How is it possible for Java HashMap to perform constant time lookup O(1) for “get” operations?

Question

I understand the basics of how a HashMap works - hm.put(obj) finds the correct bucket to place the object into, based on the obj.hashCode value. Then within that bucket if

Answer 1

Here is my two cent, friend. Think of a HashMap the way you think of an array. In fact, it is an array. If I give you index 11, you don't have to iterate through the array to find the object at index 11. You simply go there directly. That's how a HashMap works: the trick is to make the index the same as the value -- essentially.

So that is where a hashcode comes in. Let's look at the trivial case where your hash function is a unit multiplier (i.e. 1). Then if you have the values 0 to 99 and you want to store them in an array, then they will be stored at indices 0 to 99 respectively. So that a put and a get is clearly O(1) since getting and putting things in an array is O(1). Now let's imagine a less trivial hash function, say, y = x+2. So in this case the values 0 to 99 will be stored at indices 2 to 101. Here given a value, say 11, you must compute the hash to find it or put it (the hash being 11+2 =13). So okay, the hash function is doing some work to calculate the correct index given the value (in our case y = x+2= 11+2=13). But the amount of effort that goes into that work has nothing to do with how many data points you have. If I need to put 999 or 123, the amount of work for a single put or get is still the same: y= x+2: I only have to add two each time I do a put or a get: that's constant work.

Your confusion may be that you want to put n values at once. Well in that case each put is still constant c. But c multiplied by n is c*n=O(n). So the n has nothing to do with the put itself, but rather that you are making n puts all at once. I hope this explanation helps.