Space and Time Complexity

Space complexity refers to the amount of memory used by an algorithm to complete its execution, as a function of the size of the input. The space complexity of an algorithm can be affected by various factors such as the size of the input data, the data structures used in the algorithm, the number and size of temporary variables, and the recursion depth. Time complexity refers to the amount of time required by an algorithm to run as the input size grows. It is usually measured in terms of the "Big O" notation, which describes the upper bound of an algorithm's time complexity.

Why do you think a programmer should care about space and time complexity? Programmers should care about space and time complexity as it is a vital part in loading times. If we do not account for this, our program could lose its function. For example, a program whose sole purpose is to provide decent quality with quick times would fail if they tried to render all images at 8k quality. Also, if we do not account for this, the loading times might become so large that the user stops using the program, making space and time complexity very important.

Take a look at our lassen volcano example from the data compression tech talk. The first code block is the original image. In the second code block, change the baseWidth to rescale the image.

from IPython.display import Image, display
from pathlib import Path 

# prepares a series of images
def image_data(path=Path("images/"), images=None):  # path of static images is defaulted
    for image in images:
        # File to open
        image['filename'] = path / image['file']  # file with path
    return images

def image_display(images):
    for image in images:  
        display(Image(filename=image['filename']))

if __name__ == "__main__":
    lassen_volcano = image_data(images=[{'source': "Peter Carolin", 'label': "Lassen Volcano", 'file': "lassen-volcano.jpg"}])
    image_display(lassen_volcano)
    
from IPython.display import HTML, display
from pathlib import Path 
from PIL import Image as pilImage 
from io import BytesIO
import base64

# prepares a series of images
def image_data(path=Path("images/"), images=None):  # path of static images is defaulted
    for image in images:
        # File to open
        image['filename'] = path / image['file']  # file with path
    return images

def scale_image(img):
    #baseWidth = 625
    #baseWidth = 1250
    #baseWidth = 2500
    baseWidth = 5000 # see the effect of doubling or halfing the baseWidth 
    #baseWidth = 10000 
    #baseWidth = 20000
    #baseWidth = 40000
    scalePercent = (baseWidth/float(img.size[0]))
    scaleHeight = int((float(img.size[1])*float(scalePercent)))
    scale = (baseWidth, scaleHeight)
    return img.resize(scale)

def image_to_base64(img, format):
    with BytesIO() as buffer:
        img.save(buffer, format)
        return base64.b64encode(buffer.getvalue()).decode()
    
def image_management(image):  # path of static images is defaulted        
    # Image open return PIL image object
    img = pilImage.open(image['filename'])
    
    # Python Image Library operations
    image['format'] = img.format
    image['mode'] = img.mode
    image['size'] = img.size
    image['width'], image['height'] = img.size
    image['pixels'] = image['width'] * image['height']
    # Scale the Image
    img = scale_image(img)
    image['pil'] = img
    image['scaled_size'] = img.size
    image['scaled_width'], image['scaled_height'] = img.size
    image['scaled_pixels'] = image['scaled_width'] * image['scaled_height']
    # Scaled HTML
    image['html'] = '<img src="data:image/png;base64,%s">' % image_to_base64(image['pil'], image['format'])


if __name__ == "__main__":
    # Use numpy to concatenate two arrays
    images = image_data(images = [{'source': "Peter Carolin", 'label': "Lassen Volcano", 'file': "lassen-volcano.jpg"}])
    
    # Display meta data, scaled view, and grey scale for each image
    for image in images:
        image_management(image)
        print("---- meta data -----")
        print(image['label'])
        print(image['source'])
        print(image['format'])
        print(image['mode'])
        print("Original size: ", image['size'], " pixels: ", f"{image['pixels']:,}")
        print("Scaled size: ", image['scaled_size'], " pixels: ", f"{image['scaled_pixels']:,}")
        
        print("-- original image --")
        display(HTML(image['html'])) 
---- meta data -----
Lassen Volcano
Peter Carolin
JPEG
RGB
Original size:  (2792, 2094)  pixels:  5,846,448
Scaled size:  (5000, 3750)  pixels:  18,750,000
-- original image --

Do you think this is a time complexity or space complexity or both problem?

