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Better logic:
- nguyentruongtho.sg April 20, 2013A 3 element subset (3e) can be one of following form:
(a,b,c) (3 different numbers)
(a,a,b)
(a,a,a)
Now loop through the array and eliminate repeated number, mark number of appearance for them as (1, 2 or >3) ---> S1, S2, S3
List out all (a,b) which a is from S2 and b is from S1 or S2
List out all (a,b,c) which all from S1 and a<b<c
So, from that observation, you can see that we only need to separate number which appear only 1 time in the array and others which appear more than 1 in the array.
That is my idea, and the complexity can be minimal as N^3