Compute the Big O notation. Explain how you got the answer.
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.
Compute the Big O notation. Explain how you got the answer. on W NA 1 public...
b. what is the order (big -o) of this algorithm? 11. To answer this question, consider the n, consider the following algorithm: for (int i-0; i<ni i++) for (int j = 0; j <= i; j++) // three assignment statements in body of this inner loop a. (6 pts) Exactly how many assignments (in terms of n) are made in this algorithm?
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