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  • + 0 comments

    The Utopian Tree looks like such a unique and captivating piece! It's amazing how products like this can really stand out when paired with the right packaging and e-interior design services. Custom Mylar Bags would definitely elevate its presentation, offering durability and a sleek, modern feel. Desirepackaging always knows how to add that extra touch to make everything look high-end. Whether it's for retail or a special gift, thoughtful packaging really makes all the difference!

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    int utopianTree(int n) {
        if (n % 2 == 0) {
            return pow(2, n/2 + 1) - 1;
        } else {
            return pow(2, (n+3)/2) - 2;
        }
    }
    
  • + 0 comments

    //JAVA15: import java.io.*;

    class Result {

    public static int utopianTree(int n) {
        int height = 1;
        for (int i = 1; i <= n; i++) {
            if (i % 2 == 1) {
                height *= 2;
            } else {
                height += 1;
            }
        }
        return height;
    }
    

    }

    public class Solution { public static void main(String[] args) throws IOException { BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));

        // Use this for local testing:
        // BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(System.out));
    
        // Use this for HackerRank submission:
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
    
        int t = Integer.parseInt(bufferedReader.readLine().trim());
    
        for (int tItr = 0; tItr < t; tItr++) {
            int n = Integer.parseInt(bufferedReader.readLine().trim());
            int result = Result.utopianTree(n);
            bufferedWriter.write(String.valueOf(result));
            bufferedWriter.newLine();
        }
    
        bufferedReader.close();
        bufferedWriter.close();
    }
    

    }

  • + 0 comments

    My Python Solution:

    def utopianTree(n):
        # Write your code here
        if n == 0:
            return 1
    
        doubleCycle = n if n % 2 else n - 1
        treeLength = sum(2** (i+1) for i in range((doubleCycle//2)+1))
        
        return treeLength if n % 2 else treeLength + 1
    
  • + 0 comments

    C# Easy to Understand Solution

        public static int utopianTree(int n)
        {
            if (n == 0) return 1;
            int start = 1;
            for (int i = 1; i <= n; i++) {
                if (i % 2 == 0) start += 1;
                else start *= 2;
            }
            return start;
        }
    

    First of all, since the tree sapling starts at 1 height, then if "n" is 0 (no growth), then its height stays at 1.

    Then since it's always starts at Spring, then Summer, then Spring again, etc. We can think of it like an index. You might notice that each Spring happens at odd index, while each Summer happens at even index. This means you can just check if the index is odd or even. If it's an odd index (Spring), multiply the height by 2. If it's an even index (Summer), increment the height by 1.