Leetcode 1535. Find the Winner of an Array Game

Seawolf

1. Given an integer array arr of distinct integers and an integer k.
   
2. 
   
3. 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.
   
4. 
   
5. Return the integer which will win the game.
   
6. 
   
7. It is guaranteed that there will be a winner of the game.
   
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9. 
   
10. 
    
11. Example 1:
    
12. 
    
13. Input: arr = [2,1,3,5,4,6,7], k = 2
    
14. Output: 5
    
15. Explanation: Let's see the rounds of the game:
    
16. Round |       arr       | winner | win_count
    
17.   1   | [2,1,3,5,4,6,7] | 2      | 1
    
18.   2   | [2,3,5,4,6,7,1] | 3      | 1
    
19.   3   | [3,5,4,6,7,1,2] | 5      | 1
    
20.   4   | [5,4,6,7,1,2,3] | 5      | 2
    
21. So we can see that 4 rounds will be played and 5 is the winner because it wins 2 consecutive games.
    
22. Example 2:
    
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24. Input: arr = [3,2,1], k = 10
    
25. Output: 3
    
26. Explanation: 3 will win the first 10 rounds consecutively.
    
27. Example 3:
    
28. 
    
29. Input: arr = [1,9,8,2,3,7,6,4,5], k = 7
    
30. Output: 9
    
31. Example 4:
    
32. 
    
33. Input: arr = [1,11,22,33,44,55,66,77,88,99], k = 1000000000
    
34. Output: 99
    
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36. 
    
37. Constraints:
    
38. 
    
39. 2 <= arr.length <= 10^5
    
40. 1 <= arr[i] <= 10^6
    
41. arr contains distinct integers.
    
42. 1 <= k <= 10^9

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1. class Solution:
   
2.     def getWinner(self, arr: List[int], k: int) -> int:
   
3.         max_val = arr[0]
   
4.         count = 0
   
5.         for i in range(1, len(arr)):
   
6.             if arr[i] > max_val:
   
7.                 max_val = arr[i]
   
8.                 count = 1
   
9.             else:
   
10.                 count += 1
    
11.             if count == k:
    
12.                 break
    
13.         return max_val

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