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Approach: We will utilize SQL to solve this problem by taking advantage of its aggregation and date manipulation capabilities. First, we will identify each player's first login date. Then, we'll check if there's a login entry for the subsequent day after that first login date. The final step is to calculate the fraction of players who have logged in on consecutive days, starting from their first login date.
The complexity of this SQL query is primarily determined by the table scan needed for aggregation:
1SELECT ROUND(SUM(CASE WHEN NextDay IS NOT NULL THEN 1 ELSE 0 END) / COUNT(DISTINCT player_id), 2) AS fraction FROM (SELECT a.player_id, MIN(a.event_date) AS FirstLogin, (SELECT MIN(b.event_date) FROM Activity b WHERE b.player_id = a.player_id AND b.event_date > MIN(a.event_date)) AS NextDay FROM Activity a GROUP BY a.player_id) Sub;The SQL query performs the following steps:
Approach: This approach utilizes Python's data manipulation capabilities to process the table and calculate the required fraction. We will parse the data, identify each player's first login date, and check for subsequent day logins programmatically. Finally, we'll determine the desired fraction by counting players who have re-logged on the next day after their initial login.
The complexity for this approach is:
1import pandas as pd
2from datetime import timedelta
3
4data
This approach involves processing the input data in a structured manner using SQL queries to identify players who logged in on consecutive days starting from their first login date. We will use SQL window functions to handle date differences effectively and then calculate the desired fraction.
Time Complexity: O(n), where n is the number of records, as each operation scales linearly with the dataset size.
Space Complexity: O(n), as additional columns are created for processing.
1import pandas as pd
2
3data = pd
This approach involves a multi-pass strategy to handle dates and detect consecutive logins by manually checking day-by-day login activity.
Time Complexity: O(n log n) due to sorting necessary for detecting consecutive logins.
Space Complexity: O(n), where n represents distinct players and their login dates.
1#include <iostream>
2#include <vector>
3#include <unordered_map>
4#include <string>
5#include <sstream>
6#include <iomanip>
using namespace std;
struct Activity {
int player_id;
int device_id;
string event_date;
int games_played;
};
bool isNextDay(const string& cur, const string& next) {
tm cur_tm = {}, next_tm = {};
strptime(cur.c_str(), "%Y-%m-%d", &cur_tm);
strptime(next.c_str(), "%Y-%m-%d", &next_tm);
next_tm.tm_mday--;
mktime(&next_tm);
return cur_tm.tm_year == next_tm.tm_year && cur_tm.tm_mon == next_tm.tm_mon && cur_tm.tm_mday == next_tm.tm_mday;
}
string fractionConsecutiveLogins(vector<Activity> data) {
unordered_map<int, vector<string>> playerMap;
for (const auto& activity : data) {
playerMap[activity.player_id].push_back(activity.event_date);
}
int totalConsecutive = 0;
for (auto& pair : playerMap) {
auto& dates = pair.second;
sort(dates.begin(), dates.end());
for (size_t i = 1; i < dates.size(); ++i) {
if (isNextDay(dates[i - 1], dates[i])) {
totalConsecutive++;
break;
}
}
}
int totalPlayers = playerMap.size();
double fraction = static_cast<double>(totalConsecutive) / totalPlayers;
ostringstream out;
out << fixed << setprecision(2) << fraction;
return out.str();
}
int main() {
vector<Activity> data = {
{1, 2, "2016-03-01", 5},
{1, 2, "2016-03-02", 6},
{2, 3, "2017-06-25", 1},
{3, 1, "2016-03-02", 0},
{3, 4, "2018-07-03", 5}
};
cout << "Fraction: " << fractionConsecutiveLogins(data) << endl;
return 0;
}This Python code executes the following steps:
This solution leverages the Pandas library to mimic SQL-like operations. First, we calculate the first login date for each player using the 'groupby' and 'min' functions. We then merge this information back into the original data to allow comparison with the subsequent login dates. By checking if the event date matches the first login date plus one day, we determine the consecutive logins. Summing and dividing provides the result fraction.
This C++ solution uses common library functions to handle date operations while grouping data by players. It sorts the dates for each player and checks for consecutive days. The solution counts how many players meet the consecutive login criterion, resulting in the calculated fraction.