LeetCode 3680 - Generate Schedule
The problem is asking us to generate a round-robin style schedule for n teams, where each team plays every other team exactly twice, once at home and once away. Each day in the schedule can host exactly one match, and a team is not allowed to play on consecutive days.
Difficulty: 🟡 Medium
Topics: Array, Math, Greedy
Solution
Problem Understanding
The problem is asking us to generate a round-robin style schedule for n teams, where each team plays every other team exactly twice, once at home and once away. Each day in the schedule can host exactly one match, and a team is not allowed to play on consecutive days. The input is simply an integer n representing the number of teams, and the expected output is a 2D list representing the schedule with each element [home, away] for a given day.
The constraints tell us that n ranges from 2 to 50. This is small enough to allow constructive algorithms without worrying about extreme performance. However, the constraint that no team can play on consecutive days introduces a significant complexity because it prevents naive round-robin or permutation-based schedules from always being valid. Some n values, such as 3, may not allow any schedule at all due to the consecutive day restriction, so our solution must be able to detect and return an empty schedule in these cases.
Important edge cases include small values like n = 2 and n = 3, where scheduling may be impossible due to back-to-back constraints, and larger odd or even n values where generating an alternating schedule without consecutive day conflicts is feasible.
Approaches
The brute-force approach would attempt to generate all possible permutations of matches and test each permutation to see if it satisfies the consecutive-day rule. While this guarantees correctness, the factorial growth in possible permutations makes it completely impractical even for moderate n.
The optimal approach leverages structured construction. By generating matches systematically in a round-robin fashion and alternating teams carefully, we can avoid consecutive-day conflicts. For even n, we can pair teams in a mirrored pattern. For odd n, one team may be idle each round, which can also help avoid consecutive-day conflicts. The key insight is to construct the schedule iteratively while enforcing the consecutive-day constraint, instead of generating all permutations.
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force | O((n*(n-1))!) | O(n*(n-1)) | Generate all permutations of matches, test each for validity |
| Optimal | O(n^2) | O(n^2) | Construct schedule iteratively using round-robin pairing, avoid consecutive-day conflicts |
Algorithm Walkthrough
- Generate all matches where each team plays every other team twice, once as home and once as away. This will produce
n*(n-1)matches. - Initialize an empty schedule list and a list of last played days for each team.
- Iterate over the list of matches, and for each match, check if both teams have not played on the previous day. If they have not, schedule the match on the current day.
- If a match cannot be placed due to a consecutive-day conflict, postpone it and attempt to place it in the next available day.
- Repeat until all matches are scheduled or until no valid placement is possible. If some matches cannot be scheduled without violating the consecutive-day rule, return an empty list.
- Return the final schedule as a list of
[home, away]pairs.
Why it works: By keeping track of the last played day for each team, we enforce the invariant that no team plays on consecutive days. Generating all matches explicitly ensures that each pair of teams plays exactly twice.
Python Solution
from typing import List
class Solution:
def generateSchedule(self, n: int) -> List[List[int]]:
if n < 4 and n % 2 == 1:
return [] # impossible to schedule without consecutive-day conflicts for odd n < 4
matches = []
for i in range(n):
for j in range(n):
if i != j:
matches.append([i, j])
schedule = []
last_played = [-2] * n
day = 0
while matches:
placed = False
for idx, match in enumerate(matches):
home, away = match
if day - last_played[home] > 1 and day - last_played[away] > 1:
schedule.append(match)
last_played[home] = day
last_played[away] = day
matches.pop(idx)
placed = True
break
if not placed:
day += 1
else:
day += 1
return schedule
This implementation generates all matches, then iteratively places them while enforcing the no-consecutive-days constraint. last_played keeps track of the last day each team played. If a match cannot be placed on the current day, we advance the day counter until it can be placed.
Go Solution
func generateSchedule(n int) [][]int {
if n < 4 && n%2 == 1 {
return [][]int{} // impossible to schedule
}
matches := [][]int{}
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
if i != j {
matches = append(matches, []int{i, j})
}
}
}
schedule := [][]int{}
lastPlayed := make([]int, n)
for i := 0; i < n; i++ {
lastPlayed[i] = -2
}
day := 0
for len(matches) > 0 {
placed := false
for i, match := range matches {
home, away := match[0], match[1]
if day-lastPlayed[home] > 1 && day-lastPlayed[away] > 1 {
schedule = append(schedule, match)
lastPlayed[home] = day
lastPlayed[away] = day
matches = append(matches[:i], matches[i+1:]...)
placed = true
break
}
}
if !placed {
day++
} else {
day++
}
}
return schedule
}
The Go implementation mirrors the Python version closely, handling slices for matches and using an integer array to track the last day each team played. Removing an element from a slice requires appending the parts before and after the index.
Worked Examples
For n = 5, the algorithm first generates 20 matches. Using the last_played array, it schedules matches in the following order (example):
| Day | Match | last_played array after placement |
|---|---|---|
| 0 | [0,1] | [0,0,-2,-2,-2] |
| 1 | [2,3] | [0,0,1,1,-2] |
| 2 | [0,4] | [2,0,1,1,2] |
| 3 | [1,2] | [2,3,3,1,2] |
This continues until all matches are placed without consecutive-day violations.
For n = 3, no schedule is possible because there will always be a team playing consecutive days.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n^2) | Generating all matches produces O(n^2) matches and placement iterates through them. |
| Space | O(n^2) | Storing all matches and the schedule takes O(n^2) space. |
The iterative placement guarantees that each match is considered at most once per day, and the total number of days scales linearly with the number of matches.
Test Cases
sol = Solution()
# Example 1: n = 3, impossible
assert sol.generateSchedule(3) == []
# Example 2: n = 5, valid schedule
sched = sol.generateSchedule(5)
assert len(sched) == 20
teams_played = [set() for _ in range(5)]
for i, match in enumerate(sched):
home, away = match
if i > 0:
prev_home, prev_away = sched[i-1]
assert home != prev_home and home != prev_away
assert away != prev_home and away != prev_away
teams_played[home].add(away)
teams_played[away].add(home)
for i in range(5):
assert len(teams_played[i]) == 4
# Edge case: n = 2, simplest possible
sched2 = sol.generateSchedule(2)
assert len(sched2) == 2
# Edge case: n = 4, even number, possible
sched4 = sol.generateSchedule(4)
assert len(sched4) == 12
| Test | Why |
|---|---|
| n = 3 | Tests impossible small odd n scenario |
| n = 5 | Valid schedule for moderate odd n, checks consecutive-day rule |
| n = 2 | Minimal number of teams |
| n = 4 | Even number of teams, ensures algorithm works for small even n |
Edge Cases
One important edge case is n = 3, which is an odd number less than 4. Here, it is impossible to generate a schedule without violating the consecutive-day constraint. Our implementation handles this explicitly by returning an empty list early.
Another edge case is n = 2, the smallest possible number of teams. The schedule is trivial, but the consecutive-day check must still pass. Our algorithm schedules the two matches on consecutive days since each team alternates perfectly.
Finally, larger odd n like n = 5 can stress the algorithm because careful placement is