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
public class Solution {
public int GetMaximumGenerated(int n) {
if(n<2){
return n;
}
var arr = new int[n+1];
arr[0]=0;
arr[1]=1;
int i = 1;
int max = int.MinValue;
while(2*i <= n || (2*i)+1 <= n){
arr[2*i]=arr[i];
max = Math.Max(max, arr[2*i]);
if((2*i)+1 <= n){
arr[(2*i)+1]=arr[i]+arr[i+1];
max = Math.Max(max, arr[(2*i)+1]);
}
i++;
}
return max;
}
}
Time Complexity: O(n)
Space Complexity: O(n)