This is a problem with both. This is a space complexity as the space taken up for the image increased as we increased our pixel size, but the time it took to render also increased. THis indicates that this is a problem with both space and time.

Big O Notation

  • Constant O(1)
  • Linear O(n)
  • Quadratic O(n^2)
  • Logarithmic O(logn)
  • Exponential (O(2^n))
numbers = list(range(50))
print(numbers)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49]

Constant O(1)

Time

An example of a constant time algorithm is accessing a specific element in an array. It does not matter how large the array is, accessing an element in the array takes the same amount of time. Therefore, the time complexity of this operation is constant, denoted by O(1).

print(numbers[26])

ncaa_bb_ranks = {1:"Alabama",2:"Houston", 3:"Purdue", 4:"Kansas"}
#look up a value in a dictionary given a key
print(ncaa_bb_ranks[1]) 
26
Alabama

Space

This function takes two number inputs and returns their sum. The function does not create any additional data structures or variables that are dependent on the input size, so its space complexity is constant, or O(1). Regardless of how large the input numbers are, the function will always require the same amount of memory to execute.

def sum(a, b): 
  return a + b

print(sum(90,88))
print(sum(.9,.88))
178
1.78

Linear O(n)

Time

An example of a linear time algorithm is traversing a list or an array. When the size of the list or array increases, the time taken to traverse it also increases linearly with the size. Hence, the time complexity of this operation is O(n), where n is the size of the list or array being traversed.

for i in numbers:
    print(i)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49

Space

This function takes a list of elements arr as input and returns a new list with the elements in reverse order. The function creates a new list reversed_arr of the same size as arr to store the reversed elements. The size of reversed_arr depends on the size of the input arr, so the space complexity of this function is O(n). As the input size increases, the amount of memory required to execute the function also increases linearly.

def reverse_list(arr):
    n = len(arr) 
    reversed_arr = [None] * n #create a list of None based on the length or arr
    for i in range(n):
        reversed_arr[n-i-1] = arr[i] #stores the value at the index of arr to the value at the index of reversed_arr starting at the beginning for arr and end for reversed_arr 
    return reversed_arr

print(numbers)
print(reverse_list(numbers))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49]
[49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

Quadratic O(n^2)

Time

An example of a quadratic time algorithm is nested loops. When there are two nested loops that both iterate over the same collection, the time taken to complete the algorithm grows quadratically with the size of the collection. Hence, the time complexity of this operation is O(n^2), where n is the size of the collection being iterated over.

for i in numbers:
    for j in numbers:
            print(i,j)

Space

This function takes two matrices matrix1 and matrix2 as input and returns their product as a new matrix. The function creates a new matrix result with dimensions m by n to store the product of the input matrices. The size of result depends on the size of the input matrices, so the space complexity of this function is O(n^2). As the size of the input matrices increases, the amount of memory required to execute the function also increases quadratically.

Example of Matrix Multiplication

  • Main take away is that a new matrix is created.
def multiply_matrices(matrix1, matrix2):
    m = len(matrix1) 
    n = len(matrix2[0])
    result = [[0] * n] * m #this creates the new matrix based on the size of matrix 1 and 2
    for i in range(m):
        for j in range(n):
            for k in range(len(matrix2)):
                result[i][j] += matrix1[i][k] * matrix2[k][j]
    return result

print(multiply_matrices([[1,2],[3,4]], [[3,4],[1,2]]))
[[18, 28], [18, 28]]

Logarithmic O(logn)

Time

An example of a log time algorithm is binary search. Binary search is an algorithm that searches for a specific element in a sorted list by repeatedly dividing the search interval in half. As a result, the time taken to complete the search grows logarithmically with the size of the list. Hence, the time complexity of this operation is O(log n), where n is the size of the list being searched.

def binary_search(arr, low, high, target):
    while low <= high:
        mid = (low + high) // 2 #integer division
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            low = mid + 1
        else:
            high = mid - 1

target = 263
result = binary_search(numbers, 0, len(numbers) - 1, target)

print(result)
None

Space

The same algorithm above has a O(logn) space complexity. The function takes an array arr, its lower and upper bounds low and high, and a target value target. The function searches for target within the bounds of arr by recursively dividing the search space in half until the target is found or the search space is empty. The function does not create any new data structures that depend on the size of arr. Instead, the function uses the call stack to keep track of the recursive calls. Since the maximum depth of the recursive calls is O(logn), where n is the size of arr, the space complexity of this function is O(logn). As the size of arr increases, the amount of memory required to execute the function grows logarithmically.

