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@anakp07 anakp07 commented Oct 19, 2020

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Ana, well done. Take a look at my comment on find_smallest and let me know what questions you have there. Thanks for getting this in.

@@ -0,0 +1,7 @@
<?xml version="1.0" encoding="UTF-8"?>
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I suggest you add .idea to your .gitignore file.

i += 1
end
return i
end
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Suggested change
end
end

Comment on lines +10 to 13
# Time complexity: Linear or O(n) --> it will iterate over each of the elements inside array. As the input grows, the greater the amount of time it takes to perform the function.
# Space complexity: Constant of O(1) --> we only create one new variable and it doesnt depend on the size of the array.

def length(array)
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Comment on lines +22 to 24
# Time complexity: Linear or O(n) --> it will iterate over each of the elements inside array. As the input grows, the greater the amount of time it takes to perform the function.
# Space complexity: Constant of O(1) -> only one variable is created and its used throughout the entire function.
def print_array(array)
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Comment on lines +35 to 37
# Time complexity: Linear or O(n) --> it will need to iterate over all elements in worse case, which means it will grow through each iteration.
# Space complexity: Constant or O(1)
def search(array, length, value_to_find)
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Comment on lines +50 to 52
# Time complexity: Linear or O(n)
# Space complexity: Constant or O(1)
def find_largest(array, length)
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Comment on lines +81 to 83
# Time complexity: Linear or O(n)
# Space complexity: Constant or O(1)
def reverse(array, length)
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# Space complexity: Constant of O(1) - two variables are created and are used throughout the function
def find_smallest(array, length)
raise NotImplementedError
min = 0
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If the array has positive numbers starting min at 0, will cause these to fail.

Suggested change
min = 0
min = array[0]

Comment on lines +96 to 98
# Time complexity: Logarithmic or O(log n) - I remembered from lessons that binary search is 0(log n)
# Space complexity: Constant or O(1)?
def binary_search(array, length, value_to_find)
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2 participants