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CheezItMan
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| # Time complexity: n | ||
| # Space complexity: n | ||
| def factorial(n) |
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| # Time complexity: n^2 | ||
| # Space complexity: n^2 | ||
| def reverse(s) #was the answer to this supposed to be different from the one below?? |
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| # Time complexity: n | ||
| # Space complexity: n | ||
| def reverse_inplace(s) |
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👍 This is the same as the previous method and it's not in place.
Consider passing the front and rear indexes and swapping elements at each index.
def reverse_inplace(s, front = 0, last = s.length - 1)
return s if front >= last
temp = s[last]
s[last] = s[front]
s[front] = temp
return reverse_inplace(s, front + 1, last - 1)
end
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| # Time complexity: n | ||
| # Space complexity: n | ||
| def bunny(n) |
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| # Time complexity: n^2 | ||
| # Space complexity: n^2 | ||
| def nested(s) |
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| def search(array, value) |
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This works with O(n^2) space/time complexity.
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| # Time complexity: n | ||
| # Space complexity: n | ||
| def is_palindrome(s) |
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👍 But time/space complexity is O(n^2)
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| # Time complexity: n | ||
| # Space complexity: n | ||
| # | ||
| # how many calls | ||
| # how expensive is each call | ||
| def digit_match(n, m) |
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👍 This is O(log n) if n is the size of the smaller of the two numbers (basically you divide by 10 each recursive call).
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