LeetCode in Kotlin

1646. Get Maximum in Generated Array

Easy

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

class Solution {
    fun getMaximumGenerated(n: Int): Int {
        if (n == 0) {
            return 0
        }
        val nums = IntArray(n + 1)
        nums[0] = 0
        nums[1] = 1
        var max = 1
        for (i in 1..n / 2) {
            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
    }
}

This site is open source. Improve this page.