Question

2293. Min Max Game

中文文档

Description

You are given a 0-indexed integer array nums whose length is a power of 2.

Apply the following algorithm on nums:

1. Let n be the length of nums. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n / 2.
2. For every even index i where 0 <= i < n / 2, assign the value of newNums[i] as min(nums[2 * i], nums[2 * i + 1]).
3. For every odd index i where 0 <= i < n / 2, assign the value of newNums[i] as max(nums[2 * i], nums[2 * i + 1]).
4. Replace the array nums with newNums.
5. Repeat the entire process starting from step 1.

Return _the last number that remains in _nums_ after applying the algorithm._

 

Example 1:

Input: nums = [1,3,5,2,4,8,2,2]
Output: 1
Explanation: The following arrays are the results of applying the algorithm repeatedly.
First: nums = [1,5,4,2]
Second: nums = [1,4]
Third: nums = [1]
1 is the last remaining number, so we return 1.

Example 2:

Input: nums = [3]
Output: 3
Explanation: 3 is already the last remaining number, so we return 3.

 

Constraints:

• 1 <= nums.length <= 1024
• 1 <= nums[i] <= 109
• nums.length is a power of 2.

Solutions

Solution 1

class Solution:
  def minMaxGame(self, nums: List[int]) -> int:
    n = len(nums)
    while n > 1:
      n >>= 1
      for i in range(n):
        a, b = nums[i << 1], nums[i << 1 | 1]
        nums[i] = min(a, b) if i % 2 == 0 else max(a, b)
    return nums[0]

class Solution {
  public int minMaxGame(int[] nums) {
    for (int n = nums.length; n > 1;) {
      n >>= 1;
      for (int i = 0; i < n; ++i) {
        int a = nums[i << 1], b = nums[i << 1 | 1];
        nums[i] = i % 2 == 0 ? Math.min(a, b) : Math.max(a, b);
      }
    }
    return nums[0];
  }
}

class Solution {
public:
  int minMaxGame(vector<int>& nums) {
    for (int n = nums.size(); n > 1;) {
      n >>= 1;
      for (int i = 0; i < n; ++i) {
        int a = nums[i << 1], b = nums[i << 1 | 1];
        nums[i] = i % 2 == 0 ? min(a, b) : max(a, b);
      }
    }
    return nums[0];
  }
};

func minMaxGame(nums []int) int {
  for n := len(nums); n > 1; {
    n >>= 1
    for i := 0; i < n; i++ {
      a, b := nums[i<<1], nums[i<<1|1]
      if i%2 == 0 {
        nums[i] = min(a, b)
      } else {
        nums[i] = max(a, b)
      }
    }
  }
  return nums[0]
}

function minMaxGame(nums: number[]): number {
  for (let n = nums.length; n > 1; ) {
    n >>= 1;
    for (let i = 0; i < n; ++i) {
      const a = nums[i << 1];
      const b = nums[(i << 1) | 1];
      nums[i] = i % 2 == 0 ? Math.min(a, b) : Math.max(a, b);
    }
  }
  return nums[0];
}

impl Solution {
  pub fn min_max_game(mut nums: Vec<i32>) -> i32 {
    let mut n = nums.len();
    while n != 1 {
      n >>= 1;
      for i in 0..n {
        nums[i] = (if (i & 1) == 1 { i32::max } else { i32::min })(
          nums[i << 1],
          nums[(i << 1) | 1]
        );
      }
    }
    nums[0]
  }
}

#define min(a, b) (((a) < (b)) ? (a) : (b))
#define max(a, b) (((a) > (b)) ? (a) : (b))

int minMaxGame(int* nums, int numsSize) {
  while (numsSize != 1) {
    numsSize >>= 1;
    for (int i = 0; i < numsSize; i++) {
      int a = nums[i << 1];
      int b = nums[i << 1 | 1];
      nums[i] = i & 1 ? max(a, b) : min(a, b);
    }
  }
  return nums[0];
}