2.4 Space Complexity¶

"Space complexity" is used to measure the growth trend of the memory space occupied by an algorithm as the amount of data increases. This concept is very similar to time complexity, except that "running time" is replaced with "occupied memory space".

The memory space used by an algorithm during its execution mainly includes the following types.

Input Space: Used to store the input data of the algorithm.
Temporary Space: Used to store variables, objects, function contexts, and other data during the algorithm's execution.
Output Space: Used to store the output data of the algorithm.

Generally, the scope of space complexity statistics includes both "Temporary Space" and "Output Space".

Temporary space can be further divided into three parts.

Temporary Data: Used to save various constants, variables, objects, etc., during the algorithm's execution.
Stack Frame Space: Used to save the context data of the called function. The system creates a stack frame at the top of the stack each time a function is called, and the stack frame space is released after the function returns.
Instruction Space: Used to store compiled program instructions, which are usually negligible in actual statistics.

When analyzing the space complexity of a program, we typically count the Temporary Data, Stack Frame Space, and Output Data, as shown in the Figure 2-15 .

Figure 2-15 Space Types Used in Algorithms

The relevant code is as follows:

PythonC++JavaC#GoSwiftJSTSDartRustCZig

class Node:
    """Classes""""
    def __init__(self, x: int):
        self.val: int = x               # node value
        self.next: Node | None = None   # reference to the next node

def function() -> int:
    """"Functions"""""
    # Perform certain operations...
    return 0

def algorithm(n) -> int:    # input data
    A = 0                   # temporary data (constant, usually in uppercase)
    b = 0                   # temporary data (variable)
    node = Node(0)          # temporary data (object)
    c = function()          # Stack frame space (call function)
    return A + b + c        # output data

/* Structures */
struct Node {
    int val;
    Node *next;
    Node(int x) : val(x), next(nullptr) {}
};

/* Functions */
int func() {
    // Perform certain operations...
    return 0;
}

int algorithm(int n) {          // input data
    const int a = 0;            // temporary data (constant)
    int b = 0;                  // temporary data (variable)
    Node* node = new Node(0);   // temporary data (object)
    int c = func();             // stack frame space (call function)
    return a + b + c;           // output data
}

/* Classes */
class Node {
    int val;
    Node next;
    Node(int x) { val = x; }
}

/* Functions */
int function() {
    // Perform certain operations...
    return 0;
}

int algorithm(int n) {          // input data
    final int a = 0;            // temporary data (constant)
    int b = 0;                  // temporary data (variable)
    Node node = new Node(0);    // temporary data (object)
    int c = function();         // stack frame space (call function)
    return a + b + c;           // output data
}

/* Classes */
class Node {
    int val;
    Node next;
    Node(int x) { val = x; }
}

/* Functions */
int Function() {
    // Perform certain operations...
    return 0;
}

int Algorithm(int n) {  // input data
    const int a = 0;    // temporary data (constant)
    int b = 0;          // temporary data (variable)
    Node node = new(0); // temporary data (object)
    int c = Function(); // stack frame space (call function)
    return a + b + c;   // output data
}

/* Structures */
type node struct {
    val  int
    next *node
}

/* Create node structure */
func newNode(val int) *node {
    return &node{val: val}
}

/* Functions */
func function() int {
    // Perform certain operations...
    return 0
}

func algorithm(n int) int { // input data
    const a = 0             // temporary data (constant)
    b := 0                  // temporary storage of data (variable)
    newNode(0)              // temporary data (object)
    c := function()         // stack frame space (call function)
    return a + b + c        // output data
}

/* Classes */
class Node {
    var val: Int
    var next: Node?

    init(x: Int) {
        val = x
    }
}

/* Functions */
func function() -> Int {
    // Perform certain operations...
    return 0
}

func algorithm(n: Int) -> Int { // input data
    let a = 0                   // temporary data (constant)
    var b = 0                   // temporary data (variable)
    let node = Node(x: 0)       // temporary data (object)
    let c = function()          // stack frame space (call function)
    return a + b + c            // output data
}

/* Classes */
class Node {
    val;
    next;
    constructor(val) {
        this.val = val === undefined ? 0 : val; // node value
        this.next = null;                       // reference to the next node
    }
}

/* Functions */
function constFunc() {
    // Perform certain operations
    return 0;
}

function algorithm(n) {         // input data
    const a = 0;                // temporary data (constant)
    let b = 0;                  // temporary data (variable)
    const node = new Node(0);   // temporary data (object)
    const c = constFunc();      // Stack frame space (calling function)
    return a + b + c;           // output data
}

/* Classes */
class Node {
    val: number;
    next: Node | null;
    constructor(val?: number) {
        this.val = val === undefined ? 0 : val; // node value
        this.next = null;                       // reference to the next node
    }
}

/* Functions */
function constFunc(): number {
    // Perform certain operations
    return 0;
}

function algorithm(n: number): number { // input data
    const a = 0;                        // temporary data (constant)
    let b = 0;                          // temporary data (variable)
    const node = new Node(0);           // temporary data (object)
    const c = constFunc();              // Stack frame space (calling function)
    return a + b + c;                   // output data
}

/* Classes */
class Node {
  int val;
  Node next;
  Node(this.val, [this.next]);
}

/* Functions */
int function() {
  // Perform certain operations...
  return 0;
}

int algorithm(int n) {  // input data
  const int a = 0;      // temporary data (constant)
  int b = 0;            // temporary data (variable)
  Node node = Node(0);  // temporary data (object)
  int c = function();   // stack frame space (call function)
  return a + b + c;     // output data
}

use std::rc::Rc;
use std::cell::RefCell;

