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I am confused about the time complexity of hash table many articles state that they are "amortized O(1)" not true order O(1) what does this mean in real applications. What is the average time complexity of the operations in a hash table, in actual implementation not in theory, and why are the operations not true O(1)?
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Insertion and Deletion The hash key is calculated in O(1) time complexity as always, and the required location is accessed in O(1). Insertion: In the best case, the key indicates a vacant location and the element is directly inserted into the hash table. So, overall complexity is O(1).
That is O(n). The other complexities are as expected on the hash table entry. The search complexity approaches O(1) as the number of buckets increases. If at the worst case you have only one bucket in the hash table, then the search complexity is O(n).
Thus, everyone knows that hash table queries run in amortized constant time. That is, as the number of keys increases, the average time necessary to recover a key-value pair does not increase.
The time complexity for 41-step SHA-256 is 2253. 5 compression function operations and the memory requirement is 216 × 10 words.
It's impossible to know in advance how many collisions you will get with your hash function, as well as things like needing to resize. This can add an element of unpredictability to the performance of a hash table, making it not true O(1). However, virtually all hash table implementations offer O(1) on the vast, vast, vast majority of inserts. This is the same as array inserting - it's O(1) unless you need to resize, in which case it's O(n), plus the collision uncertainty.
In reality, hash collisions are very rare and the only condition in which you'd need to worry about these details is when your specific code has a very tight time window in which it must run. For virtually every use case, hash tables are O(1). More impressive than O(1) insertion is O(1) lookup.
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