algorithms/leetcode/1646-GetMaximumInGeneratedArray.go at master · freewu/algorithms · GitHub

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package main

// 1646. Get Maximum in Generated Array
// 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

import "fmt"

func getMaximumGenerated(n int) int {
    if n < 2 { return n }
    arr, res := make([]int, n + 1), 0
    arr[0], arr[1] = 0, 1
    for i := 2; i <= n; i++ {
        v := arr[i / 2]
        if i % 2 == 1 {
            v += arr[(i / 2) + 1]
        }
        arr[i] = v
        if v > res {
            res = v
        }
    }
    return res
}

func main() {
    // 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.
    fmt.Println(getMaximumGenerated(7)) // 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.
    fmt.Println(getMaximumGenerated(2)) // 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.
    fmt.Println(getMaximumGenerated(3)) // 2

    fmt.Println(getMaximumGenerated(1)) // 1
    fmt.Println(getMaximumGenerated(99)) // 21
    fmt.Println(getMaximumGenerated(100)) // 21
}