/* Structures */
struct Node {
    val: i32,
    next: Option<Rc<RefCell<Node>>>,
}

/* Creating a Node structure */
impl Node {
    fn new(val: i32) -> Self {
        Self { val: val, next: None }
    }
}

/* Functions */
fn function() -> i32 {     
    // Perform certain operations...
    return 0;
}

fn algorithm(n: i32) -> i32 {   // input data
    const a: i32 = 0;           // temporary data (constant)
    let mut b = 0;              // temporary data (variable)
    let node = Node::new(0);    // temporary data (object)
    let c = function();         // stack frame space (call function)
    return a + b + c;           // output data
}

/* Functions */
int func() {
    // Perform certain operations...
    return 0;
}

int algorithm(int n) {  // input data
    const int a = 0;    // temporary data (constant)
    int b = 0;          // temporary data (variable)
    int c = func();     // stack frame space (call function)
    return a + b + c;   // output data
}

2.4.2 Calculation Method¶

The method for calculating space complexity is roughly similar to that of time complexity, with the only change being the shift of the statistical object from "number of operations" to "size of used space".

However, unlike time complexity, we usually only focus on the worst-case space complexity. This is because memory space is a hard requirement, and we must ensure that there is enough memory space reserved under all input data.

Consider the following code, the term "worst-case" in worst-case space complexity has two meanings.

Based on the worst input data: When \(n < 10\), the space complexity is \(O(1)\); but when \(n > 10\), the initialized array nums occupies \(O(n)\) space, thus the worst-case space complexity is \(O(n)\).
Based on the peak memory used during the algorithm's execution: For example, before executing the last line, the program occupies \(O(1)\) space; when initializing the array nums, the program occupies \(O(n)\) space, hence the worst-case space complexity is \(O(n)\).

In recursive functions, stack frame space must be taken into count. Consider the following code:

PythonC++JavaC#GoSwiftJSTSDartRustCZig

def function() -> int:
    # Perform certain operations
    return 0

def loop(n: int):
    """Loop O(1)"""""
    for _ in range(n):
        function()

def recur(n: int) -> int:
    """Recursion O(n)"""""
    if n == 1: return
    return recur(n - 1)

int func() {
    // Perform certain operations
    return 0;
}
/* Cycle O(1) */
void loop(int n) {
    for (int i = 0; i < n; i++) {
        func();
    }
}
/* Recursion O(n) */
void recur(int n) {
    if (n == 1) return;
    return recur(n - 1);
}

int function() {
    // Perform certain operations
    return 0;
}
/* Cycle O(1) */
void loop(int n) {
    for (int i = 0; i < n; i++) {
        function();
    }
}
/* Recursion O(n) */
void recur(int n) {
    if (n == 1) return;
    return recur(n - 1);
}

int Function() {
    // Perform certain operations
    return 0;
}
/* Cycle O(1) */
void Loop(int n) {
    for (int i = 0; i < n; i++) {
        Function();
    }
}
/* Recursion O(n) */
int Recur(int n) {
    if (n == 1) return 1;
    return Recur(n - 1);
}

func function() int {
    // Perform certain operations
    return 0
}

/* Cycle O(1) */
func loop(n int) {
    for i := 0; i < n; i++ {
        function()
    }
}

/* Recursion O(n) */
func recur(n int) {
    if n == 1 {
        return
    }
    recur(n - 1)
}

@discardableResult
func function() -> Int {
    // Perform certain operations
    return 0
}

/* Cycle O(1) */
func loop(n: Int) {
    for _ in 0 ..< n {
        function()
    }
}

/* Recursion O(n) */
func recur(n: Int) {
    if n == 1 {
        return
    }
    recur(n: n - 1)
}

function constFunc() {
    // Perform certain operations
    return 0;
}
/* Cycle O(1) */
function loop(n) {
    for (let i = 0; i < n; i++) {
        constFunc();
    }
}
/* Recursion O(n) */
function recur(n) {
    if (n === 1) return;
    return recur(n - 1);
}

function constFunc(): number {
    // Perform certain operations
    return 0;
}
/* Cycle O(1) */
function loop(n: number): void {
    for (let i = 0; i < n; i++) {
        constFunc();
    }
}
/* Recursion O(n) */
function recur(n: number): void {
    if (n === 1) return;
    return recur(n - 1);
}

int function() {
  // Perform certain operations
  return 0;
}
/* Cycle O(1) */
void loop(int n) {
  for (int i = 0; i < n; i++) {
    function();
  }
}
/* Recursion O(n) */
void recur(int n) {
  if (n == 1) return;
  return recur(n - 1);
}

fn function() -> i32 {
    // Perform certain operations
    return 0;
}
/* Cycle O(1) */
fn loop(n: i32) {
    for i in 0..n {
        function();
    }
}
/* Recursion O(n) */
void recur(n: i32) {
    if n == 1 {
        return;
    }
    recur(n - 1);
}

int func() {
    // Perform certain operations
    return 0;
}
/* Cycle O(1) */
void loop(int n) {
    for (int i = 0; i < n; i++) {
        func();
    }
}
/* Recursion O(n) */
void recur(int n) {
    if (n == 1) return;
    return recur(n - 1);
}

The time complexity of both loop() and recur() functions is \(O(n)\), but their space complexities differ.

The loop() function calls function() \(n\) times in a loop, where each iteration's function() returns and releases its stack frame space, so the space complexity remains \(O(1)\).
The recursive function recur() will have \(n\) instances of unreturned recur() existing simultaneously during its execution, thus occupying \(O(n)\) stack frame space.