Exponential O(2^n)

for i in numbers:
    for j in numbers:
        print(i,j)

Time

An example of an O(2^n) algorithm is the recursive implementation of the Fibonacci sequence. The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding ones, starting from 0 and 1. The recursive implementation of the Fibonacci sequence calculates each number by recursively calling itself with the two preceding numbers until it reaches the base case (i.e., the first or second number in the sequence). The algorithm takes O(2^n) time in the worst case because it has to calculate each number in the sequence by making two recursive calls.

A visualization of calculating the fibonacci sequence

def fibonacci(n):
    if n <= 1:
        return n
    else:
        return fibonacci(n-1) + fibonacci(n-2)

#print(fibonacci(5))
#print(fibonacci(10))
#print(fibonacci(20))
# print(fibonacci(30))
print(fibonacci(40))
102334155

Space

This function takes a set s as input and generates all possible subsets of s. The function does this by recursively generating the subsets of the set without the first element, and then adding the first element to each of those subsets to generate the subsets that include the first element. The function creates a new list for each recursive call that stores the subsets, and each element in the list is a new list that represents a subset. The number of subsets that can be generated from a set of size n is 2^n, so the space complexity of this function is O(2^n). As the size of the input set increases, the amount of memory required to execute the function grows exponentially.

def generate_subsets(s):
    if not s:
        return [[]]
    subsets = generate_subsets(s[1:])
    return [[s[0]] + subset for subset in subsets] + subsets

print(generate_subsets([1,2,3,4,5,6]))
#print(generate_subsets(numbers))
[[1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5], [1, 2, 3, 4, 6], [1, 2, 3, 4], [1, 2, 3, 5, 6], [1, 2, 3, 5], [1, 2, 3, 6], [1, 2, 3], [1, 2, 4, 5, 6], [1, 2, 4, 5], [1, 2, 4, 6], [1, 2, 4], [1, 2, 5, 6], [1, 2, 5], [1, 2, 6], [1, 2], [1, 3, 4, 5, 6], [1, 3, 4, 5], [1, 3, 4, 6], [1, 3, 4], [1, 3, 5, 6], [1, 3, 5], [1, 3, 6], [1, 3], [1, 4, 5, 6], [1, 4, 5], [1, 4, 6], [1, 4], [1, 5, 6], [1, 5], [1, 6], [1], [2, 3, 4, 5, 6], [2, 3, 4, 5], [2, 3, 4, 6], [2, 3, 4], [2, 3, 5, 6], [2, 3, 5], [2, 3, 6], [2, 3], [2, 4, 5, 6], [2, 4, 5], [2, 4, 6], [2, 4], [2, 5, 6], [2, 5], [2, 6], [2], [3, 4, 5, 6], [3, 4, 5], [3, 4, 6], [3, 4], [3, 5, 6], [3, 5], [3, 6], [3], [4, 5, 6], [4, 5], [4, 6], [4], [5, 6], [5], [6], []]

Using the time library, we are able to see the difference in time it takes to calculate the fibonacci function above.

  • Based on what is known about the other time complexities, hypothesize the resulting elapsed time if the function is replaced.
import time

start_time = time.time()
print(fibonacci(34))
end_time = time.time()

total_time = end_time - start_time
print("Time taken:", total_time, "seconds")

start_time = time.time()
print(fibonacci(35))
end_time = time.time()

total_time = end_time - start_time
print("Time taken:", total_time, "seconds")

start_time = time.time()
print(fibonacci(36))
end_time = time.time()

total_time = end_time - start_time
print("Time taken:", total_time, "seconds")
5702887
Time taken: 1.1314542293548584 seconds
9227465
Time taken: 1.6828088760375977 seconds
14930352
Time taken: 2.7307589054107666 seconds

