Fire - Noor#37
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CheezItMan
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Nov 4, 2020
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Some issues here, and other methods are well done. Take a look at my comments and let me know what questions you have via Slack.
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| # Time complexity: O(n) | ||
| # Space complexity: O(n) | ||
| def factorial(n) |
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| # Time complexity: O(n) | ||
| # Space complexity: O(n) | ||
| def reverse(s) |
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| # Time complexity: O(n) | ||
| # Space complexity: O(n) | ||
| def reverse_inplace(s) |
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Think about if we changed the method signature to the following
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| # Time complexity: O(n) | |
| # Space complexity: O(n) | |
| def reverse_inplace(s) | |
| # Time complexity: O(n) | |
| # Space complexity: O(n) | |
| def reverse_inplace(s, low = 0, high= s.length - 1) |
Can you see a way to do this with O(n) space/time complexity now?
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| # Time complexity: O(n) | ||
| # Space complexity: O(n) | ||
| def bunny(n) |
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| if s[i+1] == "(" && s[-i] == ")" | ||
| return true |
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| if s[i+1] == "(" && s[-i] == ")" | |
| return true | |
| if s[i-1] == "(" && s[-i] == ")" | |
| return nested(s, i + 1) |
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| # Time complexity: O(n) | ||
| # Space complexity: O(n) | ||
| def search(array, value, i = 0) |
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| # Time complexity: O(n) | ||
| # Space complexity: O(n) | ||
| def is_palindrome(s) |
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👍 Because of the slice you are doing with each recursive call, it's O(n^2)
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| if n[i] == m[i] | ||
| return count + 1 | ||
| else | ||
| return count | ||
| end | ||
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| return digit_match(n, m, i + 1, count) |
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| if n[i] == m[i] | |
| return count + 1 | |
| else | |
| return count | |
| end | |
| return digit_match(n, m, i + 1, count) | |
| if n[i] == m[i] | |
| count += 1 | |
| end | |
| return digit_match(n, m, i + 1, count) |
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| n.to_s | ||
| m.to_s |
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| n.to_s | |
| m.to_s | |
| n = n.to_s | |
| m = m.to_s |
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| # Time complexity: O(n) | ||
| # Space complexity: O(n) | ||
| def digit_match(n, m, i = 0, count = 0) |
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