Average case costs with separate chaining

• Assume a table with load factor α = N/M
• There are N items total distributed over M linked lists (some of which may be empty), so the average number of items per linked list is:
  
• In unsuccessful find/insert, one of the linked lists in the table must be exhaustively searched, and the average length of a linked list in the table is α . So the average number of comparisons for insert or unsuccessful find with separate chaining is

• In successful find, the linked list containing the target key will be searched. There are on average (N-1)/M keys in that list besides the target key; on average half of them will be searched before finding the target. So the average number of comparisons for successful find with separate chaining is

• These are less than 1 and 1.5 respectively, when α < 1
• And these remain O(1), independent of N, even when α exceeds 1.