Space Complexity with O(n^2)

Question

So, Everbody is familiar with constant O(1) or Linear O(N) Space Complexity.

But I have a question that, is there any case when Space Complexity of an Algorithm is proportional to O(NLogn) or O(N^2). If that's possible, what is the advantage of it.

P.S.- I have researched over various sites but didn't got any satisfactory solution.

Any algorithm which returns a result of size `f(N)` takes at least `O(f(N))` additional space (provided of course the result does not share space with the input). — Mo B., Jan 08 '21 at 14:15

score 2 · Accepted Answer · answered Jan 08 '21 at 12:28

Almost any algorithm can be made to use O(N^2) memory. Consider some f(a,b) where 0 < a,b < N and f is expensive to calculate. To reduce runtime an obvious solution is to use a look up table of size N * N with pre calculated results. Such kind of trade-off between runtime and memory usage can be made very often.

In general, algorithms that are using matrices typically take N*N memory to store the matrices. For example to rotate a point in N=3 dimensions, you can use a 3x3 rotation matrix.

Space Complexity with O(n^2)

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