Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
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.).
The tournament consists of multiple rounds (starting from round number 1). In each round, the ith player from the front of the row competes against the ith player from the end of the row, and the winner advances to the next round. When the number of players is odd for the current round, the player in the middle automatically advances to the next round.
1, 2, 4, 6, 7
1 competes against player 7.2 competes against player 6.4 automatically advances to the next round.After each round is over, the winners are lined back up in the row based on the original ordering assigned to them initially (ascending order).
The players numbered firstPlayer and secondPlayer are the best in the tournament. They can win against any other player before they compete against each other. If any two other players compete against each other, either of them might win, and thus you may choose the outcome of this round.
Given the integers n, firstPlayer, and secondPlayer, return an integer array containing two values, the earliest possible round number and the latest possible round number in which these two players will compete against each other, respectively.
Example 1:
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
Example 2:
Input: n = 5, firstPlayer = 1, secondPlayer = 5 Output: [1,1] Explanation: The players numbered 1 and 5 compete in the first round. There is no way to make them compete in any other round.
Constraints:
2 <= n <= 281 <= firstPlayer < secondPlayer <= nProblem summary: 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.). The tournament consists of multiple rounds (starting from round number 1). In each round, the ith player from the front of the row competes against the ith player from the end of the row, and the winner advances to the next round. When the number of players is odd for the current round, the player in the middle automatically advances to the next round. For example, if the row consists of players 1, 2, 4, 6, 7 Player 1 competes against player 7. Player 2 competes against player 6. Player 4 automatically advances to the next round. After each round is over, the winners are lined back up in the row based on the original ordering
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
11 2 4
5 1 5
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1900: The Earliest and Latest Rounds Where Players Compete
class Solution {
static int[][][] f = new int[30][30][31];
public int[] earliestAndLatest(int n, int firstPlayer, int secondPlayer) {
return dfs(firstPlayer - 1, secondPlayer - 1, n);
}
private int[] dfs(int l, int r, int n) {
if (f[l][r][n] != 0) {
return decode(f[l][r][n]);
}
if (l + r == n - 1) {
f[l][r][n] = encode(1, 1);
return new int[] {1, 1};
}
int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE;
int m = n >> 1;
for (int i = 0; i < (1 << m); i++) {
boolean[] win = new boolean[n];
for (int j = 0; j < m; j++) {
if (((i >> j) & 1) == 1) {
win[j] = true;
} else {
win[n - 1 - j] = true;
}
}
if ((n & 1) == 1) {
win[m] = true;
}
win[n - 1 - l] = win[n - 1 - r] = false;
win[l] = win[r] = true;
int a = 0, b = 0, c = 0;
for (int j = 0; j < n; j++) {
if (j == l) {
a = c;
}
if (j == r) {
b = c;
}
if (win[j]) {
c++;
}
}
int[] t = dfs(a, b, c);
min = Math.min(min, t[0] + 1);
max = Math.max(max, t[1] + 1);
}
f[l][r][n] = encode(min, max);
return new int[] {min, max};
}
private int encode(int x, int y) {
return (x << 8) | y;
}
private int[] decode(int val) {
return new int[] {val >> 8, val & 255};
}
}
// Accepted solution for LeetCode #1900: The Earliest and Latest Rounds Where Players Compete
var f [30][30][31]int
func earliestAndLatest(n int, firstPlayer int, secondPlayer int) []int {
return dfs(firstPlayer-1, secondPlayer-1, n)
}
func dfs(l, r, n int) []int {
if f[l][r][n] != 0 {
return decode(f[l][r][n])
}
if l+r == n-1 {
f[l][r][n] = encode(1, 1)
return []int{1, 1}
}
min, max := 1<<30, -1<<31
m := n >> 1
for i := 0; i < (1 << m); i++ {
win := make([]bool, n)
for j := 0; j < m; j++ {
if (i>>j)&1 == 1 {
win[j] = true
} else {
win[n-1-j] = true
}
}
if n&1 == 1 {
win[m] = true
}
win[n-1-l] = false
win[n-1-r] = false
win[l] = true
win[r] = true
a, b, c := 0, 0, 0
for j := 0; j < n; j++ {
if j == l {
a = c
}
if j == r {
b = c
}
if win[j] {
c++
}
}
t := dfs(a, b, c)
if t[0]+1 < min {
min = t[0] + 1
}
if t[1]+1 > max {
max = t[1] + 1
}
}
f[l][r][n] = encode(min, max)
return []int{min, max}
}
func encode(x, y int) int {
return (x << 8) | y
}
func decode(val int) []int {
return []int{val >> 8, val & 255}
}
# Accepted solution for LeetCode #1900: The Earliest and Latest Rounds Where Players Compete
@cache
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)
// Accepted solution for LeetCode #1900: The Earliest and Latest Rounds Where Players Compete
/**
* [1900] The Earliest and Latest Rounds Where Players Compete
*
* 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.).
