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usaco-guide/content/3_Bronze/Time_Comp.mdx
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---
id: time-comp
title: "Time Complexity"
author: Darren Yao, Benjamin Qi
description: Measuring how long your algorithm takes to run in terms of the input size.
---
## Additional Resources
<resources>
<resource source="CPH" title="2 - Time Complexity" starred>good intro and examples</resource>
<resource source="PAPS" title="5 - Time Complexity">more in-depth</resource>
</resources>
# Time Complexity
In programming contests, your program needs to finish running within a certain timeframe in order to receive credit. For USACO, this limit is $4$ seconds for Java submissions. A conservative estimate for the number of operations the grading server can handle per second is $10^8$ (but could be closer to $5 \cdot 10^8$ given good constant factors).
(define time complexity?)
## Complexity Calculations
We want a method of how many operations it takes to run each algorithm, in terms of the input size $n$. Fortunately, this can be done relatively easily using [Big O Notation](https://en.wikipedia.org/wiki/Big_O_notation), which expresses worst-case time complexity as a function of $n$ as $n$ gets arbitrarily large. Complexity is an upper bound for the number of steps an algorithm requires as a function of the input size. In Big O notation, we denote the complexity of a function as $O(f(n))$, where $f(n)$ is a function without constant factors or lower-order terms. We'll see some examples of how this works, as follows.
(ben - formal definition of big O? constant factors don't **have** to be omitted)
The following code is $O(1)$, because it executes a constant number of operations.
```cpp
int a = 5;
int b = 7;
int c = 4;
int d = a + b + c + 153;
```
Input and output operations are also assumed to be $O(1)$.
In the following examples, we assume that the code inside the loops is $O(1)$.
The time complexity of loops is the number of iterations that the loop runs. For example, the following code examples are both $O(n)$.
```cpp
for(int i = 1; i <= n; i++){
// constant time code here
}
```
```cpp
int i = 0;
while(i < n){
// constant time node here
i++;
}
```
Because we ignore constant factors and lower order terms, the following examples are also $O(n)$:
```cpp
for(int i = 1; i <= 5*n + 17; i++){
// constant time code here
}
```
```cpp
for(int i = 1; i <= n + 457737; i++){
// constant time code here
}
```
We can find the time complexity of multiple loops by multiplying together the time complexities of each loop. This example is $O(nm)$, because the outer loop runs $O(n)$ iterations and the inner loop $O(m)$.
```cpp
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
// constant time code here
}
}
```
In this example, the outer loop runs $O(n)$ iterations, and the inner loop runs anywhere between $1$ and $n$ iterations (which is a maximum of $n$). Since Big O notation calculates worst-case time complexity, we must (?) take the factor of $n$ from the inner loop. Thus, this code is $O(n^2)$.
```cpp
for(int i = 1; i <= n; i++){
for(int j = i; j <= n; j++){
// constant time code here
}
}
```
If an algorithm contains multiple blocks, then its time complexity is the worst time complexity out of any block. For example, the following code is $O(n^2)$.
```cpp
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
// constant time code here
}
}
for(int i = 1; i <= n + 58834; i++){
// more constant time code here
}
```
The following code is $O(n^2 + nm)$, because it consists of two blocks of complexity $O(n^2)$ and $O(nm)$, and neither of them is a lower order function with respect to the other.
```cpp
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
// constant time code here
}
}
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
// more constant time code here
}
}
```
## Common Complexities and Constraints
Complexity factors that come from some common algorithms and data structures are as follows:
- Mathematical formulas that just calculate an answer: $O(1)$
- Unordered set/map: $O(1)$ per operation
- Binary search: $O(\log n)$
- Ordered set/map or priority queue: $O(\log n)$ per operation
- Prime factorization of an integer, or checking primality or compositeness of an integer naively: $O(\sqrt{n})$
- Reading in $n$ items of input: $O(n)$
- Iterating through an array or a list of $n$ elements: $O(n)$
- Sorting: usually $O(n \log n)$ for default sorting algorithms (mergesort, for example `Collections.sort` or `Arrays.sort` on objects)
- Java Quicksort `Arrays.sort` function on primitives: $O(n^2)$
- See "Introduction to Sorting" for details.
- Iterating through all subsets of size $k$ of the input elements: $O(n^k)$. For example, iterating through all triplets is $O(n^3)$.
- Iterating through all subsets: $O(2^n)$
- Iterating through all permutations: $O(n!)$
Here are conservative upper bounds on the value of $n$ for each time complexity. You can probably get away with more than this, but this should allow you to quickly check whether an algorithm is viable.
<center>
| $n$ | Possible complexities |
| --------------------- | ----------------------------------- |
| $n \le 10$ | $O(n!)$, $O(n^7)$, $O(n^6)$ |
| $n \le 20$ | $O(2^n \cdot n)$, $O(n^5)$ |
| $n \le 80$ | $O(n^4)$ |
| $n \le 400$ | $O(n^3)$ |
| $n \le 7500$ | $O(n^2)$ |
| $n \le 7 \cdot 10^4$ | $O(n \sqrt n)$ |
| $n \le 5 \cdot 10^5$ | $O(n \log n)$ |
| $n \le 5 \cdot 10^6$ | $O(n)$ |
| $n \le 10^{18}$ | $O(\log^2 n)$, $O(\log n)$, $O(1)$ |
</center>