You can do it with two heaps. Not sure if there’s a less ‘contrived’ solution, but this one provides O(logn) time complexity and heaps are also included in standard libraries of most programming languages.
First heap (heap A) contains smallest 75% elements, another heap (heap B) – the rest (biggest 25%). First one has biggest element on the top, second one – smallest.
- Adding element.
See if new element x is <= max(A). If it is, add it to heap A, otherwise – to heap B.
Now, if we added x to heap A and it became too big (holds more than 75% of elements), we need to remove biggest element from A (O(logn)) and add it to heap B (also O(logn)).
Similar if heap B became too big.
- Finding “0.75 median”
Just take the largest element from A (or smallest from B). Requires O(logn) or O(1) time, depending on heap implementation.
edit
As Dolphin noted, we need to specify precisely how big each heap should be for every n (if we want precise answer). For example, if size(A) = floor(n * 0.75) and size(B) is the rest, then, for every n > 0, array[array.size * 3/4] = min(B).