Amortized analysis

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

Amortized analysis spreads occasional expensive operations over an entire sequence. It gives a deterministic sequence bound without random inputs. Aggregate, accounting, and potential arguments describe the resource budget.

Work through an example

A doubling array copies 1+2+4+… slots. Across n appends this sum is below 2n, so writes plus copies total O(n), despite some Θ(n) appends.

Time and space complexity

Geometric growth gives amortized O(1) append and O(n) storage.

Why it works

A resize creates enough spare slots to pay for copying before the next resize.

Common mistakes

Growing capacity by one causes quadratic total copies.
Shrinking and growing at one threshold can cause thrashing.

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