Given this naive recursive function:
def largest_tower(heights_and_widths, prev_height=1000, prev_width=1000): if not heights_and_widths: return 0 results = {0} for i in range(len(heights_and_widths)): current_height, current_width = heights_and_widths[i] if current_height < prev_height and current_width < prev_width: subarray = heights_and_widths[:i] + heights_and_widths[i+1:] results.add(1 + largest_tower(subarray, current_height, current_width)) return max(results) At first sight, I would have said that the time complexity of this function is O(n^2). But, if I’m not wrong, explained in plain words would be: This function may call itself (n-1) + (n-2) + ... + 1 times, being that O(n^2), but a total of n times, so the actual time complexity is O(n^3). Am I right?
And for the memoized version:
def largest_tower_memoized(heights_and_widths, prev_height=1000, prev_width=1000, cache=None): if not heights_and_widths: return 0 if cache is None: cache = {} results = {0} for i in range(len(heights_and_widths)): current_height, current_width = heights_and_widths[i] if current_height < prev_height and current_width < prev_width: subarray = tuple(heights_and_widths[:i] + heights_and_widths[i+1:]) if (current_height, current_width) not in cache: cache[(current_height, current_width)] = largest_tower_memoized( subarray, current_height, current_width, cache ) results.add(1 + cache[(current_height, current_width)]) return max(results) I find this more complicated to explain. My intuition tells me that it’s O(n), since we the recursion tree for each input is generated only once, but I’m not sure.
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