1646. Get Maximum in Generated Array

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

Solution (Java)

class Solution {
    public int getMaximumGenerated(int n) {
        if (n == 0) {
            return 0;
        }
        int[] nums = new int[n + 1];
        nums[0] = 0;
        nums[1] = 1;
        int max = 1;
        for (int i = 1; i <= n / 2; i++) {
            nums[(i * 2)] = nums[i];
            max = Math.max(max, nums[i]);
            if ((i * 2) + 1 <= n) {
                nums[(i * 2) + 1] = nums[i] + nums[i + 1];
                max = Math.max(max, nums[(i * 2) + 1]);
            }
        }
        return max;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).