1646 Get Maximum in Generated Array - Easy

Problem:

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

Problem Analysis:

Solutions:

Straightforward solution based on requirements

class Solution(object):
    def getMaximumGenerated(self, n):
        """
        :type n: int
        :rtype: int
        """
        tmp = []

        for i in range(n+1):
            if i == 0:
                tmp.append(0)
            elif i == 1:
                tmp.append(1)
            elif i%2 == 0:
                tmp.append(tmp[i//2])
            else: 
                tmp.append(tmp[i//2] + tmp[i//2+1])


        return max(tmp)

class Solution(object):
    def getMaximumGenerated(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n == 0 or n == 1: 
            return n

        max_val = 0
        #initialize array
        tmp = [0] * (n + 1)

        tmp[1] = 1            
        for i in range(2,n+1):  
            if i%2 == 0:
                tmp[i] = (tmp[i//2])
            else: 
                tmp[i]= (tmp[i//2] + tmp[i//2+1])
            max_val = max(max_val, tmp[i])
        return max_val

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