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1646. Get Maximum in Generated Array

Description

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

Solutions

  • class Solution {
        public int getMaximumGenerated(int n) {
            if (n < 2) {
                return n;
            }
            int[] nums = new int[n + 1];
            nums[1] = 1;
            for (int i = 2; i <= n; ++i) {
                nums[i] = i % 2 == 0 ? nums[i >> 1] : nums[i >> 1] + nums[(i >> 1) + 1];
            }
            return Arrays.stream(nums).max().getAsInt();
        }
    }
    
  • class Solution {
    public:
        int getMaximumGenerated(int n) {
            if (n < 2) {
                return n;
            }
            int nums[n + 1];
            nums[0] = 0;
            nums[1] = 1;
            for (int i = 2; i <= n; ++i) {
                nums[i] = i % 2 == 0 ? nums[i >> 1] : nums[i >> 1] + nums[(i >> 1) + 1];
            }
            return *max_element(nums, nums + n + 1);
        }
    };
    
  • class Solution:
        def getMaximumGenerated(self, n: int) -> int:
            if n < 2:
                return n
            nums = [0] * (n + 1)
            nums[1] = 1
            for i in range(2, n + 1):
                nums[i] = nums[i >> 1] if i % 2 == 0 else nums[i >> 1] + nums[(i >> 1) + 1]
            return max(nums)
    
    
  • func getMaximumGenerated(n int) int {
    	if n < 2 {
    		return n
    	}
    	nums := make([]int, n+1)
    	nums[1] = 1
    	for i := 2; i <= n; i++ {
    		if i%2 == 0 {
    			nums[i] = nums[i/2]
    		} else {
    			nums[i] = nums[i/2] + nums[i/2+1]
    		}
    	}
    	return slices.Max(nums)
    }
    
  • function getMaximumGenerated(n: number): number {
        if (n === 0) {
            return 0;
        }
        const nums: number[] = new Array(n + 1).fill(0);
        nums[1] = 1;
        for (let i = 2; i < n + 1; ++i) {
            nums[i] = i % 2 === 0 ? nums[i >> 1] : nums[i >> 1] + nums[(i >> 1) + 1];
        }
        return Math.max(...nums);
    }
    
    

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