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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

• Java
• C++
• Go
• TypeScript
• Python

• 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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