carloscn
commented
11 months ago Analysis
pub fn min_max_game(nums: Vec<i32>) -> i32
{
let mut ret:i32 = 0;
if nums.len() == 0 {
return ret;
}
let mut array:Vec<i32> = nums.clone();
while array.len() != 1 {
let mut j:usize = 0;
let mut i:usize = 1;
let mut t_buf:Vec<i32> = vec![];
while i < array.len() {
let e:i32;
if (j & 0x1) == 0 {
e = array[i].min(array[i - 1]);
} else {
e = array[i].max(array[i - 1]);
}
t_buf.push(e);
j += 1;
i = 2 * i + 1;
}
array = t_buf;
}
ret = array[0];
return ret;
}
Description
You are given a 0-indexed integer array nums whose length is a power of 2.
Apply the following algorithm on nums:
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. 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]). 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]). Replace the array nums with newNums. 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.