Arrays exercise submission @ Water class-Hanh Solo #36
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seattlefurby17
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CheezItMan
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Sep 22, 2020
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| # Time complexity: There is only one while loop where we iterate over each | ||
| # element of the array. Time complexity will be O(n) time where n is how many | ||
| # times the loop will execute. | ||
| # Space complexity: The method used one local integer variable | ||
| # no matter the input size, so the space complexity is constant or O(1). | ||
| def length(array) |
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| # Time complexity: There is only one while loop where we iterate over each | ||
| # element of the array. Time complexity will be n time where n is how many | ||
| # times the loop will execute. | ||
| # Space complexity: constant or O(1) | ||
| def print_array(array) |
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| # Time complexity:O(n) There is only one while loop where we iterate over each | ||
| # element of the array. Time complexity will be n time where n is how many | ||
| # times the loop will execute. | ||
| # Space complexity: constant or O(1) | ||
| def search(array, length, value_to_find) |
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| # Time complexity:This method will iterate through the entire array. Thus if the | ||
| # array size doubles the loop will take twice as long. So the time complexity is O(n) | ||
| # Space complexity: This method only creates two additional variables, | ||
| # index and max which does not change no matter how large the array gets. | ||
| # Therefore the space complexity does not change as the size of the array increases. | ||
| # This means the space complexity is O(1). | ||
| def find_largest(array, length) |
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| # Time complexity: This method will iterate through the entire array. Thus if the | ||
| # array size doubles the loop will take twice as long. So the time complexity is O(n) | ||
| # Space complexity: This method only creates two additional variables, | ||
| # index and min which does not change no matter how large the array gets. | ||
| # Therefore the space complexity does not change as the size of the array increases. | ||
| # This means the space complexity is O(1). | ||
| def find_smallest(array, length) |
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| # Time complexity:O(n) this method only iterate through half of the array however | ||
| # it is depends on the size of the input. The longer the array, the longer it takes or | ||
| # linear complexity. | ||
| # Space complexity: constant | ||
| def reverse(array, length) |
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| # Time complexity: Log2 n means at each step the value reduces by half of the remaining. | ||
| # It also means that as the size of the array doubles, | ||
| # the number of iterations only increases by 1. | ||
| # The main loop in binary search runs log2 n number of times where | ||
| # n is the number of elements in the input array. | ||
| # Space complexity: constant | ||
| def binary_search(array, length, value_to_find) |
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