Welcome to Subscribe On Youtube
1904. The Number of Full Rounds You Have Played
Description
You are participating in an online chess tournament. There is a chess round that starts every 15
minutes. The first round of the day starts at 00:00
, and after every 15
minutes, a new round starts.
 For example, the second round starts at
00:15
, the fourth round starts at00:45
, and the seventh round starts at01:30
.
You are given two strings loginTime
and logoutTime
where:
loginTime
is the time you will login to the game, andlogoutTime
is the time you will logout from the game.
If logoutTime
is earlier than loginTime
, this means you have played from loginTime
to midnight and from midnight to logoutTime
.
Return the number of full chess rounds you have played in the tournament.
Note: All the given times follow the 24hour clock. That means the first round of the day starts at 00:00
and the last round of the day starts at 23:45
.
Example 1:
Input: loginTime = "09:31", logoutTime = "10:14" Output: 1 Explanation: You played one full round from 09:45 to 10:00. You did not play the full round from 09:30 to 09:45 because you logged in at 09:31 after it began. You did not play the full round from 10:00 to 10:15 because you logged out at 10:14 before it ended.
Example 2:
Input: loginTime = "21:30", logoutTime = "03:00" Output: 22 Explanation: You played 10 full rounds from 21:30 to 00:00 and 12 full rounds from 00:00 to 03:00. 10 + 12 = 22.
Constraints:
loginTime
andlogoutTime
are in the formathh:mm
.00 <= hh <= 23
00 <= mm <= 59
loginTime
andlogoutTime
are not equal.
Solutions
Solution 1: Convert to Minutes
We can convert the input strings to minutes $a$ and $b$. If $a > b$, it means that it crosses midnight, so we need to add one day’s minutes $1440$ to $b$.
Then we round $a$ up to the nearest multiple of $15$, and round $b$ down to the nearest multiple of $15$. Finally, we return the difference between $b$ and $a$. Note that we should take the larger value between $0$ and $b  a$.
The time complexity is $O(1)$, and the space complexity is $O(1)$.

class Solution { public int numberOfRounds(String loginTime, String logoutTime) { int a = f(loginTime), b = f(logoutTime); if (a > b) { b += 1440; } return Math.max(0, b / 15  (a + 14) / 15); } private int f(String s) { int h = Integer.parseInt(s.substring(0, 2)); int m = Integer.parseInt(s.substring(3, 5)); return h * 60 + m; } }

class Solution { public: int numberOfRounds(string loginTime, string logoutTime) { auto f = [](string& s) { int h, m; sscanf(s.c_str(), "%d:%d", &h, &m); return h * 60 + m; }; int a = f(loginTime), b = f(logoutTime); if (a > b) { b += 1440; } return max(0, b / 15  (a + 14) / 15); } };

class Solution: def numberOfRounds(self, loginTime: str, logoutTime: str) > int: def f(s: str) > int: return int(s[:2]) * 60 + int(s[3:]) a, b = f(loginTime), f(logoutTime) if a > b: b += 1440 a, b = (a + 14) // 15, b // 15 return max(0, b  a)

func numberOfRounds(loginTime string, logoutTime string) int { f := func(s string) int { var h, m int fmt.Sscanf(s, "%d:%d", &h, &m) return h*60 + m } a, b := f(loginTime), f(logoutTime) if a > b { b += 1440 } return max(0, b/15(a+14)/15) }

function numberOfRounds(startTime: string, finishTime: string): number { const f = (s: string): number => { const [h, m] = s.split(':').map(Number); return h * 60 + m; }; let [a, b] = [f(startTime), f(finishTime)]; if (a > b) { b += 1440; } return Math.max(0, Math.floor(b / 15)  Math.ceil(a / 15)); }