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.
Constraints:
2 <= arr.length <= 1051 <= arr[i] <= 106arr contains distinct integers.1 <= k <= 109Problem Overview: You are given an integer array where the first two elements compete in rounds. The larger value wins and stays at the front, while the smaller moves to the end. The first number that wins k consecutive rounds becomes the winner. The task is to simulate this process and return the winning element.
Approach 1: Simulated Rounds Approach (O(n * k) time, O(n) space)
This method directly simulates the rules of the game using a queue-like process. Compare the first two elements, move the smaller value to the end, and keep the larger value at the front while tracking consecutive wins. Each round performs constant operations but can repeat many times if k is large. A queue or simple array rotation can model the process. This approach mirrors the problem description exactly and is useful for understanding the mechanics of the game.
Approach 2: Break Condition Approach (O(n) time, O(1) space)
The key observation is that the largest number in the array will eventually dominate every comparison. Instead of fully simulating rotations, iterate once through the array while maintaining a current champion and its consecutive win count. If the next value is larger, it becomes the new champion and the streak resets. Otherwise the champion's win streak increases. Once the streak reaches k, you can stop early. If no element reaches k, the maximum element of the array becomes the winner by default.
This approach works because the champion only changes when a larger element appears, which can happen at most n times. It avoids repeated array rotations and keeps only two variables: the current winner and win count.
Recommended for interviews: The Break Condition Approach is what interviewers typically expect. It shows you recognized the core insight that the maximum element will eventually dominate and that full simulation is unnecessary. Mentioning the simulated approach first demonstrates understanding of the game rules, while optimizing to the linear scan shows algorithmic reasoning. The problem mainly tests understanding of array traversal and efficient simulation patterns commonly used in competitive coding.
This approach involves simulating the game as described. In each round, compare the first two elements. The larger element stays at the front, while the smaller one is moved to the end. Track the number of consecutive wins for the current maximum element. The game terminates once an element wins k consecutive games, or we can conclude the maximum element will always win once it has dominated for enough rounds due to its magnitude relative to other elements.
The code starts by assuming the first element is the maximum. It then iterates through the rest of the elements. If the current max is greater than the current element, increase the win count; otherwise, update the current max and reset the win count. The loop runs until an element wins k consecutive times or until the end of the array, returning the max.
Time Complexity: O(n) where n is the length of the array.
Space Complexity: O(1) as we use only constant space.
This variant uses a concept very similar to the simulation approach but adds a breaking condition for improved efficiency. It exploits the game's characteristics by noting if the current maximum wins continuously even before counting to k, it will naturally reach the conclusion. This allows early termination if the maximum element becomes apparent quickly.
This C language solution enhances the standard simulation method with a conditional break. It re-evaluates upon encountering the largest possible number, indicated by iterating long enough to guarantee the outcome irrespective of k, hence optimizing rounds required for final decision.
Time Complexity: O(n), potentially terminating faster.
Space Complexity: O(1).
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 n-1 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:
| Approach | Complexity |
|---|---|
| Simulated Rounds Approach | Time Complexity: O(n) where n is the length of the array. |
| Break Condition Approach | Time Complexity: O(n), potentially terminating faster. |
| Quick Thinking | — |
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Simulated Rounds Approach | O(n * k) | O(n) | When implementing the game rules exactly or demonstrating a straightforward simulation. |
| Break Condition Approach | O(n) | O(1) | Best for interviews and large inputs since it avoids repeated rotations and stops early when a champion reaches k wins. |
Find the Winner of an Array Game | Simulation | Easy Explanation | GOOGLE | Leetcode - 1535 • codestorywithMIK • 4,569 views views
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