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3175. Find The First Player to win K Games in a Row
Description
A competition consists of n
players numbered from 0
to n  1
.
You are given an integer array skills
of size n
and a positive integer k
, where skills[i]
is the skill level of player i
. All integers in skills
are unique.
All players are standing in a queue in order from player 0
to player n  1
.
The competition process is as follows:
 The first two players in the queue play a game, and the player with the higher skill level wins.
 After the game, the winner stays at the beginning of the queue, and the loser goes to the end of it.
The winner of the competition is the first player who wins k
games in a row.
Return the initial index of the winning player.
Example 1:
Input: skills = [4,2,6,3,9], k = 2
Output: 2
Explanation:
Initially, the queue of players is [0,1,2,3,4]
. The following process happens:
 Players 0 and 1 play a game, since the skill of player 0 is higher than that of player 1, player 0 wins. The resulting queue is
[0,2,3,4,1]
.  Players 0 and 2 play a game, since the skill of player 2 is higher than that of player 0, player 2 wins. The resulting queue is
[2,3,4,1,0]
.  Players 2 and 3 play a game, since the skill of player 2 is higher than that of player 3, player 2 wins. The resulting queue is
[2,4,1,0,3]
.
Player 2 won k = 2
games in a row, so the winner is player 2.
Example 2:
Input: skills = [2,5,4], k = 3
Output: 1
Explanation:
Initially, the queue of players is [0,1,2]
. The following process happens:
 Players 0 and 1 play a game, since the skill of player 1 is higher than that of player 0, player 1 wins. The resulting queue is
[1,2,0]
.  Players 1 and 2 play a game, since the skill of player 1 is higher than that of player 2, player 1 wins. The resulting queue is
[1,0,2]
.  Players 1 and 0 play a game, since the skill of player 1 is higher than that of player 0, player 1 wins. The resulting queue is
[1,2,0]
.
Player 1 won k = 3
games in a row, so the winner is player 1.
Constraints:
n == skills.length
2 <= n <= 10^{5}
1 <= k <= 10^{9}
1 <= skills[i] <= 10^{6}
 All integers in
skills
are unique.
Solutions
Solution 1: Quick Thinking
We notice that each time the first two elements of the array are compared, regardless of the result, the next comparison will always be between the next element in the array and the current winner. Therefore, if we have looped $n1$ times, the final winner must be the maximum element in the array. Otherwise, if an element has won consecutively $k$ times, then this element is the final winner.
The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
Similar problems:

class Solution { public int findWinningPlayer(int[] skills, int k) { int n = skills.length; k = Math.min(k, n  1); int i = 0, cnt = 0; for (int j = 1; j < n; ++j) { if (skills[i] < skills[j]) { i = j; cnt = 1; } else { ++cnt; } if (cnt == k) { break; } } return i; } }

class Solution { public: int findWinningPlayer(vector<int>& skills, int k) { int n = skills.size(); k = min(k, n  1); int i = 0, cnt = 0; for (int j = 1; j < n; ++j) { if (skills[i] < skills[j]) { i = j; cnt = 1; } else { ++cnt; } if (cnt == k) { break; } } return i; } };

class Solution: def findWinningPlayer(self, skills: List[int], k: int) > int: n = len(skills) k = min(k, n  1) i = cnt = 0 for j in range(1, n): if skills[i] < skills[j]: i = j cnt = 1 else: cnt += 1 if cnt == k: break return i

func findWinningPlayer(skills []int, k int) int { n := len(skills) k = min(k, n1) i, cnt := 0, 0 for j := 1; j < n; j++ { if skills[i] < skills[j] { i = j cnt = 1 } else { cnt++ } if cnt == k { break } } return i }

function findWinningPlayer(skills: number[], k: number): number { const n = skills.length; k = Math.min(k, n  1); let [i, cnt] = [0, 0]; for (let j = 1; j < n; ++j) { if (skills[i] < skills[j]) { i = j; cnt = 1; } else { ++cnt; } if (cnt === k) { break; } } return i; }