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