I found a similar solution looking somewhere else.
The solution for this problem is:
a. There can be atmost 3 elements (only 3 distinct elements with each
repeating n/3 times) -- check for this and done. -- O(n) time.
b. There can be atmost 2 elements if not above case.
1. Find the median of the input. O(N)
2. Check if median element is repeated N/3 times or more -- O(n) - *{mark
for output if yes}*
3. partition the array based on median found above. - O(n) -- {partition
is single step in quicksort}
4. find median in left partition from (3). - O(n)
5. check if median element is repeared n/3 times or more - O(n) *{mark for
output if yes}*
6. find median in right partition from (3). - O(n)
7. check if median element is repeared n/3 times or more - O(n) *{mark
for output if yes}*
its 7*O(N) => O(N) solution. Constant space.
we need not check further down or recursively. {why? reason it.. its simple}
On Wednesday, 27 June 2012 18:35:12 UTC-4, Navin Kumar wrote:
>
>
> Design an algorithm that, given a list of n elements in an array, finds
> all the elements that appear more than n/3 times in the list. The algorithm
> should run in linear time
>
> ( n >=0 ).
>
> You are expected to use comparisons and achieve linear time. No
> hashing/excessive space/ and don't use standard linear time deterministic
> selection algo.
>
>
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