Best, average, and worst cases

What are time and space complexity?

Time complexity describes how the amount of work grows when the input gets larger. Space complexity describes how much memory an algorithm needs. They describe growth, rather than an exact number of seconds or bytes. Here, n usually means the number of input items.

Example: Reading one known array position takes O(1), or constant work. Visiting every item takes O(n), or work proportional to the number of items. Keeping a second copy of n items takes O(n) extra space.

Helpful before this chapter

If these ideas are new, begin with:

Programming foundations

Understand the idea

Costs vary with input arrangement. Best and worst cases consider extrema among inputs of size n. Average-case analysis requires a stated distribution; expected cost can also come from randomized algorithm choices.

Work through an example

Linear search takes one comparison at the first position, n at the last or when absent. Uniformly distributed successful positions require (n+1)/2 comparisons on average.

Time and space complexity

Best Θ(1), worst Θ(n); average depends on the probability model.

Why it works

Search returns after finding equality or exhausting the sequence.

Common mistakes

Do not assume actual input is uniformly random.
Amortized guarantees are not probability statements.

Chapter lab · Python implementation

Analyze growth before optimizing constants. The lab separates direct work, nested work, and repeated halving. Count iterations explicitly so asymptotic claims can be checked. Unless stated otherwise, n is input length, arithmetic fits a machine word, and auxiliary space excludes input but includes temporary collections.

This shared lab connects the chapter's topics. Follow the current lesson's worked example first, then trace the complete implementation. Additional algorithms described in topic exercises are variations you can implement independently.

def counts(n):
    if n < 0:
        raise ValueError('n must be nonnegative')
    linear = sum(1 for _ in range(n))
    pairs = sum(1 for i in range(n) for j in range(i))
    halvings, x = 0, n
    while x > 1:
        x //= 2
        halvings += 1
    return linear, pairs, halvings

def doubling_copies(n):
    size, capacity, copied = 0, 1, 0
    for _ in range(n):
        if size == capacity:
            copied += size
            capacity *= 2
        size += 1
    return copied

Run and test the lab

Copy the implementation and these assertions into a Python 3 file and run it. An assertion failure means the actual result differs from the expectation. Add the current topic's exercise and boundary tests to deepen your understanding.

assert counts(0) == (0,0,0)
assert counts(8) == (8,28,3)
assert counts(5) == (5,10,2)
assert doubling_copies(9) == 15
assert all(doubling_copies(n) < 2*n for n in range(1,100))

CHECK YOUR UNDERSTANDING