2.4.3 Common Types¶

Let the size of the input data be \(n\), the following chart displays common types of space complexities (arranged from low to high).

\[ \begin{aligned} O(1) < O(\log n) < O(n) < O(n^2) < O(2^n) \newline \text{Constant Order} < \text{Logarithmic Order} < \text{Linear Order} < \text{Quadratic Order} < \text{Exponential Order} \end{aligned} \]

Figure 2-16 Common Types of Space Complexity

1. Constant Order \(O(1)\)¶

Constant order is common in constants, variables, objects that are independent of the size of input data \(n\).

Note that memory occupied by initializing variables or calling functions in a loop, which is released upon entering the next cycle, does not accumulate over space, thus the space complexity remains \(O(1)\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig

space_complexity.py

def function() -> int:
    """函数"""
    # 执行某些操作
    return 0

def constant(n: int):
    """常数阶"""
    # 常量、变量、对象占用 O(1) 空间
    a = 0
    nums = [0] * 10000
    node = ListNode(0)
    # 循环中的变量占用 O(1) 空间
    for _ in range(n):
        c = 0
    # 循环中的函数占用 O(1) 空间
    for _ in range(n):
        function()

space_complexity.cpp

/* 函数 */
int func() {
    // 执行某些操作
    return 0;
}

/* 常数阶 */
void constant(int n) {
    // 常量、变量、对象占用 O(1) 空间
    const int a = 0;
    int b = 0;
    vector<int> nums(10000);
    ListNode node(0);
    // 循环中的变量占用 O(1) 空间
    for (int i = 0; i < n; i++) {
        int c = 0;
    }
    // 循环中的函数占用 O(1) 空间
    for (int i = 0; i < n; i++) {
        func();
    }
}

space_complexity.java

/* 函数 */
int function() {
    // 执行某些操作
    return 0;
}

/* 常数阶 */
void constant(int n) {
    // 常量、变量、对象占用 O(1) 空间
    final int a = 0;
    int b = 0;
    int[] nums = new int[10000];
    ListNode node = new ListNode(0);
    // 循环中的变量占用 O(1) 空间
    for (int i = 0; i < n; i++) {
        int c = 0;
    }
    // 循环中的函数占用 O(1) 空间
    for (int i = 0; i < n; i++) {
        function();
    }
}

space_complexity.cs

/* 函数 */
int Function() {
    // 执行某些操作
    return 0;
}

/* 常数阶 */
void Constant(int n) {
    // 常量、变量、对象占用 O(1) 空间
    int a = 0;
    int b = 0;
    int[] nums = new int[10000];
    ListNode node = new(0);
    // 循环中的变量占用 O(1) 空间
    for (int i = 0; i < n; i++) {
        int c = 0;
    }
    // 循环中的函数占用 O(1) 空间
    for (int i = 0; i < n; i++) {
        Function();
    }
}

space_complexity.go

/* 函数 */
func function() int {
    // 执行某些操作...
    return 0
}

/* 常数阶 */
func spaceConstant(n int) {
    // 常量、变量、对象占用 O(1) 空间
    const a = 0
    b := 0
    nums := make([]int, 10000)
    node := newNode(0)
    // 循环中的变量占用 O(1) 空间
    var c int
    for i := 0; i < n; i++ {
        c = 0
    }
    // 循环中的函数占用 O(1) 空间
    for i := 0; i < n; i++ {
        function()
    }
    b += 0
    c += 0
    nums[0] = 0
    node.val = 0
}

space_complexity.swift

/* 函数 */
@discardableResult
func function() -> Int {
    // 执行某些操作
    return 0
}

/* 常数阶 */
func constant(n: Int) {
    // 常量、变量、对象占用 O(1) 空间
    let a = 0
    var b = 0
    let nums = Array(repeating: 0, count: 10000)
    let node = ListNode(x: 0)
    // 循环中的变量占用 O(1) 空间
    for _ in 0 ..< n {
        let c = 0
    }
    // 循环中的函数占用 O(1) 空间
    for _ in 0 ..< n {
        function()
    }
}

space_complexity.js

/* 函数 */
function constFunc() {
    // 执行某些操作
    return 0;
}

/* 常数阶 */
function constant(n) {
    // 常量、变量、对象占用 O(1) 空间
    const a = 0;
    const b = 0;
    const nums = new Array(10000);
    const node = new ListNode(0);
    // 循环中的变量占用 O(1) 空间
    for (let i = 0; i < n; i++) {
        const c = 0;
    }
    // 循环中的函数占用 O(1) 空间
    for (let i = 0; i < n; i++) {
        constFunc();
    }
}

space_complexity.ts

/* 函数 */
function constFunc(): number {
    // 执行某些操作
    return 0;
}

/* 常数阶 */
function constant(n: number): void {
    // 常量、变量、对象占用 O(1) 空间
    const a = 0;
    const b = 0;
    const nums = new Array(10000);
    const node = new ListNode(0);
    // 循环中的变量占用 O(1) 空间
    for (let i = 0; i < n; i++) {
        const c = 0;
    }
    // 循环中的函数占用 O(1) 空间
    for (let i = 0; i < n; i++) {
        constFunc();
    }
}