* The tournament consists of multiple rounds (starting from round number 1). In each round, the i^th player from the front of the row competes against the i^th player from the end of the row, and the winner advances to the next round. When the number of players is odd for the current round, the player in the middle automatically advances to the next round.
*
* For example, if the row consists of players 1, 2, 4, 6, 7
*
* Player 1 competes against player 7.
* Player 2 competes against player 6.
* Player 4 automatically advances to the next round.
*
*
*
* After each round is over, the winners are lined back up in the row based on the original ordering assigned to them initially (ascending order).
* The players numbered firstPlayer and secondPlayer are the best in the tournament. They can win against any other player before they compete against each other. If any two other players compete against each other, either of them might win, and thus you may choose the outcome of this round.
* Given the integers n, firstPlayer, and secondPlayer, return an integer array containing two values, the earliest possible round number and the latest possible round number in which these two players will compete against each other, respectively.
*
* Example 1:
*
* 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
*
* Example 2:
*
* Input: n = 5, firstPlayer = 1, secondPlayer = 5
* Output: [1,1]
* Explanation: The players numbered 1 and 5 compete in the first round.
* There is no way to make them compete in any other round.
*
*
* Constraints:
*
* 2 <= n <= 28
* 1 <= firstPlayer < secondPlayer <= n
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/the-earliest-and-latest-rounds-where-players-compete/
// discuss: https://leetcode.com/problems/the-earliest-and-latest-rounds-where-players-compete/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
pub fn earliest_and_latest(n: i32, first_player: i32, second_player: i32) -> Vec<i32> {
vec![]
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
#[ignore]
fn test_1900_example_1() {
let n = 11;
let first_player = 2;
let second_player = 4;
let result = vec![3, 4];
assert_eq!(
Solution::earliest_and_latest(n, first_player, second_player),
result
);
}
#[test]
#[ignore]
fn test_1900_example_2() {
let n = 5;
let first_player = 1;
let second_player = 5;
let result = vec![1, 1];
assert_eq!(
Solution::earliest_and_latest(n, first_player, second_player),
result
);
}
}
// Accepted solution for LeetCode #1900: The Earliest and Latest Rounds Where Players Compete
function earliestAndLatest(n: number, firstPlayer: number, secondPlayer: number): number[] {
return dfs(firstPlayer - 1, secondPlayer - 1, n);
}
const f: number[][][] = Array.from({ length: 30 }, () =>
Array.from({ length: 30 }, () => Array(31).fill(0)),
);
function dfs(l: number, r: number, n: number): number[] {
if (f[l][r][n] !== 0) {
return decode(f[l][r][n]);
}
if (l + r === n - 1) {
f[l][r][n] = encode(1, 1);
return [1, 1];
}
let min = Number.MAX_SAFE_INTEGER;
let max = Number.MIN_SAFE_INTEGER;
const m = n >> 1;
for (let i = 0; i < 1 << m; i++) {
const win: boolean[] = Array(n).fill(false);
for (let j = 0; j < m; j++) {
if ((i >> j) & 1) {
win[j] = true;
} else {
win[n - 1 - j] = true;
}
}
if (n & 1) {
win[m] = true;
}
win[n - 1 - l] = false;
win[n - 1 - r] = false;
win[l] = true;
win[r] = true;
let a = 0,
b = 0,
c = 0;
for (let j = 0; j < n; j++) {
if (j === l) a = c;
if (j === r) b = c;
if (win[j]) c++;
}
const t = dfs(a, b, c);
min = Math.min(min, t[0] + 1);
max = Math.max(max, t[1] + 1);
}
f[l][r][n] = encode(min, max);
return [min, max];
}
function encode(x: number, y: number): number {
return (x << 8) | y;
}
function decode(val: number): number[] {
return [val >> 8, val & 255];
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.