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 &lt;= 2 * i &lt;= n
	nums[2 * i + 1] = nums[i] + nums[i + 1] when 2 &lt;= 2 * i + 1 &lt;= n


Return the maximum integer in the array nums​​​.

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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.


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Constraints:


	0 &lt;= n &lt;= 100


class Solution:
    def getMaximumGenerated(self, n):
        nums = [0]*(n+2)
        nums[1] = 1
        for i in range(2, n+1):
            nums[i] = nums[i//2] + nums[(i//2)+1] * (i%2)
    
        return max(nums[:n+1])
class Solution {
    public int getMaximumGenerated(int n) {
        if(n==0 || n==1) return n;

        int nums[]=new int [n+1];

        nums[0]=0;
        nums[1]=1;
        int max=Integer.MIN_VALUE;

        for(int i=2;i<=n;i++){
            if(i%2==0){
                nums[i]=nums[i/2];
            }
            else{
                nums[i]=nums[i/2]+nums[i/2 + 1];
            }
            max=Math.max(max,nums[i]);
        }
        return max;
    }
}
class Solution {
public:
    int getMaximumGenerated(int n) {
        // base cases
        if (n < 2) return n;
        // support variables
        int arr[n + 1], m;
        arr[0] = 0, arr[1] = 1;
        // building arr
        for (int i = 2; i <= n; i++) {
            if (i % 2) arr[i] = arr[i / 2] + arr[i / 2 + 1];
            else arr[i] = arr[i / 2];
            // updating m
            m = max(arr[i], m);
        }
        return m;
    }
};
var getMaximumGenerated = function(n) {
    if (n === 0) return 0;
    if (n === 1) return 1;
    let arr = [0, 1];
    let max = 0;
    for (let i = 0; i < n; i++) {
        if (2 <= 2 * i && 2 * i <= n) {
            arr[2 * i] = arr[i]
            if (arr[i] > max) max = arr[i];
        }
        if (2 <= 2 * i && 2 * i + 1 <= n) {
            arr[2 * i + 1] = arr[i] + arr[i + 1]
            if (arr[i] + arr[i + 1] > max) max = arr[i] + arr[i + 1];
        };
    }
    return max;
};
Get Maximum in Generated Array