Sieve of Eratosthenes
The Sieve of Eratosthenes finds all primes up to n by marking multiples of each discovered prime as composite.
Start with this mental model
The Sieve of Eratosthenes finds all primes up to n by marking multiples of each discovered prime as composite.
How it works, without unnecessary jargon
Start with all candidates marked prime.
For each prime p up to √n, mark multiples starting at p²; smaller multiples were already handled.
Read the algorithm as a sequence of decisions
State the mathematical identity/property.
Choose numeric representation and constraints.
Translate the property into an iterative/recursive computation and analyze precision.
Read a tiny input by hand and write down the state after every important operation. For Sieve of Eratosthenes, the goal is to make the invariant visible rather than to memorize code.
function primesUpTo(n) {
const prime = Array(n + 1).fill(true);
prime[0] = prime[1] = false;
for (let p = 2; p * p <= n; p++) {
if (!prime[p]) continue;
for (let x = p * p; x <= n; x += p) prime[x] = false;
}
return prime.flatMap((isPrime, i) => isPrime ? [i] : []);
}Do not memorize the syntax. Follow the state that changes on each line and connect it to the invariant described above.
O(n log log n)
Do not quote this complexity mechanically. Ask which operation or input dimension it describes, then derive the cost from the loops, recursion depth, data-structure operations and any preprocessing assumptions.
Good fit
- Use the direct mathematical property when it removes unnecessary brute force.
What usually goes wrong
- Ignoring floating-point precision or integer range.
- Using a formula outside its valid assumptions.
A professional answer shape
Mention O(n log log n) time and O(n) space, and explain why marking can start from p².
What problem does Sieve of Eratosthenes solve?
The Sieve of Eratosthenes finds all primes up to n by marking multiples of each discovered prime as composite.
What should I mention about complexity?
Its typical cost here is O(n log log n), but a complete answer should connect that cost to the representation, input constraints and the dominant operations.
What is the most common mistake?
Ignoring floating-point precision or integer range.