[Solved] Python reverse list with k steps, in-place operator vs standard operator [closed]

Question

The easiest way is to check how nums looks like after every operation (with print(nums) in scratch or calling it (nums) in terminal). Next time, try it yourself.

In A:
With the first line (nums[:k] = nums[len(nums)-k:]): nums = [5, 6, 7, 4, 5, 6, 7] – you’ve substituted nums[0:3] (1, 2, 3) with nums[4:7] (5, 6, 7).

Then, with the second line (nums[k:] = nums[:len(nums)-k]): nums = [5, 6, 7, 5, 6, 7, 4] – you’ve substituted nums[3:7] (4, 5, 6, 7) with nums[0:4] (5, 6, 7, 4).

In B:
You’ve done this operations “in place”, without changing the nums between them. You’ve “grabbed” 1, 2, 3, 4 and 5, 6, 7, made list [5,6,7,1,2,3,4] and assigned it to nums overwriting its value ([1,2,3,4,5,6,7]).

Accepted Answer

The easiest way is to check how nums looks like after every operation (with print(nums) in scratch or calling it (nums) in terminal). Next time, try it yourself.

In A:
With the first line (nums[:k] = nums[len(nums)-k:]): nums = [5, 6, 7, 4, 5, 6, 7] – you’ve substituted nums[0:3] (1, 2, 3) with nums[4:7] (5, 6, 7).

Then, with the second line (nums[k:] = nums[:len(nums)-k]): nums = [5, 6, 7, 5, 6, 7, 4] – you’ve substituted nums[3:7] (4, 5, 6, 7) with nums[0:4] (5, 6, 7, 4).

In B:
You’ve done this operations “in place”, without changing the nums between them. You’ve “grabbed” 1, 2, 3, 4 and 5, 6, 7, made list [5,6,7,1,2,3,4] and assigned it to nums overwriting its value ([1,2,3,4,5,6,7]).