Task Examples with Different Complexity Levels

3.1 Tasks with Constant Complexity O(1).

Accessing an array element by index:

Accessing an element in an array by its index is done in constant time, because the element's address is calculated directly.

def get_element(arr, index):
    return arr[index]

Inserting an element at the start of a list (Deque):

Using a double-ended queue (deque) allows inserting an element at the start of a list in constant time.

from collections import deque

def insert_element(dq, element):
    dq.appendleft(element)

3.2 Tasks with Linear Complexity O(n).

Linear search in an array:

Searching for an element in an unsorted array is done in linear time because you might need to check each element.

def linear_search(arr, target):
    for i in range(len(arr)):
        if arr[i] == target:
            return i
    return -1

Counting the number of elements in an array:

Traversing all elements in an array to count them takes linear time.

def count_elements(arr):
    count = 0
    for element in arr:
        count += 1
    return count

3.3 Tasks with Logarithmic Complexity O(log n).

Binary search:

Searching for an element in a sorted array using binary search is done in logarithmic time.

def binary_search(arr, target):
    left, right = 0, len(arr) - 1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

Inserting an element in a binary search tree:

Inserting an element into a balanced binary search tree (BST) takes logarithmic time.

class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.val = key
        
def insert(root, key):
    if root is None:
        return Node(key)
    if key < root.val:
        root.left = insert(root.left, key)
    else:
        root.right = insert(root.right, key)
    return root

3.4 Tasks with Quadratic Complexity O(n^2).

Bubble sort:

Sorting an array using bubble sort is quadratic in time.

def bubble_sort(arr):
    n = len(arr)
    for i in range(n):
        for j in range(0, n - i - 1):
            if arr[j] > arr[j + 1]:
                arr[j], arr[j + 1] = arr[j + 1], arr[j]

Checking for duplicates using a double loop:

Checking an array for duplicates using a double loop takes quadratic time.

def has_duplicates(arr):
    n = len(arr)
    for i in range(n):
        for j in range(i + 1, n):
            if arr[i] == arr[j]:
                return True
    return False

3.5 Tasks with Exponential Complexity O(2^n).

Tower of Hanoi problem:

Solving the Tower of Hanoi problem takes exponential time due to the need to move each disk.

def hanoi(n, source, target, auxiliary):
    if n == 1:
        print(f"Move disk 1 from {source} to {target}")
        return
    hanoi(n - 1, source, auxiliary, target)
    print(f"Move disk {n} from {source} to {target}")
    hanoi(n - 1, auxiliary, target, source)

Generating all subsets of a set:

Generating all subsets of a set takes exponential time since each subset must be considered.

def generate_subsets(s):
    result = []
    subset = []

    def backtrack(index):
        if index == len(s):
            result.append(subset[:])
            return
        subset.append(s[index])
        backtrack(index + 1)
        subset.pop()
        backtrack(index + 1)
        
    backtrack(0)
    return result
        
print(generate_subsets([1, 2, 3]))

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Task

Python SELF EN, level 61, lesson 2

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Number of elements

Number of elements

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2

Task

Python SELF EN, level 61, lesson 2

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Towers of Hanoi

Towers of Hanoi

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