Restricted Arrays #40
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CheezItMan
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Sep 22, 2020
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| # Time complexity: 0(n) Linear: iteration depends on length, we have to go through the array until we find the termination character nil. | ||
| # Space complexity: O(1); uses constant memory, doesn't matter how big array is, we have one counter, no new data structures being created | ||
| def length(array) |
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| # Time complexity: O(n) linear, in order to print every element in array we have to go through entire array. | ||
| # Space complexity: O(1) uses constant amount of memory, prints elements in array. No matter how array changes, | ||
| # new memory is not taken up for this method. | ||
| def print_array(array) |
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| # Time complexity: O(n) linear, have to iterate through array until we find it. | ||
| # Space complexity: O(1), constant memory, | ||
| def search(array, length, value_to_find) |
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| # Time complexity: O(n) linear: since it's unsorted, will have to go through the entire array until the largest value is found. | ||
| # Space complexity: O(1) memory usage does not change for this method, it is constant. | ||
| def find_largest(array, length) |
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| # Time complexity: O(n) linear: since it's unsorted, will have to go through the entire array until the smallest value is found. | ||
| # Space complexity: O(1) memory usage does not change for this method, it is constant. | ||
| def find_smallest(array, length) |
lib/using_restricted_array.rb
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| # Time complexity: O(n) execute in linear time as compared to the size of the input | ||
| # Space complexity: O(1), since it is in place. New memory is not needed to store the reversed array. | ||
| # Amount of memory used does not change as the size of the input array changes. | ||
| def reverse(array, length) |
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CheezItMan
reviewed
Dec 6, 2020
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| # Time complexity: O(log n), the data set size affects the efficiency of the algorithm in a logarithmic fashion | ||
| # Space complexity: O(1), memory used does not change as the size of the input array changes | ||
| def binary_search(array, length, value_to_find) |
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