README.md

1646. Get Maximum in Generated Array

You are given an integer n. An 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 3.

Example 2:

Input: n = 2
Output: 1
Explanation: According to the given rules, the maximum between nums[0], nums[1], and nums[2] is 1.

Example 3:

Input: n = 3
Output: 2
Explanation: According to the given rules, the maximum between nums[0], nums[1], nums[2], and nums[3] is 2.

Constraints:

• 0 <= n <= 100

Solutions (Rust)

1. Dynamic Programming

impl Solution {
    pub fn get_maximum_generated(n: i32) -> i32 {
        let mut nums = vec![0; n as usize + 1];

        for j in 1..nums.len() {
            nums[j] = match j / 2 {
                0 => 1,
                i if j % 2 == 0 => nums[i],
                i => nums[i] + nums[i + 1],
            };
        }

        nums.into_iter().max().unwrap()
    }
}