space_complexity.dart

/* 函数 */
int function() {
  // 执行某些操作
  return 0;
}

/* 常数阶 */
void constant(int n) {
  // 常量、变量、对象占用 O(1) 空间
  final int a = 0;
  int b = 0;
  List<int> nums = List.filled(10000, 0);
  ListNode node = ListNode(0);
  // 循环中的变量占用 O(1) 空间
  for (var i = 0; i < n; i++) {
    int c = 0;
  }
  // 循环中的函数占用 O(1) 空间
  for (var i = 0; i < n; i++) {
    function();
  }
}

space_complexity.rs

/* 函数 */
fn function() ->i32 {
    // 执行某些操作
    return 0;
}

/* 常数阶 */
#[allow(unused)]
fn constant(n: i32) {
    // 常量、变量、对象占用 O(1) 空间
    const A: i32 = 0;
    let b = 0;
    let nums = vec![0; 10000];
    let node = ListNode::new(0);
    // 循环中的变量占用 O(1) 空间
    for i in 0..n {
        let c = 0;
    }
    // 循环中的函数占用 O(1) 空间
    for i in 0..n {
        function();
    }
}

space_complexity.c

/* 函数 */
int func() {
    // 执行某些操作
    return 0;
}

/* 常数阶 */
void constant(int n) {
    // 常量、变量、对象占用 O(1) 空间
    const int a = 0;
    int b = 0;
    int nums[1000];
    ListNode *node = newListNode(0);
    free(node);
    // 循环中的变量占用 O(1) 空间
    for (int i = 0; i < n; i++) {
        int c = 0;
    }
    // 循环中的函数占用 O(1) 空间
    for (int i = 0; i < n; i++) {
        func();
    }
}

space_complexity.zig

// 函数
fn function() i32 {
    // 执行某些操作
    return 0;
}

// 常数阶
fn constant(n: i32) void {
    // 常量、变量、对象占用 O(1) 空间
    const a: i32 = 0;
    var b: i32 = 0;
    var nums = [_]i32{0}**10000;
    var node = inc.ListNode(i32){.val = 0};
    var i: i32 = 0;
    // 循环中的变量占用 O(1) 空间
    while (i < n) : (i += 1) {
        var c: i32 = 0;
        _ = c;
    }
    // 循环中的函数占用 O(1) 空间
    i = 0;
    while (i < n) : (i += 1) {
        _ = function();
    }
    _ = a;
    _ = b;
    _ = nums;
    _ = node;
}

Code Visualization

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2. Linear Order \(O(n)\)¶

Linear order is common in arrays, linked lists, stacks, queues, etc., where the number of elements is proportional to \(n\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig

space_complexity.py

def linear(n: int):
    """线性阶"""
    # 长度为 n 的列表占用 O(n) 空间
    nums = [0] * n
    # 长度为 n 的哈希表占用 O(n) 空间
    hmap = dict[int, str]()
    for i in range(n):
        hmap[i] = str(i)

space_complexity.cpp

/* 线性阶 */
void linear(int n) {
    // 长度为 n 的数组占用 O(n) 空间
    vector<int> nums(n);
    // 长度为 n 的列表占用 O(n) 空间
    vector<ListNode> nodes;
    for (int i = 0; i < n; i++) {
        nodes.push_back(ListNode(i));
    }
    // 长度为 n 的哈希表占用 O(n) 空间
    unordered_map<int, string> map;
    for (int i = 0; i < n; i++) {
        map[i] = to_string(i);
    }
}

space_complexity.java

/* 线性阶 */
void linear(int n) {
    // 长度为 n 的数组占用 O(n) 空间
    int[] nums = new int[n];
    // 长度为 n 的列表占用 O(n) 空间
    List<ListNode> nodes = new ArrayList<>();
    for (int i = 0; i < n; i++) {
        nodes.add(new ListNode(i));
    }
    // 长度为 n 的哈希表占用 O(n) 空间
    Map<Integer, String> map = new HashMap<>();
    for (int i = 0; i < n; i++) {
        map.put(i, String.valueOf(i));
    }
}

space_complexity.cs

/* 线性阶 */
void Linear(int n) {
    // 长度为 n 的数组占用 O(n) 空间
    int[] nums = new int[n];
    // 长度为 n 的列表占用 O(n) 空间
    List<ListNode> nodes = [];
    for (int i = 0; i < n; i++) {
        nodes.Add(new ListNode(i));
    }
    // 长度为 n 的哈希表占用 O(n) 空间
    Dictionary<int, string> map = [];
    for (int i = 0; i < n; i++) {
        map.Add(i, i.ToString());
    }
}

space_complexity.go

/* 线性阶 */
func spaceLinear(n int) {
    // 长度为 n 的数组占用 O(n) 空间
    _ = make([]int, n)
    // 长度为 n 的列表占用 O(n) 空间
    var nodes []*node
    for i := 0; i < n; i++ {
        nodes = append(nodes, newNode(i))
    }
    // 长度为 n 的哈希表占用 O(n) 空间
    m := make(map[int]string, n)
    for i := 0; i < n; i++ {
        m[i] = strconv.Itoa(i)
    }
}

space_complexity.swift

/* 线性阶 */
func linear(n: Int) {
    // 长度为 n 的数组占用 O(n) 空间
    let nums = Array(repeating: 0, count: n)
    // 长度为 n 的列表占用 O(n) 空间
    let nodes = (0 ..< n).map { ListNode(x: $0) }
    // 长度为 n 的哈希表占用 O(n) 空间
    let map = Dictionary(uniqueKeysWithValues: (0 ..< n).map { ($0, "\($0)") })
}

