, vector<transaction*>> trans;
for(string &st : transactions) {
istringstream ss(st);
string token = "";
getline(ss, token, ',');
string name = token;
getline(ss, token, ',');
int time = stoi(token);
getline(ss, token, ',');
int amount = stoi(token);
getline(ss, token, ',');
string city = token;
transaction *t = new transaction(time, amount, city, st);
trans[name].push_back(t);
}
vector<string> result;
for(auto &p : trans) {
sort(p.second.begin(), p.second.end(), [](const transaction* a, const transaction* b) {
return a->time < b->time;
});
for(int i=0; i<p.second.size(); i++) {
if(p.second[i]->amount > 1000) {
result.push_back(p.second[i]->tran);
p.second[i]->marked = true;
}
if(i > 0 && p.second[i]->time <= p.second[i-1]->time + 60) {
int r = i-1;
while(r >= 0 && p.second[r]->time >= p.second[i]->time - 60) {
if(p.second[i]->city != p.second[r]->city) {
if(!p.second[r]->marked) {
result.push_back(p.second[r]->tran);
p.second[r]->marked = true;
}
if(!p.second[i]->marked) {
p.second[i]->marked = true;
result.push_back(p.second[i]->tran);
}
}
r--;
}
}
}
}
return result;
}
};
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