Submission - wilson-romero #10
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0xp4blo
suggested changes
May 23, 2020
solutions/wilson_romero/solution.py
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| while True: | ||
| try: | ||
| next_index = arr.index(0, next_index) | ||
| arr.insert(next_index + 1, 0) |
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Watch out with this operation. Your solution isn't optimum, its O(n^2) in time and O(n) in space. There's a O(n) in time and O(1) in space solution.
Contributor
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Think about what happens to the array when you do arr[step + 1:] = arr[step:-1]What happens what the input array is |
0xp4blo
suggested changes
May 26, 2020
Comment on lines
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| zeros = [] | ||
| for step, value in enumerate(arr): | ||
| if value == 0 and step >= index: | ||
| arr[step + 1:] = arr[step:-1] | ||
| index = step + 2 | ||
| if value == 0: | ||
| zeros.append(step) |
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This is creating additional space, i.e.: for an array full of 0s of length n, you'll be creating an auxiliary array of size n containing its indices.
solutions/wilson_romero/solution.py
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Comment on lines
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| for step, value in enumerate(zeros): | ||
| arr[value + 1:] = arr[value:-1] |
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The operation performed with slices is highly inefficient. This for loop itself have time complexity O(n^2).
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