Data Structures- Space and Time Complexity
Observing the time complexity of different algorithms
- Space and Time Complexity
- Constant O(1)
- Linear O(n)
- Quadratic O(n^2)
- Logarithmic O(logn)
- Exponential O(2^n)
- Hacks
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']))
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.
numbers = list(range(50))
print(numbers)
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])
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))
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)
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))
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.
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]]))
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)
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.
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.
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))
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))
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")
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>
</span> Complete the Time and Space Complexity analysis questions linked below.
Practice </p>
Code: 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> </p>
Code: 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> </p>
Code: 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> </p>
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> </p>
Code: 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> </p>
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> </p>
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> </p>
Code: 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> </p>
Code: 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. </span>
RED FOR WRONG GREEN FOR RIGHT
</div>
</div>
</div>
1
a = 0
b = 0
for i in range(N):
a = a + random()
for i in range(M):
b= b + random()
2
a = 0;
for i in range(N):
for j in reversed(range(i,N)):
a = a + i + j;
3
k = 0;
for i in range(n//2,n):
for j in range(2,n,pow(2,j)):
k = k + n / 2;
4. What does it mean when we say that an algorithm X is asymptotically more efficient than Y?
5
a = 0
i = N
while (i > 0):
a += i
i //= 2
6. Which of the following best describes the useful criterion for comparing the efficiency of algorithms?
7. How is time complexity measured?
8
for(var i=0;i<n;i++)
i*=k
9
value = 0;
for i in range(n):
for j in range(i):
value=value+1