Counting operations

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

Define input size and the dominant operation first. Count how often it executes, then simplify growth. Sequential loops add their work; dependent nested loops require summing their lengths.

Work through an example

For i from 0 to n-1 and j from 0 to i-1, iterations total n(n-1)/2. At n=8 this is 28, not 64, but both expressions grow quadratically.

Time and space complexity

The triangular loop is Θ(n²) time and O(1) auxiliary state when counting directly.

Why it works

Every legal pair (j,i) with j<i is visited once.

Common mistakes

Two adjacent loops are not automatically quadratic.
A library call inside a loop may hide linear work.

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