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Two for loops O(N^2) solution
softest mike
Nov 27, 2024
class Solution: def findSubarrays(self, arr, target): result = [] n = len(arr) for i in range(n): subarray_starting_at_i = [] curr_subarr_i = [] if arr[i] < target: curr_subarr_i.append(arr[i]) subarray_starting_at_i.append(curr_subarr_i.copy()) product = arr[i] for j in range(i+1, n): product = product * arr[j] if product < target: curr_subarr_i.append(arr[j]) subarray_starting_at_i.append(curr_subarr_i.copy()) else: break result.extend(subarray_starting_at_i) else: continue return result
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softest mikea year ago
In the worst case, there will be n-i runs at each iterator position. That gives 1+2+3+...+N. The sum of the series will result in O(N^2)