Algorithm Analysis

Big-Oh Notation

$$f(n) \leq cg(n), \quad for \; n\leq n_0$$

It is called as "f(n) is big-Oh of g(n)".

For example

• $8n+5$ is $O(n)$

• $5n^4+3n^3+2n^2+4n+1$ is $O(n^4)$

• $a_0+a_1n+\cdots+a_dn^d$ is $O(n^d)$

• $2n+100logn$ is $O(n)$

def find_max(data):
    # Return the maximum element from a nonempty Python list
    biggest = data[0] # The initial value to beat
    for val in data: # For each value:
        if val > biggest: # if it is greater than the best so far,
            biggest = val # we have found a new best (so far)
    return biggest # When loop ends, biggest is the max
    

• initialization: O(1)

• loop: O(n)

• return: O(1)

To sum up, this algorithm has O(n) time complexity.

def prefix_average1(S):
    # Return list such tath, for all j, A[j] equals average of S[0], ..., S[j]
    n = len(S)
    A = [0]*n     # Create new list of n zeros
    for j in range(n):
        total = 0    # begin computing S[0]+...+S[j]
        for i in range(j+1):
            total += S[i]
        A[j] = total / (j+1)    # record the average
    return A

The running time of prefix_average1 is $O(n^2)$

def prefix_average2(S):
    n = len(S)
    A = [0]*n
    for j in range(n):
        A[j] = sum(S[0:j])/(j+1)
    return A

This big-Oh notation is used widely to characterize running times and space bounds in terms of some parameter n. (prefix_average2 is also $O(n^2)$)

def prefix_average3(S):
    n = len(S)
    A = [0] * n
    total = 0
    for j in range(n):
        total += S[j]
        A[j] = total / (j+1)
    return A

The above expression only has $O(n)$time complexity.

Time complexity in Python

• len(data): O(1)

• data[j]: O(1)

Python's lists are implemented as array-based sequences.