submitting currently solved methods - marj #46
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CheezItMan
reviewed
Sep 23, 2020
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| # Time complexity: Time complexity is O(1) because the arry has a fixed length of 20 | ||
| # Space complexity: I think also O(1) since it is a fixed amount that it can hold | ||
| def length(array) |
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👍 Although the time complexity is O(n) because you have to travel the length of the array and it will vary by array size.
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| # Time complexity: if array length does not got past 20, o(1), but if it increases o(n) because the more/less values the more/less time to process | ||
| # Space complexity: pretty much the same as above, there are no new values added past 20 or new arrays created | ||
| def print_array(array) |
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You need a loop like length.
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| # Time complexity: o(n) because it's not sorted so the more values it has to go through to find a specific value, the longer it takes | ||
| # Space complexity: o(1) just sorting and not adding values | ||
| def search(array, length, value_to_find) |
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| # Time complexity: for each value that is not the largest value, the more time it will take - o(n) | ||
| # Space complexity: no new arrays - o(1) | ||
| def find_largest(array, length) |
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| # Time complexity: same as for largest integer in unsorted array - or each value that is not the largest value, the more time it will take - o(n) | ||
| # Space complexity: same - no new arrays - o(1) | ||
| def find_smallest(array, length) |
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| # Time complexity: for set length O(1), for any array length O(n) | ||
| # Space complexity: O(1) because reverses in place, no new array is made | ||
| def reverse(array, length) |
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👍 Although the time complexity is O(n) because you have to travel half the length of the array and it will vary by array size.
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| # Time complexity: o(nlog n) - since doing the opposite of loop and cutting array in half | ||
| # Space complexity: o(1) since no new array is being created | ||
| def binary_search(array, length, value_to_find) |
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Looks like you have a good start, just not a finished result.
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