Question

Big O Exact Execution speed: Problem 2: for (int i 0 i< n; it+) Some Statement for (int j 0;j<n;j++) SomeStatement public cla

Determine the Big 0 provide the order (Big O) of the execution speed also determine the exact execution speed.

public class CountIt {
public long SnippetNestedLoop(long n) {

long i, j, x = 0; i=0; x++; while(i<n){ x++;

//i<n
// SomeStatement // j = 0;
// j < n
// SomeStatement // j++;

// Can you explain why is this here? // i++;

// Can you explain why is this here? Ans: i < n

} }

}
x++; return x;

x++;
j = 0; x++; while (j < n) { x++;

x++; j++; x++;

}
x++;
i++; x++;

}
public static void main(String[] args) {

CountIt r = new CountIt();
Long t = System.currentTimeMillis();
System.out.println("x = " + r.SnippetNestedLoop(100)); System.out.println ("Time:" + (System.currentTimeMillis() - t));

0 0
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Answer #1

Problem 2)

The outer loop for i runs n times and the inner loop for j runs n times for each i.

Hence, the total time complexity of this code snippet is O(n2).

Similarly, the time taken by the class CountIt is given as:

T(n) = n2 + C where C is a positive constant

Hence, T(n) = O(n2)

The execution speed of this code is given as:

x = 30502

Time:1

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