Today I have done leetcode's May LeetCoding Challenge with cpp.

May LeetCoding Challenge 31

Description

Design Underground System

An underground railway system is keeping track of customer travel times between different stations. They are using this data to calculate the average time it takes to travel from one station to another.

Implement the UndergroundSystem class:

void checkIn(int id, string stationName, int t)

<ul>
	<li>A customer with a card ID equal to <code>id</code>, checks in at the station <code>stationName</code> at time <code>t</code>.</li>
	<li>A customer can only be checked into one place at a time.</li>
</ul>
</li>
<li><code>void checkOut(int id, string stationName, int t)</code>
<ul>
	<li>A customer with a card ID equal to <code>id</code>, checks out from the station <code>stationName</code> at time <code>t</code>.</li>
</ul>
</li>
<li><code>double getAverageTime(string startStation, string endStation)</code>
<ul>
	<li>Returns the average time it takes to travel from <code>startStation</code> to <code>endStation</code>.</li>
	<li>The average time is computed from all the previous traveling times from <code>startStation</code> to <code>endStation</code> that happened <strong>directly</strong>, meaning a check in at <code>startStation</code> followed by a check out from <code>endStation</code>.</li>
	<li>The time it takes to travel from <code>startStation</code> to <code>endStation</code> <strong>may be different</strong> from the time it takes to travel from <code>endStation</code> to <code>startStation</code>.</li>
	<li>There will be at least one customer that has traveled from <code>startStation</code> to <code>endStation</code> before <code>getAverageTime</code> is called.</li>
</ul>
</li>

You may assume all calls to the checkIn and checkOut methods are consistent. If a customer checks in at time t₁ then checks out at time t₂, then t₁ < t₂. All events happen in chronological order.

Example 1:

Input
["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"]
[[],[45,"Leyton",3],[32,"Paradise",8],[27,"Leyton",10],[45,"Waterloo",15],[27,"Waterloo",20],[32,"Cambridge",22],["Paradise","Cambridge"],["Leyton","Waterloo"],[10,"Leyton",24],["Leyton","Waterloo"],[10,"Waterloo",38],["Leyton","Waterloo"]]

Output
[null,null,null,null,null,null,null,14.00000,11.00000,null,11.00000,null,12.00000]

Explanation
UndergroundSystem undergroundSystem = new UndergroundSystem();
undergroundSystem.checkIn(45, "Leyton", 3);
undergroundSystem.checkIn(32, "Paradise", 8);
undergroundSystem.checkIn(27, "Leyton", 10);
undergroundSystem.checkOut(45, "Waterloo", 15);  // Customer 45 "Leyton" -> "Waterloo" in 15-3 = 12
undergroundSystem.checkOut(27, "Waterloo", 20);  // Customer 27 "Leyton" -> "Waterloo" in 20-10 = 10
undergroundSystem.checkOut(32, "Cambridge", 22); // Customer 32 "Paradise" -> "Cambridge" in 22-8 = 14
undergroundSystem.getAverageTime("Paradise", "Cambridge"); // return 14.00000. One trip "Paradise" -> "Cambridge", (14) / 1 = 14
undergroundSystem.getAverageTime("Leyton", "Waterloo");    // return 11.00000. Two trips "Leyton" -> "Waterloo", (10 + 12) / 2 = 11
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo");    // return 11.00000
undergroundSystem.checkOut(10, "Waterloo", 38);  // Customer 10 "Leyton" -> "Waterloo" in 38-24 = 14
undergroundSystem.getAverageTime("Leyton", "Waterloo");    // return 12.00000. Three trips "Leyton" -> "Waterloo", (10 + 12 + 14) / 3 = 12

Example 2:

Input
["UndergroundSystem","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime","checkIn","checkOut","getAverageTime"]
[[],[10,"Leyton",3],[10,"Paradise",8],["Leyton","Paradise"],[5,"Leyton",10],[5,"Paradise",16],["Leyton","Paradise"],[2,"Leyton",21],[2,"Paradise",30],["Leyton","Paradise"]]

Output
[null,null,null,5.00000,null,null,5.50000,null,null,6.66667]

Explanation
UndergroundSystem undergroundSystem = new UndergroundSystem();
undergroundSystem.checkIn(10, "Leyton", 3);
undergroundSystem.checkOut(10, "Paradise", 8); // Customer 10 "Leyton" -> "Paradise" in 8-3 = 5
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.00000, (5) / 1 = 5
undergroundSystem.checkIn(5, "Leyton", 10);
undergroundSystem.checkOut(5, "Paradise", 16); // Customer 5 "Leyton" -> "Paradise" in 16-10 = 6
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 5.50000, (5 + 6) / 2 = 5.5
undergroundSystem.checkIn(2, "Leyton", 21);
undergroundSystem.checkOut(2, "Paradise", 30); // Customer 2 "Leyton" -> "Paradise" in 30-21 = 9
undergroundSystem.getAverageTime("Leyton", "Paradise"); // return 6.66667, (5 + 6 + 9) / 3 = 6.66667

Constraints:

1 <= id, t <= 10⁶
1 <= stationName.length, startStation.length, endStation.length <= 10
All strings consist of uppercase and lowercase English letters and digits.
There will be at most 2 * 10⁴ calls in total to checkIn, checkOut, and getAverageTime.
Answers within 10^-5 of the actual value will be accepted.

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
struct Time {
  long long total;
  double count;
  Time(): total(0), count(0) {}
};
class UndergroundSystem {
  unordered_map<string, unordered_map<string, Time>> time;
  unordered_map<int, int> checkInTime;
  unordered_map<int, string> checkInStation;
public:
  UndergroundSystem(){}

  void checkIn(int id, string stationName, int t) {
    checkInTime[id] = t;
    checkInStation[id] = stationName;
  }

  void checkOut(int id, string end, int t) {
    int interval = t - checkInTime[id];
    string start = checkInStation[id];
    time[start][end].total += interval;
    time[start][end].count += 1;
  }

  double getAverageTime(string start, string end) {
    return time[start][end].total / time[start][end].count;
  }
};

// Accepted
// 56/56 cases passed (127 ms)
// Your runtime beats 99.46 % of cpp submissions
// Your memory usage beats 70.27 % of cpp submissions (58.6 MB)

2023-05-31 Daily Challenge

May LeetCoding Challenge 31

Description

Solution