Two sum: 
Brute force approach: time: O(n^2) and space: O(1)
Two pointer approach (sorting the array): time: O(n log2(n)) and space: O(log2(n))
Set-based approach: time: O(n); space: O(n)

1	3	2		target: 3

set: 
1
3

Median finder (LeetCode 295):

1	2	3 -> middle element: 2 (odd number of values)
1	2	2	4 -> sum of two middle elements / 2 (even number of values)

3 2 4

PQ ---> Lower (Max-heap) 	and PQ ---> highers (Min-heap)

0 and 1 and 2 and 3		4 and 5

LeetCode 380: RandomizedSet problem