space_complexity.js

/* 线性阶 */
function linear(n) {
    // 长度为 n 的数组占用 O(n) 空间
    const nums = new Array(n);
    // 长度为 n 的列表占用 O(n) 空间
    const nodes = [];
    for (let i = 0; i < n; i++) {
        nodes.push(new ListNode(i));
    }
    // 长度为 n 的哈希表占用 O(n) 空间
    const map = new Map();
    for (let i = 0; i < n; i++) {
        map.set(i, i.toString());
    }
}

space_complexity.ts

/* 线性阶 */
function linear(n: number): void {
    // 长度为 n 的数组占用 O(n) 空间
    const nums = new Array(n);
    // 长度为 n 的列表占用 O(n) 空间
    const nodes: ListNode[] = [];
    for (let i = 0; i < n; i++) {
        nodes.push(new ListNode(i));
    }
    // 长度为 n 的哈希表占用 O(n) 空间
    const map = new Map();
    for (let i = 0; i < n; i++) {
        map.set(i, i.toString());
    }
}

space_complexity.dart

/* 线性阶 */
void linear(int n) {
  // 长度为 n 的数组占用 O(n) 空间
  List<int> nums = List.filled(n, 0);
  // 长度为 n 的列表占用 O(n) 空间
  List<ListNode> nodes = [];
  for (var i = 0; i < n; i++) {
    nodes.add(ListNode(i));
  }
  // 长度为 n 的哈希表占用 O(n) 空间
  Map<int, String> map = HashMap();
  for (var i = 0; i < n; i++) {
    map.putIfAbsent(i, () => i.toString());
  }
}

space_complexity.rs

/* 线性阶 */
#[allow(unused)]
fn linear(n: i32) {
    // 长度为 n 的数组占用 O(n) 空间
    let mut nums = vec![0; n as usize];
    // 长度为 n 的列表占用 O(n) 空间
    let mut nodes = Vec::new();
    for i in 0..n {
        nodes.push(ListNode::new(i))
    }
    // 长度为 n 的哈希表占用 O(n) 空间
    let mut map = HashMap::new();
    for i in 0..n {
        map.insert(i, i.to_string());
    }
}

space_complexity.c

/* 哈希表 */
typedef struct {
    int key;
    int val;
    UT_hash_handle hh; // 基于 uthash.h 实现
} HashTable;

/* 线性阶 */
void linear(int n) {
    // 长度为 n 的数组占用 O(n) 空间
    int *nums = malloc(sizeof(int) * n);
    free(nums);

    // 长度为 n 的列表占用 O(n) 空间
    ListNode **nodes = malloc(sizeof(ListNode *) * n);
    for (int i = 0; i < n; i++) {
        nodes[i] = newListNode(i);
    }
    // 内存释放
    for (int i = 0; i < n; i++) {
        free(nodes[i]);
    }
    free(nodes);

    // 长度为 n 的哈希表占用 O(n) 空间
    HashTable *h = NULL;
    for (int i = 0; i < n; i++) {
        HashTable *tmp = malloc(sizeof(HashTable));
        tmp->key = i;
        tmp->val = i;
        HASH_ADD_INT(h, key, tmp);
    }

    // 内存释放
    HashTable *curr, *tmp;
    HASH_ITER(hh, h, curr, tmp) {
        HASH_DEL(h, curr);
        free(curr);
    }
}

space_complexity.zig

// 线性阶
fn linear(comptime n: i32) !void {
    // 长度为 n 的数组占用 O(n) 空间
    var nums = [_]i32{0}**n;
    // 长度为 n 的列表占用 O(n) 空间
    var nodes = std.ArrayList(i32).init(std.heap.page_allocator);
    defer nodes.deinit();
    var i: i32 = 0;
    while (i < n) : (i += 1) {
        try nodes.append(i);
    }
    // 长度为 n 的哈希表占用 O(n) 空间
    var map = std.AutoArrayHashMap(i32, []const u8).init(std.heap.page_allocator);
    defer map.deinit();
    var j: i32 = 0;
    while (j < n) : (j += 1) {
        const string = try std.fmt.allocPrint(std.heap.page_allocator, "{d}", .{j});
        defer std.heap.page_allocator.free(string);
        try map.put(i, string);
    }
    _ = nums;
}

Code Visualization

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As shown below, this function's recursive depth is \(n\), meaning there are \(n\) instances of unreturned linear_recur() function, using \(O(n)\) size of stack frame space:

PythonC++JavaC#GoSwiftJSTSDartRustCZig

space_complexity.py

def linear_recur(n: int):
    """线性阶（递归实现）"""
    print("递归 n =", n)
    if n == 1:
        return
    linear_recur(n - 1)

space_complexity.cpp

/* 线性阶（递归实现） */
void linearRecur(int n) {
    cout << "递归 n = " << n << endl;
    if (n == 1)
        return;
    linearRecur(n - 1);
}

space_complexity.java

/* 线性阶（递归实现） */
void linearRecur(int n) {
    System.out.println("递归 n = " + n);
    if (n == 1)
        return;
    linearRecur(n - 1);
}

space_complexity.cs

/* 线性阶（递归实现） */
void LinearRecur(int n) {
    Console.WriteLine("递归 n = " + n);
    if (n == 1) return;
    LinearRecur(n - 1);
}

space_complexity.go

/* 线性阶（递归实现） */
func spaceLinearRecur(n int) {
    fmt.Println("递归 n =", n)
    if n == 1 {
        return
    }
    spaceLinearRecur(n - 1)
}

space_complexity.swift

/* 线性阶（递归实现） */
func linearRecur(n: Int) {
    print("递归 n = \(n)")
    if n == 1 {
        return
    }
    linearRecur(n: n - 1)
}

