Please help me understand, what is the Order of Growth?
What is the Big O notation for this program
What i did so far?
I wrote this problem.
What is the Order of Growth according to me?
There are two access to array
- once during this check, c[jumpLevel2]==0 , Worst case, n time access
- Another during this, c[jumpLevel1]==0 , Worst case, n time access
So, 2N access
Therefore O(n)
Please correct me if I am wrong.
import java.io.;
import java.math.;
import java.security.;
import java.text.;
import java.util.;
import java.util.concurrent.;
import java.util.regex.*;
public class Solution {
// Complete the jumpingOnClouds function below. static int jumpingOnClouds(int[] c) { int i=0; int path = 0; int clouds = c.length; while (i<clouds){ int jumpLevel2 = i+2; int jumpLevel1 = i+1; if ((jumpLevel2<clouds) && c[jumpLevel2]==0){ path+=1; i=jumpLevel2; } else if ((jumpLevel1<clouds) && c[jumpLevel1]==0){ path+=1; i=jumpLevel1; } else { return path; } } return path; } private static final Scanner scanner = new Scanner(System.in); public static void main(String[] args) throws IOException { BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); int n = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); int[] c = new int[n]; String[] cItems = scanner.nextLine().split(" "); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); for (int i = 0; i < n; i++) { int cItem = Integer.parseInt(cItems[i]); c[i] = cItem; } int result = jumpingOnClouds(c); bufferedWriter.write(String.valueOf(result)); bufferedWriter.newLine(); bufferedWriter.close(); scanner.close(); } }
Note: Code from Hackerrank, for my personal practise.
I am using dev.to platform to help understand my understanding. haha
Top comments (1)
Hi dev.toer's
I want suggestion on my conclusion about the Order of growth as 2N which becomes O(N)
thanks in advance