Wrote recursive methods #33
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CheezItMan
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Nov 4, 2020
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| # Time complexity: O(n) - it takes one recursion for every increase in n | ||
| # Space complexity: O(n) - because of the system stack | ||
| def factorial(n) |
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| # Time complexity: O(n) - It takes n recursions for each character | ||
| # Space complexity: O(n^2) - n is number of characters | ||
| # String objects with length n * (n-1) * (n-2) ... are made | ||
| def reverse(s) |
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👍 , however your space/time complexity are both O(n^2)
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| # Time complexity: O(n) - It will take n/2 swaps to reverse the string | ||
| # Space complexity: O(n) - For the system stack | ||
| def reverse_inplace(s) |
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| # Time complexity: O(n) - one recursion for each bunny | ||
| # Space complexity: O(n) - system stack | ||
| def bunny(n) |
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| # Time complexity: O(n) - has to look through each character | ||
| # Space complexity: O(n) - system stack | ||
| def nested(s) |
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This is working if the number of "(" and ")" match, but what about ")("?
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| # Time complexity: O(n) - Linear search | ||
| # Space complexity: O(n) - system stack | ||
| def search(array, value) |
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| # Time complexity: O(n) - n/2 recursions | ||
| # Space complexity: O(n) - system stack | ||
| def is_palindrome(s) |
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| # Time complexity: O(logn) - Where n is the larger of n or m. | ||
| # Reduces n by a factor of 10 every recursion | ||
| # Space complexity: O(logn) - System stack | ||
| def digit_match(n, m) |
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| # Time complexity: O(n) - One recursion for every n | ||
| # Space compexlity: O(n) - System stack | ||
| def fib(n, f0 = 0, f1 = 1, index = 1) |
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Nice work on a tail recursive version of Fibonacci.
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