Hacks

  • Record your findings when testing the time elapsed of the different algorithms.
  • Although we will go more in depth later, time complexity is a key concept that relates to the different sorting algorithms. Do some basic research on the different types of sorting algorithms and their time complexity.
  • Why is time and space complexity important when choosing an algorithm? Space complexity is important as you must find the perfect balance between the size of an object and its quality. Too high of a quality and it my take up too much space and be in-effective, too low of a quality and nobody will enjoy it.
  • Should you always use a constant time algorithm / Should you never use an exponential time algorithm? Explain? It depends, an exponential time algorithm is useful for things like solving exponential functions. But, this also increases the time exponentially so it is not always useful to use an exponential time algorithm.
  • What are some general patterns that you noticed to determine each algorithm's time and space complexity? As the size of the image increased, the time that it took to render grew exponentially. Eventually, the program would just crash if the iamge was too large.

Notes

</p>

  • Time increases exponentially as quality increases
  • Different devices take drastically different times to render.
  • Logaritm growth can quickly break your computer by adding just one extra variable.
  • Going from printing 1-6 subset to 1-7 subset makes it take twice as long to render, even though there is just one new variable, there are way more new permutations.

</span>

Complete the Time and Space Complexity analysis questions linked below. Practice

RED FOR WRONG GREEN FOR RIGHT

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</p>

1

Code:

a = 0
b = 0
for i in range(N):
  a = a + random()

for i in range(M):
  b= b + random()

Choices: O(N M) time, O(1) space O(N + M) time, O(N + M) space O(N + M) time, O(1) space O(N M) time, O(N + M) space

Answer: The answer is the third option because the time is from variables M and N. Those are both independent from each other and therefore must be linked by using + where both are be used by O therefore time must be O(N+M). Now, for space, the sapce does not change as the outputs are variables so it must be O(1) therefore the answer is 3.

</span>

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</p>

2

Code:

a = 0;
for i in range(N):
  for j in reversed(range(i,N)):
    a = a + i + j;

Choices: O(N) O(Nlog(N)) O(N Sqrt(N)) O(N*N)

Answer: The answer is the last option because we are trying to find the time complexity. Since it prints all N values and then for each of those prints it in reverse, the time value relates to N*N.

</span>

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</p>

3

Code:

k = 0;
for i in range(n//2,n):
  for j in range(2,n,pow(2,j)):
        k = k + n / 2;

Choices: O(n) O(N log N) O(n^2) O(n^2Logn)

Answer: The answer is the third option because we are finding the values for i and j for N values. THis means the time complexity relates to N^2 variables so 3.

</span>

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</p>

4. What does it mean when we say that an algorithm X is asymptotically more efficient than Y?

Choices: X will always be a better choice for small inputs X will always be a better choice for large inputs Y will always be a better choice for small inputs X will always be a better choice for all inputs

Answer: I am not sure, I know that it means that X will be a better input for some things. But because the first and third options must both be true if one is true, and I know the last is false, it must be the third. I got this one wrong it is actually number 2 because x being larger doesnt force y to be the best for small inputs

</span>

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</p>

5

Code:

a = 0
i = N
while (i > 0):
  a += i
  i //= 2

Choices: O(N) O(Sqrt(N)) O(N / 2) O(log N)

Answer: The answer is the fourth one because for each value, we are searching for a value n where we are searching while the value i is greater than o so 4.

</span>

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</p>

6. Which of the following best describes the useful criterion for comparing the efficiency of algorithms?

Time Memory Both of the above None of the above

Answer: Both because when you compare efficiency you are trying to find which is the best for memory and fir time. </span>

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</p>

7. How is time complexity measured?

By counting the number of algorithms in an algorithm. By counting the number of primitive operations performed by the algorithm on a given input size. By counting the size of data input to the algorithm. None of the above

Answer: By counting the number of primitive operations performed by the algorithm on a given input size.

</span>

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</p>

8

Code:

for(var i=0;i<n;i++)
    i*=k

Choices: O(n) O(k) O(logkn) O(lognk)

Answer: Either 3 or 4: Correction it is 3 because it loops for k^n-1. </span>

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</p>

9

Code:

value = 0;
for i in range(n):
  for j in range(i):
    value=value+1

Choices: n (n+1) n(n-1) n(n+1)

Answer: The answer is the third option because the i un range n gives us n. But because the j is in range i, it needs to be one less so n-1 so n(n-1) therefore the answer is third option.

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