README.md

1535. Find the Winner of an Array Game

Given an integer array arr of distinct integers and an integer k.

A game will be played between the first two elements of the array (i.e. arr[0] and arr[1]). In each round of the game, we compare arr[0] with arr[1], the larger integer wins and remains at position 0 and the smaller integer moves to the end of the array. The game ends when an integer wins k consecutive rounds.

Return the integer which will win the game.

It is guaranteed that there will be a winner of the game.

Example 1:

Input: arr = [2,1,3,5,4,6,7], k = 2
Output: 5
Explanation: Let's see the rounds of the game:
Round |       arr       | winner | win_count
  1   | [2,1,3,5,4,6,7] | 2      | 1
  2   | [2,3,5,4,6,7,1] | 3      | 1
  3   | [3,5,4,6,7,1,2] | 5      | 1
  4   | [5,4,6,7,1,2,3] | 5      | 2
So we can see that 4 rounds will be played and 5 is the winner because it wins 2 consecutive games.

Example 2:

Input: arr = [3,2,1], k = 10
Output: 3
Explanation: 3 will win the first 10 rounds consecutively.

Example 3:

Input: arr = [1,9,8,2,3,7,6,4,5], k = 7
Output: 9

Example 4:

Input: arr = [1,11,22,33,44,55,66,77,88,99], k = 1000000000
Output: 99

Constraints:

• 2 <= arr.length <= 10^5
• 1 <= arr[i] <= 10^6
• arr contains distinct integers.
• 1 <= k <= 10^9

Solutions (Ruby)

1. Solution

# @param {Integer[]} arr
# @param {Integer} k
# @return {Integer}
def get_winner(arr, k)
  winner = arr[0]
  wins = 0

  arr[1..].each do |x|
    if x > winner
      winner = x
      wins = 0
    end
    wins += 1
    break if wins == k
  end

  winner
end

Solutions (Rust)

1. Solution

impl Solution {
    pub fn get_winner(arr: Vec<i32>, k: i32) -> i32 {
        let mut winner = arr[0];
        let mut wins = 0;

        for &x in &arr[1..] {
            if x > winner {
                winner = x;
                wins = 0;
            }
            wins += 1;
            if wins == k {
                break;
            }
        }

        winner
    }
}