[yxhuvud]

Because it takes n time to access n units of memory.

Heavily simplified due to caches etc. To the point where people sometimes measure in cache misses instead as that is usually what actually matters.

[ajuc]

Let's say you have a lookup table and your algorithm looks up a value and returns it. O(n) space O(1) in time.

[yxhuvud]

n is some value depending on the size of the input, so if you have a look up table that is O(n), then that memory needs to be initialized somehow. If you have a fixed size lookup table then it is O(1), even if it is big.

[ajuc]

Compile time fill a lookup table. Run time read 1 value from it. I call it O(1) in time, O(n) in memory. You call it O(1) in both, right?

Now move the compile time code to runtime.

I call it O(n) in time and memory. What do you call it?

If your say O(n) in time and memory - why is moving the code changing its complexity?

If you say still O(1) - then everything is O(1), thus proving P=NP among other things :)

Either way your interpretation isn't very useful.

[xigoi]

What if you allocate a huge chunk of memory and only use a small part of it? For example, checking if a list of numbers contains duplicates using a boolean array.

[Sharlin]

If your algorithm requires O(n) memory, any O(1) amount of memory can never be enough, no matter how huge. That's the entire point of O notation.

And if your implementation of an algorithm allocates more space in the big-oh sense than it can actually touch (eg. O(n) space for O(log n) time or whatever), that's just a wasteful implementation. Doesn't make the algorithm itself require more space than it has time to actually use.