[LeetCode] 1646. Get Maximum in Generated Array

- Problem

1646. Get Maximum in Generated Array
You are given an integer n. A 0-indexed integer array nums of length n + 1 is generated in the following way:

• nums[0] = 0
• nums[1] = 1
• nums[2 * i] = nums[i] when 2 <= 2 * i <= n
• nums[2 * i + 1] = nums[i] + nums[i + 1] when 2 <= 2 * i + 1 <= n

Return the maximum integer in the array nums.

Example 1:

Input: n = 7
Output: 3
Explanation: According to the given rules:
  nums[0] = 0
  nums[1] = 1
  nums[(1 * 2) = 2] = nums[1] = 1
  nums[(1 * 2) + 1 = 3] = nums[1] + nums[2] = 1 + 1 = 2
  nums[(2 * 2) = 4] = nums[2] = 1
  nums[(2 * 2) + 1 = 5] = nums[2] + nums[3] = 1 + 2 = 3
  nums[(3 * 2) = 6] = nums[3] = 2
  nums[(3 * 2) + 1 = 7] = nums[3] + nums[4] = 2 + 1 = 3
Hence, nums = [0,1,1,2,1,3,2,3], and the maximum is max(0,1,1,2,1,3,2,3) = 3.

Example 2:

Input: n = 2
Output: 1
Explanation: According to the given rules, nums = [0,1,1]. The maximum is max(0,1,1) = 1.

Example 3:

Input: n = 3
Output: 2
Explanation: According to the given rules, nums = [0,1,1,2]. The maximum is max(0,1,1,2) = 2.

Constraints:

• 0 <= n <= 100

- 내 풀이

class Solution:
    def getMaximumGenerated(self, n: int) -> int:
        if n == 0:
            return 0
        elif n == 1:
            return 1
        
        dp = [0] * (n + 1)
        dp[1] = 1
        i = 2
        
        while i <= n:
            if i % 2 == 0:
                dp[i] = dp[i//2]
            else:
                dp[i] = dp[i//2] + dp[(i//2)+1]
            i += 1
                
        return max(dp)

- 결과