159 lines
7.1 KiB
Text
159 lines
7.1 KiB
Text
---
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id: time-comp
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title: "Time Complexity"
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author: Darren Yao, Benjamin Qi
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description: Measuring how long your algorithm takes to run in terms of the input size.
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---
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## Additional Resources
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<resources>
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<resource source="CPH" title="2 - Time Complexity" starred>Intro and examples</resource>
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<resource source="PAPS" title="5 - Time Complexity" starred>More in-depth. In particular, 5.2 gives a formal definition of Big O.</resource>
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</resources>
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# Time Complexity
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In programming contests, your program needs to finish running within a certain timeframe in order to receive credit. For USACO, this limit is $2$ seconds for C++ submissions, and $4$ seconds for Java/Python submissions. A conservative estimate for the number of <TextTooltip content={`You can think of an "operation" as one instruction for the computer. For example, multiplying two numbers together, reading in a number from the input, or outputting "Hello World" would all be considered "operations." `}>operations</TextTooltip> the grading server can handle per second is $10^8$, but it could be closer to $5 \cdot 10^8$ given good constant factors<Asterisk>If you don't know what constant factors are, don't worry -- we'll explain them below.</Asterisk>.
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## Complexity Calculations
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We want a method to calculate 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 constant factors and lower-order terms are generally omitted from $f(n)$. We'll see some examples of how this works, as follows.
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The following code is $O(1)$, because it executes a constant number of operations.
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```cpp
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int a = 5;
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int b = 7;
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int c = 4;
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int d = a + b + c + 153;
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```
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Input and output operations are also assumed to be $O(1)$. In the following examples, we assume that the code inside the loops is $O(1)$.
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The time complexity of loops is the number of iterations that the loop runs. For example, the following code examples are both $O(n)$.
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```cpp
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for(int i = 1; i <= n; i++){
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// constant time code here
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}
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```
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<div />
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```cpp
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int i = 0;
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while(i < n){
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// constant time node here
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i++;
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}
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```
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Because we ignore constant factors and lower order terms, the following examples are also $O(n)$:
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```cpp
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for(int i = 1; i <= 5*n + 17; i++){
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// constant time code here
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}
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```
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<div />
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```cpp
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for(int i = 1; i <= n + 457737; i++){
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// constant time code here
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}
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```
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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 $O(m)$.
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```cpp
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for(int i = 1; i <= n; i++){
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for(int j = 1; j <= m; j++){
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// constant time code here
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}
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}
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```
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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 treat the inner loop as a factor of $n$.<Asterisk>We can also do some math to calculate exactly how many times the code runs: 1+2+...+n = n*(n-1)/2 = (n^2 - n)/2 = O(n^2)</Asterisk> Thus, this code is $O(n^2)$.
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```cpp
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for(int i = 1; i <= n; i++){
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for(int j = i; j <= n; j++){
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// constant time code here
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}
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}
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```
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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 $O(n^2)$.
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```cpp
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for(int i = 1; i <= n; i++){
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for(int j = 1; j <= n; j++){
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// constant time code here
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}
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}
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for(int i = 1; i <= n + 58834; i++){
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// more constant time code here
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}
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```
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The following code is $O(n^2 + m)$, because it consists of two blocks of complexity $O(n^2)$ and $O(m)$, and neither of them is a lower order function with respect to the other.
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```cpp
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for(int i = 1; i <= n; i++){
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for(int j = 1; j <= n; j++){
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// constant time code here
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}
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}
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for(int j = 1; j <= m; j++){
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// more constant time code here
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}
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```
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## Constant Factor
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The "Constant Factor" of an algorithm refers to the coefficient of the complexity of an algorithm. If an algorithm runs in $O(kn)$ time, where $k$ is a constant and $n$ is the input size, then the "constant factor" would be $k$.
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Normally when using the big-O notation, we ignore the constant factor: $O(3n) = O(n)$. This is fine most of the time, but sometimes we have an algorithm that just barely gets TLE, perhaps by just a few hundred milliseconds. When this happens, it is worth optimizing the constant factor of our algorithm. For example, if our code currently runs in $O(n^2)$ time, perhaps we can modify our code to make it run in $O(n^2/32)$ by using a bitset. (Of course, with big-O notation, $O(n^2) = O(n^2/32)$.)
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For now, don't worry about how to optimize constant factors -- just be aware of them.
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## Common Complexities and Constraints
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Complexity factors that come from some common algorithms and data structures are as follows:
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- Mathematical formulas that just calculate an answer: $O(1)$
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- Unordered set/map: $O(1)$ per operation
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- Binary search: $O(\log n)$
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- Ordered set/map or priority queue: $O(\log n)$ per operation
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- Prime factorization of an integer, or checking primality or compositeness of an integer naively: $O(\sqrt{n})$
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- Reading in $n$ items of input: $O(n)$
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- Iterating through an array or a list of $n$ elements: $O(n)$
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- Sorting: usually $O(n \log n)$ for default sorting algorithms (mergesort, `Collections.sort`, `Arrays.sort`)
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- Java Quicksort `Arrays.sort` function on primitives: $O(n^2)$
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- See "Introduction to Data Structures" for details.
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- Iterating through all subsets of size $k$ of the input elements: $O(n^k)$. For example, iterating through all triplets is $O(n^3)$.
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- Iterating through all subsets: $O(2^n)$
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- Iterating through all permutations: $O(n!)$
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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.
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<center>
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| $n$ | Possible complexities |
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| --------------------- | ----------------------------------- |
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| $n \le 10$ | $O(n!)$, $O(n^7)$, $O(n^6)$ |
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| $n \le 20$ | $O(2^n \cdot n)$, $O(n^5)$ |
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| $n \le 80$ | $O(n^4)$ |
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| $n \le 400$ | $O(n^3)$ |
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| $n \le 7500$ | $O(n^2)$ |
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| $n \le 7 \cdot 10^4$ | $O(n \sqrt n)$ |
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| $n \le 5 \cdot 10^5$ | $O(n \log n)$ |
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| $n \le 5 \cdot 10^6$ | $O(n)$ |
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| $n \le 10^{18}$ | $O(\log^2 n)$, $O(\log n)$, $O(1)$ |
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</center>
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