Question

Compute the Big O notation. Explain how you got the answer.

on W NA 1 public String modify (String str) { if (str.length() <= 1) return ""; int half = str.length() / 2; modify(str.substring(half)); 5}

Answer 1

1. O(logN)(base 2) because everytime we are reducing the size of the problem by half i.e. with a factor of 2.

2. First loop runs from i = 0 to n, therefore, its complexity is O(N).

Second loop runs from j = 0 to 5, i.e for some constant time, its complexity is O(5).

The third loop again runs from 0 to N for k, complexity being O(N).

Since these are the nested for loops, we get the complexity of snippet to be O(5N^2), here 5 is a constant, ignore this, complexity is O(N^2).

3. First loop runs from n to 0, again it is O(N).

Inside the first loop, we are changing k as i*N, in the following patter, N^2, (N-1)N, (N-2)N, .... 1(N).

The inner loop runs from j = K but the condition being j < N, and we clearly from the above pattern for k see that it can never be less tha 0, hence this loop never executes.

The overall time complexity is thus O(N) because of the outside loop.