space_complexity.js

/* 线性阶（递归实现） */
function linearRecur(n) {
    console.log(`递归 n = ${n}`);
    if (n === 1) return;
    linearRecur(n - 1);
}

space_complexity.ts

/* 线性阶（递归实现） */
function linearRecur(n: number): void {
    console.log(`递归 n = ${n}`);
    if (n === 1) return;
    linearRecur(n - 1);
}

space_complexity.dart

/* 线性阶（递归实现） */
void linearRecur(int n) {
  print('递归 n = $n');
  if (n == 1) return;
  linearRecur(n - 1);
}

space_complexity.rs

/* 线性阶（递归实现） */
fn linear_recur(n: i32) {
    println!("递归 n = {}", n);
    if n == 1 {return};
    linear_recur(n - 1);
}

space_complexity.c

/* 线性阶（递归实现） */
void linearRecur(int n) {
    printf("递归 n = %d\r\n", n);
    if (n == 1)
        return;
    linearRecur(n - 1);
}

space_complexity.zig

// 线性阶（递归实现）
fn linearRecur(comptime n: i32) void {
    std.debug.print("递归 n = {}\n", .{n});
    if (n == 1) return;
    linearRecur(n - 1);
}

Code Visualization

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Figure 2-17 Recursive Function Generating Linear Order Space Complexity

3. Quadratic Order \(O(n^2)\)¶

Quadratic order is common in matrices and graphs, where the number of elements is quadratic to \(n\):

PythonC++JavaC#GoSwiftJSTSDartRustCZig

space_complexity.py

def quadratic(n: int):
    """平方阶"""
    # 二维列表占用 O(n^2) 空间
    num_matrix = [[0] * n for _ in range(n)]

space_complexity.cpp

/* 平方阶 */
void quadratic(int n) {
    // 二维列表占用 O(n^2) 空间
    vector<vector<int>> numMatrix;
    for (int i = 0; i < n; i++) {
        vector<int> tmp;
        for (int j = 0; j < n; j++) {
            tmp.push_back(0);
        }
        numMatrix.push_back(tmp);
    }
}

space_complexity.java

/* 平方阶 */
void quadratic(int n) {
    // 矩阵占用 O(n^2) 空间
    int[][] numMatrix = new int[n][n];
    // 二维列表占用 O(n^2) 空间
    List<List<Integer>> numList = new ArrayList<>();
    for (int i = 0; i < n; i++) {
        List<Integer> tmp = new ArrayList<>();
        for (int j = 0; j < n; j++) {
            tmp.add(0);
        }
        numList.add(tmp);
    }
}

space_complexity.cs

/* 平方阶 */
void Quadratic(int n) {
    // 矩阵占用 O(n^2) 空间
    int[,] numMatrix = new int[n, n];
    // 二维列表占用 O(n^2) 空间
    List<List<int>> numList = [];
    for (int i = 0; i < n; i++) {
        List<int> tmp = [];
        for (int j = 0; j < n; j++) {
            tmp.Add(0);
        }
        numList.Add(tmp);
    }
}

space_complexity.go

/* 平方阶 */
func spaceQuadratic(n int) {
    // 矩阵占用 O(n^2) 空间
    numMatrix := make([][]int, n)
    for i := 0; i < n; i++ {
        numMatrix[i] = make([]int, n)
    }
}

space_complexity.swift

/* 平方阶 */
func quadratic(n: Int) {
    // 二维列表占用 O(n^2) 空间
    let numList = Array(repeating: Array(repeating: 0, count: n), count: n)
}

space_complexity.js

/* 平方阶 */
function quadratic(n) {
    // 矩阵占用 O(n^2) 空间
    const numMatrix = Array(n)
        .fill(null)
        .map(() => Array(n).fill(null));
    // 二维列表占用 O(n^2) 空间
    const numList = [];
    for (let i = 0; i < n; i++) {
        const tmp = [];
        for (let j = 0; j < n; j++) {
            tmp.push(0);
        }
        numList.push(tmp);
    }
}

space_complexity.ts

/* 平方阶 */
function quadratic(n: number): void {
    // 矩阵占用 O(n^2) 空间
    const numMatrix = Array(n)
        .fill(null)
        .map(() => Array(n).fill(null));
    // 二维列表占用 O(n^2) 空间
    const numList = [];
    for (let i = 0; i < n; i++) {
        const tmp = [];
        for (let j = 0; j < n; j++) {
            tmp.push(0);
        }
        numList.push(tmp);
    }
}

space_complexity.dart

/* 平方阶 */
void quadratic(int n) {
  // 矩阵占用 O(n^2) 空间
  List<List<int>> numMatrix = List.generate(n, (_) => List.filled(n, 0));
  // 二维列表占用 O(n^2) 空间
  List<List<int>> numList = [];
  for (var i = 0; i < n; i++) {
    List<int> tmp = [];
    for (int j = 0; j < n; j++) {
      tmp.add(0);
    }
    numList.add(tmp);
  }
}

space_complexity.rs

/* 平方阶 */
#[allow(unused)]
fn quadratic(n: i32) {
    // 矩阵占用 O(n^2) 空间
    let num_matrix = vec![vec![0; n as usize]; n as usize];
    // 二维列表占用 O(n^2) 空间
    let mut num_list = Vec::new();
    for i in 0..n {
        let mut tmp = Vec::new();
        for j in 0..n {
            tmp.push(0);
        }
        num_list.push(tmp);
    }
}

space_complexity.c

/* 平方阶 */
void quadratic(int n) {
    // 二维列表占用 O(n^2) 空间
    int **numMatrix = malloc(sizeof(int *) * n);
    for (int i = 0; i < n; i++) {
        int *tmp = malloc(sizeof(int) * n);
        for (int j = 0; j < n; j++) {
            tmp[j] = 0;
        }
        numMatrix[i] = tmp;
    }

