If an operation is $$O(n!)$$, it means that the time complexity of the operation grows factorially with respect to the input size $$n$$. In other words, the number of operations required to complete the task increases extremely rapidly as $$n$$ increases.
Explanation
Factorial Growth: The factorial of $$n$$ (denoted as $$n!$$) is the product of all positive integers up to $$n$$. For example, $$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$. This means that for an input size of 5, the operation would require 120 steps.
Rapid Increase: Factorial growth is much faster than polynomial or exponential growth. For instance, while $$n^2$$ and $$2^n$$ grow quickly, $$n!$$ grows even faster. For example, $$10! = 3,628,800$$, which is significantly larger than $$10^2 = 100$$ or $$2^{10} = 1,024$$.
Practical Implications: Algorithms with $$O(n!)$$ complexity are generally impractical for large input sizes because the number of operations becomes infeasible very quickly. Such algorithms are often used in scenarios where the input size is small or where an exact solution is required despite the high computational cost.
Examples
Traveling Salesman Problem (TSP): A classic example of an $$O(n!)$$ algorithm is the brute-force solution to the Traveling Salesman Problem, where the algorithm tries every possible permutation of cities to find the shortest possible route. For $$n$$ cities, there are $$n!$$ permutations to check.
Permutations: Generating all permutations of a set of $$n$$ elements also has a time complexity of $$O(n!)$$. For example, generating all permutations of the set $${1, 2, 3}$$ results in $$3! = 6$$ permutations.
Formal Definition
Formally, a function $$f(n)$$ is $$O(n!)$$ if there exist constants $$c > 0$$ and $$n_0 \geq 0$$ such that for all $$n \geq n_0$$, $$f(n) \leq c \cdot n!$$. This means that beyond a certain point $...
