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Weekly Contest 200 Q2
Array
Simulation

中文文档

Description

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 <= 105
  • 1 <= arr[i] <= 106
  • arr contains distinct integers.
  • 1 <= k <= 109

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 $n-1$ 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:

Python3

class Solution:
    def getWinner(self, arr: List[int], k: int) -> int:
        mx = arr[0]
        cnt = 0
        for x in arr[1:]:
            if mx < x:
                mx = x
                cnt = 1
            else:
                cnt += 1
            if cnt == k:
                break
        return mx

Java

class Solution {
    public int getWinner(int[] arr, int k) {
        int mx = arr[0];
        for (int i = 1, cnt = 0; i < arr.length; ++i) {
            if (mx < arr[i]) {
                mx = arr[i];
                cnt = 1;
            } else {
                ++cnt;
            }
            if (cnt == k) {
                break;
            }
        }
        return mx;
    }
}

C++

class Solution {
public:
    int getWinner(vector<int>& arr, int k) {
        int mx = arr[0];
        for (int i = 1, cnt = 0; i < arr.size(); ++i) {
            if (mx < arr[i]) {
                mx = arr[i];
                cnt = 1;
            } else {
                ++cnt;
            }
            if (cnt == k) {
                break;
            }
        }
        return mx;
    }
};

Go

func getWinner(arr []int, k int) int {
	mx, cnt := arr[0], 0
	for _, x := range arr[1:] {
		if mx < x {
			mx = x
			cnt = 1
		} else {
			cnt++
		}
		if cnt == k {
			break
		}
	}
	return mx
}

TypeScript

function getWinner(arr: number[], k: number): number {
    let mx = arr[0];
    let cnt = 0;
    for (const x of arr.slice(1)) {
        if (mx < x) {
            mx = x;
            cnt = 1;
        } else {
            ++cnt;
        }
        if (cnt === k) {
            break;
        }
    }
    return mx;
}

C#

public class Solution {
    public int GetWinner(int[] arr, int k) {
        int maxElement = arr[0], count = 0;
        for (int i = 1; i < arr.Length; i++) {
            if (maxElement < arr[i]) {
                maxElement = arr[i];
                count = 1;
            } else {
                count++;
            }
            if (count == k) {
                break;
            }
        }
        return maxElement;
    }
}