State The Big O Of The Following Fragment:For(I= 0; I < N; I++){ For(J = 1; J < N; J = J*2){ Cout << I << " "; }}
Question
State The Big O Of The Following Fragment:For(I= 0; I < N; I++){ For(J = 1; J < N; J = J*2){ Cout << I << " "; }}
Solution
The outer loop runs N times and the inner loop runs log(N) times because J doubles with each iteration. Therefore, the time complexity of the fragment is O(N log N).
Similar Questions
What is the time complexity of the following code fragment?for(i=1; i<n; i= i*2) a[i]=0;
What is the time complexity of the following code? for(int i = 0; i < n; i++) { for(int j = 0; j < i; j++) { System.out.println(i * j); } } a. O(n) b. O(2^n) c. O(n^2) d. O(n log n)
What is the time complexity of the following code :int i, j, k = 0;for (i = n/2; i <= n; i++) { for (j = 2; j <= n; j = j * 2) { k = k + n/2; }}
The time complexity of the following code snippet isc=0; while (n>=1) {for(i=1;i<=n;i++){ c++;}n=n/2;}
Using big-O notation in terms of its parameter n, what is the running time of the below method in its worst case? Give the tightest and simplest bound possible, and justify your answer.public static int mystery(int n) { int total = 0; for (int i = 1; i < n; i *= 2) { for (int j = i; j < 2 * i; j++) { total += j; } } return total;}
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.