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2293. Min Max Game

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

  • 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]; } }; 
  • 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] 
  • 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] } }

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