LeetCode 1900: The Earliest and Latest Rounds Where Players Compete Solution
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1900. The Earliest and Latest Rounds Where Players Compete
Problem Explanation
Explanation:
To solve this problem, we need to find the earliest and latest rounds in which the two given players, firstPlayer and secondPlayer, will compete against each other in the tournament.
The key observation is that the players will compete against each other in the earliest round when they are adjacent in the sequence of players, and in the latest round when the players are as far away from each other as possible in the sequence.
We can simulate the tournament rounds to determine the earliest and latest rounds in which the two players will compete against each other. We keep track of the range of players between firstPlayer and secondPlayer after each round, and update the range accordingly based on the outcomes of the matches.
Solution Code
class Solution {
public int[] earliestAndLatest(int n, int firstPlayer, int secondPlayer) {
int minRound = 1, maxRound = 1;
int distance = secondPlayer - firstPlayer;
while (n > 1) {
if (secondPlayer == firstPlayer + 1 && firstPlayer % 2 == 1 && n % 2 == 0) {
break;
}
if (firstPlayer % 2 != 0) {
firstPlayer = (firstPlayer + 1) / 2;
} else {
firstPlayer /= 2;
}
if (secondPlayer % 2 != 0) {
secondPlayer = (secondPlayer + 1) / 2;
} else {
secondPlayer /= 2;
}
n = (n + 1) / 2;
minRound++;
maxRound++;
}
return new int[]{minRound, maxRound};
}
}Try It Yourself
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