Leaves_Ga-Young #27
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Leaves_Ga-Young #27
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CheezItMan
reviewed
Sep 19, 2019
| counter = 0 | ||
| list.length.times do |num| | ||
| if list[counter] == list[counter + 1] | ||
| list.delete_at(counter + 1) |
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Since delete_at shifts all subsequent element left one index, it has a runtime of O(n). Using in your loop here gives you a method time complexity of O(n2)
| end | ||
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| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(n) where n is the strings array |
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If the strings are small yes, otherwise you can say O(n * m) where m is the min length of the strings.
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