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Formatted question description: https://leetcode.ca/all/1646.html

 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

Algorithm

Just add up following problem desciption.

Code

  • 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);
        }
    };
    
  • func getMaximumGenerated(n int) (ans 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]
    		}
    		ans = max(ans, nums[i])
    	}
    	return
    }
    
    func max(a, b int) int {
    	if a > b {
    		return a
    	}
    	return b
    }
    
  • 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);
    }
    
    
  • 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)
    
    

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