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From the outset, it is obvious that the memory complexity is $\mathcal{O}(n)$  The cost of initializing the memory is: $n$  The  algorithm will do $n$ fills. It will also do: $i - 1$ lookup for each $i$, so the total complexity is:  $T(n) = n + n + \displaystyle\sum_{i=0}^{n} [i-1]$