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When left neighbour is present, instead of just updating (N-1)->F->L, you should update the L value for every number between (N-1)->F and N. For instance, suppose 1, 2, 3 are present in the hashtable and now 4 is inserted, you need to update the L value for all 1, 2, 3 to 4.
- alphayoung August 03, 2011Thus the time complexity is O(nL) with L being the maximum length of consecutive numbers.