Solution Explanation

This problem asks us to generate an array nums based on specific rules and then find the maximum value within that array. The rules for generating the array are:

1. nums[0] = 0
2. nums[1] = 1
3. nums[2 * i] = nums[i] (even indices)
4. nums[2 * i + 1] = nums[i] + nums[i + 1] (odd indices)

The constraints specify that 0 <= n <= 100, meaning the array will have at most 101 elements.

Approach

The solution involves iteratively building the nums array according to the rules. We initialize the array with the first two elements (nums[0] = 0, nums[1] = 1). Then, we iterate from i = 2 to n. For each i, we check if it's even or odd:

• Even: nums[i] = nums[i/2] (rule 3). We use bitwise right shift (>> 1) for efficient integer division by 2.
• Odd: nums[i] = nums[i/2] + nums[i/2 + 1] (rule 4). Again, bitwise right shift is used.

After generating the array, we find the maximum element using built-in functions like max() in Python or Arrays.stream().max().getAsInt() in Java, max_element in C++, slices.Max in Go, and Math.max(...nums) in Typescript.

Time and Space Complexity Analysis

Time Complexity: O(n)

The algorithm iterates through the array once to generate it and then iterates (or uses a built-in function which has O(n) complexity) once to find the maximum element. Therefore, the overall time complexity is linear with respect to the input n.

Space Complexity: O(n)

The space complexity is also linear because we need to store the nums array, which has a size of n + 1.

Code Explanation (Python)

class Solution:
    def getMaximumGenerated(self, n: int) -> int:
        if n < 2:
            return n
        nums = [0] * (n + 1)
        nums[1] = 1
        for i in range(2, n + 1):
            nums[i] = nums[i >> 1] if i % 2 == 0 else nums[i >> 1] + nums[(i >> 1) + 1]
        return max(nums)

1. Base Case: If n is less than 2, we return n directly because the maximum element will be n itself.
2. Initialization: We create an array nums of size n + 1 and initialize nums[0] to 0 and nums[1] to 1.
3. Iteration: We iterate from i = 2 to n, applying the rules to populate the nums array using conditional logic and bitwise right shift for efficiency.
4. Maximum Element: Finally, the max() function efficiently finds the maximum value in the nums array.

The other language implementations follow the same approach, differing only in syntax and built-in functions used for array manipulation and finding the maximum element.