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| Original file line number | Diff line number | Diff line change |
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| @@ -1,8 +1,20 @@ | ||
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| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(n) | ||
| # Space Complexity: O(1) | ||
| def max_sub_array(nums) | ||
| # you calculated | ||
| return 0 if nums == nil | ||
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| raise NotImplementedError, "Method not implemented yet!" | ||
| current_max = nums[0] | ||
| global_max = nums[0] | ||
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| i = 1 | ||
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| while i < nums.length | ||
| current_max = [nums[i], (current_max + nums[i])].max | ||
| global_max = [global_max, current_max].max | ||
| i += 1 | ||
| end | ||
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| return global_max | ||
| end | ||
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,7 +1,41 @@ | ||
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| # Time complexity: ? | ||
| # Space Complexity: ? | ||
| # Time complexity: O(n) | ||
| # Space Complexity: O(1) | ||
| def newman_conway(num) | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. How would you have O(1) space complexity? You're building an array. Otherwise well done. |
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| raise NotImplementedError, "newman_conway isn't implemented" | ||
| end | ||
| raise ArgumentError, "num must be >= 0" if num <= 0 | ||
| return "1" if num == 1 | ||
| return "1 1" if num == 2 | ||
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| # I made a new premade array that was the length plus one of num that was passed in | ||
| array = Array.new(num + 1){} | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Maybe call it something other than just generically |
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| # Build a memo of subproblems | ||
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| # value at index 1 and 2 are set two zero, but why is value at index 0 nil? How did everyone initialize | ||
| # their array? | ||
| array[1] = 1 | ||
| array[2] = 1 | ||
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| # initialize i to 3 because values at index 1 and 2 are already set. | ||
| i = 3 | ||
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| while i < array.length | ||
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| # I used the sum of previous values in the array to get the next value. So I initialized i to index 3 since that is the next | ||
| # value that needs to be calculated. As a result of plugging 3 into that equation, I got 2. So the value at index 3 is 2. And you'd just keep | ||
| # doing that until you go past the length of the array. | ||
| if i > 2 | ||
| array[i] = array[array[i - 1]] + array[i - array[i - 1]] | ||
| end | ||
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| i += 1 | ||
| end | ||
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| # used shift to pop off first value which was nil | ||
| array.shift() | ||
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| # joined them so answer would come out as string instead of array to pass the test | ||
| return array.join(" ") | ||
| end | ||
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👍 This is really elegantly done.