    // 内存释放
    for (int i = 0; i < n; i++) {
        free(numMatrix[i]);
    }
    free(numMatrix);
}

space_complexity.zig

// 平方阶
fn quadratic(n: i32) !void {
    // 二维列表占用 O(n^2) 空间
    var nodes = std.ArrayList(std.ArrayList(i32)).init(std.heap.page_allocator);
    defer nodes.deinit();
    var i: i32 = 0;
    while (i < n) : (i += 1) {
        var tmp = std.ArrayList(i32).init(std.heap.page_allocator);
        defer tmp.deinit();
        var j: i32 = 0;
        while (j < n) : (j += 1) {
            try tmp.append(0);
        }
        try nodes.append(tmp);
    }
}

Code Visualization

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As shown below, the recursive depth of this function is \(n\), and in each recursive call, an array is initialized with lengths \(n\), \(n-1\), \(\dots\), \(2\), \(1\), averaging \(n/2\), thus overall occupying \(O(n^2)\) space:

PythonC++JavaC#GoSwiftJSTSDartRustCZig

space_complexity.py

def quadratic_recur(n: int) -> int:
    """平方阶（递归实现）"""
    if n <= 0:
        return 0
    # 数组 nums 长度为 n, n-1, ..., 2, 1
    nums = [0] * n
    return quadratic_recur(n - 1)

space_complexity.cpp

/* 平方阶（递归实现） */
int quadraticRecur(int n) {
    if (n <= 0)
        return 0;
    vector<int> nums(n);
    cout << "递归 n = " << n << " 中的 nums 长度 = " << nums.size() << endl;
    return quadraticRecur(n - 1);
}

space_complexity.java

/* 平方阶（递归实现） */
int quadraticRecur(int n) {
    if (n <= 0)
        return 0;
    // 数组 nums 长度为 n, n-1, ..., 2, 1
    int[] nums = new int[n];
    System.out.println("递归 n = " + n + " 中的 nums 长度 = " + nums.length);
    return quadraticRecur(n - 1);
}

space_complexity.cs

/* 平方阶（递归实现） */
int QuadraticRecur(int n) {
    if (n <= 0) return 0;
    int[] nums = new int[n];
    Console.WriteLine("递归 n = " + n + " 中的 nums 长度 = " + nums.Length);
    return QuadraticRecur(n - 1);
}

space_complexity.go

/* 平方阶（递归实现） */
func spaceQuadraticRecur(n int) int {
    if n <= 0 {
        return 0
    }
    nums := make([]int, n)
    fmt.Printf("递归 n = %d 中的 nums 长度 = %d \n", n, len(nums))
    return spaceQuadraticRecur(n - 1)
}

space_complexity.swift

/* 平方阶（递归实现） */
@discardableResult
func quadraticRecur(n: Int) -> Int {
    if n <= 0 {
        return 0
    }
    // 数组 nums 长度为 n, n-1, ..., 2, 1
    let nums = Array(repeating: 0, count: n)
    print("递归 n = \(n) 中的 nums 长度 = \(nums.count)")
    return quadraticRecur(n: n - 1)
}

space_complexity.js

/* 平方阶（递归实现） */
function quadraticRecur(n) {
    if (n <= 0) return 0;
    const nums = new Array(n);
    console.log(`递归 n = ${n} 中的 nums 长度 = ${nums.length}`);
    return quadraticRecur(n - 1);
}

space_complexity.ts

/* 平方阶（递归实现） */
function quadraticRecur(n: number): number {
    if (n <= 0) return 0;
    const nums = new Array(n);
    console.log(`递归 n = ${n} 中的 nums 长度 = ${nums.length}`);
    return quadraticRecur(n - 1);
}

space_complexity.dart

/* 平方阶（递归实现） */
int quadraticRecur(int n) {
  if (n <= 0) return 0;
  List<int> nums = List.filled(n, 0);
  print('递归 n = $n 中的 nums 长度 = ${nums.length}');
  return quadraticRecur(n - 1);
}

space_complexity.rs

/* 平方阶（递归实现） */
fn quadratic_recur(n: i32) -> i32 {
    if n <= 0 {return 0};
    // 数组 nums 长度为 n, n-1, ..., 2, 1
    let nums = vec![0; n as usize];
    println!("递归 n = {} 中的 nums 长度 = {}", n, nums.len());
    return quadratic_recur(n - 1);
}

space_complexity.c

/* 平方阶（递归实现） */
int quadraticRecur(int n) {
    if (n <= 0)
        return 0;
    int *nums = malloc(sizeof(int) * n);
    printf("递归 n = %d 中的 nums 长度 = %d\r\n", n, n);
    int res = quadraticRecur(n - 1);
    free(nums);
    return res;
}

space_complexity.zig

// 平方阶（递归实现）
fn quadraticRecur(comptime n: i32) i32 {
    if (n <= 0) return 0;
    var nums = [_]i32{0}**n;
    std.debug.print("递归 n = {} 中的 nums 长度 = {}\n", .{n, nums.len});
    return quadraticRecur(n - 1);
}

Code Visualization

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Figure 2-18 Recursive Function Generating Quadratic Order Space Complexity

