Blogs Hub

by Sumit Chourasia | Nov 08, 2020 | Category :coding | Tags : algorithm array data-structure easy leetcode

Get Maximum in Generated Array - Array - Easy - LeetCode

Get Maximum in Generated Array - Array - Easy - LeetCode

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)