Leetcode #1900: The Earliest and Latest Rounds Where Players Compete
In this guide, we solve Leetcode #1900 The Earliest and Latest Rounds Where Players Compete in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There is a tournament where n players are participating. The players are standing in a single row and are numbered from 1 to n based on their initial standing position (player 1 is the first player in the row, player 2 is the second player in the row, etc.).
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Memoization, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: n = 11, firstPlayer = 2, secondPlayer = 4
Output: [3,4]
Explanation:
One possible scenario which leads to the earliest round number:
First round: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Second round: 2, 3, 4, 5, 6, 11
Third round: 2, 3, 4
One possible scenario which leads to the latest round number:
First round: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Second round: 1, 2, 3, 4, 5, 6
Third round: 1, 2, 4
Fourth round: 2, 4
Python Solution
def dfs(l: int, r: int, n: int):
if l + r == n - 1:
return [1, 1]
res = [inf, -inf]
m = n >> 1
for i in range(1 << m):
win = [False] * n
for j in range(m):
if i >> j & 1:
win[j] = True
else:
win[n - 1 - j] = True
if n & 1:
win[m] = True
win[n - 1 - l] = win[n - 1 - r] = False
win[l] = win[r] = True
a = b = c = 0
for j in range(n):
if j == l:
a = c
if j == r:
b = c
if win[j]:
c += 1
x, y = dfs(a, b, c)
res[0] = min(res[0], x + 1)
res[1] = max(res[1], y + 1)
return res
class Solution:
def earliestAndLatest(
self, n: int, firstPlayer: int, secondPlayer: int
) -> List[int]:
return dfs(firstPlayer - 1, secondPlayer - 1, n)
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.