4. Exponential Order \(O(2^n)\)¶

Exponential order is common in binary trees. Observe the below image, a "full binary tree" with \(n\) levels has \(2^n - 1\) nodes, occupying \(O(2^n)\) space:

PythonC++JavaC#GoSwiftJSTSDartRustCZig

space_complexity.py

def build_tree(n: int) -> TreeNode | None:
    """指数阶（建立满二叉树）"""
    if n == 0:
        return None
    root = TreeNode(0)
    root.left = build_tree(n - 1)
    root.right = build_tree(n - 1)
    return root

space_complexity.cpp

/* 指数阶（建立满二叉树） */
TreeNode *buildTree(int n) {
    if (n == 0)
        return nullptr;
    TreeNode *root = new TreeNode(0);
    root->left = buildTree(n - 1);
    root->right = buildTree(n - 1);
    return root;
}

space_complexity.java

/* 指数阶（建立满二叉树） */
TreeNode buildTree(int n) {
    if (n == 0)
        return null;
    TreeNode root = new TreeNode(0);
    root.left = buildTree(n - 1);
    root.right = buildTree(n - 1);
    return root;
}

space_complexity.cs

/* 指数阶（建立满二叉树） */
TreeNode? BuildTree(int n) {
    if (n == 0) return null;
    TreeNode root = new(0) {
        left = BuildTree(n - 1),
        right = BuildTree(n - 1)
    };
    return root;
}

space_complexity.go

/* 指数阶（建立满二叉树） */
func buildTree(n int) *TreeNode {
    if n == 0 {
        return nil
    }
    root := NewTreeNode(0)
    root.Left = buildTree(n - 1)
    root.Right = buildTree(n - 1)
    return root
}

space_complexity.swift

/* 指数阶（建立满二叉树） */
func buildTree(n: Int) -> TreeNode? {
    if n == 0 {
        return nil
    }
    let root = TreeNode(x: 0)
    root.left = buildTree(n: n - 1)
    root.right = buildTree(n: n - 1)
    return root
}

space_complexity.js

/* 指数阶（建立满二叉树） */
function buildTree(n) {
    if (n === 0) return null;
    const root = new TreeNode(0);
    root.left = buildTree(n - 1);
    root.right = buildTree(n - 1);
    return root;
}

space_complexity.ts

/* 指数阶（建立满二叉树） */
function buildTree(n: number): TreeNode | null {
    if (n === 0) return null;
    const root = new TreeNode(0);
    root.left = buildTree(n - 1);
    root.right = buildTree(n - 1);
    return root;
}

space_complexity.dart

/* 指数阶（建立满二叉树） */
TreeNode? buildTree(int n) {
  if (n == 0) return null;
  TreeNode root = TreeNode(0);
  root.left = buildTree(n - 1);
  root.right = buildTree(n - 1);
  return root;
}

space_complexity.rs

/* 指数阶（建立满二叉树） */
fn build_tree(n: i32) -> Option<Rc<RefCell<TreeNode>>> {
    if n == 0 {return None};
    let root = TreeNode::new(0);
    root.borrow_mut().left = build_tree(n - 1);
    root.borrow_mut().right = build_tree(n - 1);
    return Some(root);
}

space_complexity.c

/* 指数阶（建立满二叉树） */
TreeNode *buildTree(int n) {
    if (n == 0)
        return NULL;
    TreeNode *root = newTreeNode(0);
    root->left = buildTree(n - 1);
    root->right = buildTree(n - 1);
    return root;
}

space_complexity.zig

// 指数阶（建立满二叉树）
fn buildTree(mem_allocator: std.mem.Allocator, n: i32) !?*inc.TreeNode(i32) {
    if (n == 0) return null;
    const root = try mem_allocator.create(inc.TreeNode(i32));
    root.init(0);
    root.left = try buildTree(mem_allocator, n - 1);
    root.right = try buildTree(mem_allocator, n - 1);
    return root;
}

Code Visualization

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Figure 2-19 Full Binary Tree Generating Exponential Order Space Complexity

5. Logarithmic Order \(O(\log n)\)¶

Logarithmic order is common in divide-and-conquer algorithms. For example, in merge sort, an array of length \(n\) is recursively divided in half each round, forming a recursion tree of height \(\log n\), using \(O(\log n)\) stack frame space.

Another example is converting a number to a string. Given a positive integer \(n\), its number of digits is \(\log_{10} n + 1\), corresponding to the length of the string, thus the space complexity is \(O(\log_{10} n + 1) = O(\log n)\).

2.4.4 Balancing Time and Space¶

Ideally, we aim for both time complexity and space complexity to be optimal. However, in practice, optimizing both simultaneously is often difficult.

Lowering time complexity usually comes at the cost of increased space complexity, and vice versa. The approach of sacrificing memory space to improve algorithm speed is known as "space-time tradeoff"; the reverse is known as "time-space tradeoff".

The choice depends on which aspect we value more. In most cases, time is more precious than space, so "space-time tradeoff" is often the more common strategy. Of course, controlling space complexity is also very important when dealing with large volumes of data.

2.4 Space Complexity¶

2.4.1 Space Related to Algorithms¶

2.4.2 Calculation Method¶

2.4.3 Common Types¶

1. Constant Order \(O(1)\)¶

2. Linear Order \(O(n)\)¶

3. Quadratic Order \(O(n^2)\)¶

4. Exponential Order \(O(2^n)\)¶

5. Logarithmic Order \(O(\log n)\)¶

2.4.4 Balancing Time and Space¶

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