Why hashmap lookup is O(1) i.e. constant time?

Under the appropriate assumptions on the hash function being used, we can say that hash table lookups take expected O(1) time (assuming you're using a standard hashing scheme like linear probing or chained hashing). This means that on average, the amount of work that a hash table does to perform a lookup is at most some constant.

Intuitively, if you have a "good" hash function, you would expect that elements would be distributed more or less evenly throughout the hash table, meaning that the number of elements in each bucket would be close to the number of elements divided by the number of buckets. If the hash table implementation keeps this number low (say, by adding more buckets every time the ratio of elements to buckets exceeds some constant), then the expected amount of work that gets done ends up being some baseline amount of work to choose which bucket should be scanned, then doing "not too much" work looking at the elements there, because on expectation there will only be a constant number of elements in that bucket.

This doesn't mean that hash tables have guaranteed O(1) behavior. In fact, in the worst case, the hashing scheme will degenerate and all elements will end up in one bucket, making lookups take time Θ(n) in the worst case. This is why it's important to design good hash functions.

For more information, you might want to read an algorithms textbook to see the formal derivation of why hash tables support lookups so efficiently. This is usually included as part of a typical university course on algorithms and data structures, and there are many good resources online.

Fun fact: there are certain types of hash tables (cuckoo hash tables, dynamic perfect hash tables) where the worst case lookup time for an element is O(1). These hash tables work by guaranteeing that each element can only be in one of a few fixed positions, with insertions sometimes scrambling around elements to try to make everything fit.

Hope this helps!

The key is in this statement in the docs:

If many mappings are to be stored in a HashMap instance, creating it with a sufficiently large capacity will allow the mappings to be stored more efficiently than letting it perform automatic rehashing as needed to grow the table.

and

The load factor is a measure of how full the hash table is allowed to get before its capacity is automatically increased. When the number of entries in the hash table exceeds the product of the load factor and the current capacity, the hash table is rehashed (that is, internal data structures are rebuilt) so that the hash table has approximately twice the number of buckets.

http://docs.oracle.com/javase/6/docs/api/java/util/HashMap.html

The internal bucket structure will actually be rebuilt if the load factor is exceeded, allowing for the amortized cost of get and put to be O(1).

Note that if the internal structure is rebuilt, that introduces a performance penalty that is likely to be O(N), so quite a few get and put may be required before the amortized cost approaches O(1) again. For that reason, plan the initial capacity and load factor appropriately, so that you neither waste space, nor trigger avoidable rebuilding of the internal structure.