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 arraynums.

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

implSolution{pubfnget_maximum_generated(n:i32) -> i32{letmut nums = vec![0; n asusize + 1];for j in1..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()}}