The solution immediately simulates the array game by comparing the first two elements repeatedly, tracking consecutive wins. The larger element stays in front, and the smaller moves to the end. If k is greater than or equal to the array length, the overall maximum element is returned directly, ensuring correctness and efficiency in all edge cases.
Problem Statement
Given a distinct integer array arr and an integer k, play a game comparing the first two elements repeatedly. In each round, the larger of arr[0] and arr[1] remains at the front, while the smaller moves to the array's end. The game continues until an element wins k rounds consecutively.
Return the integer that wins the game. If k is greater than or equal to the array length, the winner is guaranteed to be the largest element, as it will eventually win k consecutive rounds regardless of initial order.
Examples
Example 1
Input: arr = [2,1,3,5,4,6,7], k = 2
Output: 5
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
3 will win the first 10 rounds consecutively.
Constraints
- 2 <= arr.length <= 105
- 1 <= arr[i] <= 106
- arr contains distinct integers.
- 1 <= k <= 109
Solution Approach
Simulate the Game Iteratively
Maintain a counter for consecutive wins and compare the first two elements of the array in each round. Move the smaller element to the end and increment the winner's count. Reset the count if the first element loses. Continue until the counter reaches k.
Handle Large k Efficiently
If k is equal to or larger than the array length, immediately return the maximum element. This avoids unnecessary simulation since the largest element will inevitably win k consecutive rounds.
Optimize for Space and Time
Use constant space by updating the array in-place and track only the current winner and consecutive wins. This maintains O(n) time complexity and O(1) space, avoiding extra data structures.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | O(n) |
| Space | O(1) |
Time complexity is O(n) because each element can be compared at most once before the maximum element secures k wins. Space complexity is O(1) since the array is updated in-place and only counters are maintained.
What Interviewers Usually Probe
- Will you simulate each round or find a shortcut based on k?
- How do you handle extremely large k values without full simulation?
- Can you optimize the solution to constant space without extra data structures?
Common Pitfalls or Variants
Common pitfalls
- Failing to reset the consecutive win count when the leading element loses.
- Not handling cases where k exceeds array length, leading to unnecessary iterations.
- Using extra space unnecessarily instead of tracking only the current winner and count.
Follow-up variants
- Find the first element to reach m consecutive wins in a modified array game.
- Determine the winner when the comparison rule is based on modulo values instead of direct magnitude.
- Extend the game to allow multiple elements to compete in each round and track the first to reach k wins.
How GhostInterview Helps
- Quickly simulates each round with precise counter tracking to identify the winner efficiently.
- Provides immediate handling for edge cases like k >= arr.length without wasting computation.
- Optimizes space and time by focusing on the winner element and consecutive win logic rather than full array rotations.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the winner in 'Find the Winner of an Array Game' if k is very large?
If k is equal to or exceeds the array length, the winner is the maximum element, as it will inevitably win k consecutive rounds.
Can I simulate the game using extra arrays or queues?
Yes, but it's unnecessary. Tracking only the current winner and consecutive wins achieves O(1) space, which is more efficient.
How does the array plus simulation pattern apply here?
The problem directly uses pairwise comparisons in an array to simulate the game's progress, embodying the array plus simulation pattern.
What if multiple elements tie in consecutive wins?
The rules guarantee distinct integers, so ties cannot occur in consecutive win counts, simplifying simulation logic.
Is there a shortcut instead of full simulation for small arrays?
Yes, for small arrays or when k >= arr.length, returning the maximum element immediately avoids full iteration.
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