Fire - Anna #28
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Fire - Anna #28
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CheezItMan
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Sep 22, 2020
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CheezItMan
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Well done Anna, you hit the learning goals here. Nice work!
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| # Complexity assessment: | ||
| # Note: where n = length of the input array | ||
| # Time complexity: O(n); linear; at most will parse through each array elem once | ||
| # Space complexity: O(1); constant; excluding the input space, | ||
| # the variables included in the function are all constants | ||
| def length(array) |
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| # Complexity assessment: | ||
| # Note: where n = length of the input array | ||
| # Time complexity: O(n); linear; prints each elem of array | ||
| # Space complexity: O(1); constant; space for index (int) | ||
| def print_array(array) |
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| # Complexity assessment: | ||
| # Note: where n = length of the input array | ||
| # Time complexity: O(n); linear; worst case goes through array from beginning to end to find value | ||
| # Space complexity: O(1); constant; doesn't require more space than input array and index constant | ||
| def search(array, length, value_to_find) |
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| # Complexity assessment: | ||
| # Note: where n = length of the input array | ||
| # Time complexity: O(n); linear; worst case will parse through array from beginning to end to find max | ||
| # Space complexity: O(1); constant; does not require new vars aside from index (int) and max value (int) | ||
| def find_largest(array, length) |
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| # Complexity assessment: | ||
| # Note: where n = length of the input array | ||
| # Time complexity: O(n); linear; worst case will parse through array from beginning to end to find min | ||
| # Space complexity: O(1); constant; does not require new vars aside from index (int) and min value (int) | ||
| def find_smallest(array, length) |
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| # Complexity assessment: | ||
| # Note: where n = length of the input array | ||
| # Time complexity: O(n); linear; traverse through half of array O(n/2) --> O(n) | ||
| # Space complexity: O(1); constant; does not require new vars aside from index (int) and reverse_ind (int) | ||
| def reverse(array, length) |
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| # Complexity assessment: | ||
| # Note: where n = length of the input array | ||
| # Time complexity: O(log n); logarithmic; half of array is eliminated with each search: | ||
| # doubling size of array only increases search count by 1 | ||
| # Space complexity: O(1); constant; does not require new vars aside from begin_ind (int), end_ind (int), and midpoint (int) | ||
| def binary_search(array, length, value_to_find) |
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