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take two buckets
- priyank.rastogi1 September 20, 2012first element in bucket1
second element if different from first one then in bucket 2
else in bucket 1
if 3 rd element is different from both buckets then pop both buckets and forget 3 rd element
continue this till array ends.
finally the elements in the buckets may be the elements so we have to lookup two scans for the elements in buckets finally.
So time=3n ie O(n)