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§Sonar Sweep
The built in windows method comes in handy for this solution. For part 1 a straightforward
sliding window of size 2 allows us to compare each 2 consecutive values.
For part 2 we can use a trick to simplify. If we consider the first 2 windows of 3 elements each:
A1 A2 A3
B1 B2 B3
then the middle 2 elements are always in common, so the subsequent window is greater only if the last element is greater than the first. This means we can pick a sliding window of size 4 and compare the first and last elements, without having to sum intermediate elements.