Divide and conquer - Plural Array
I want to design an algorithm that determines whether an array A is plural, and return that element.
An array is plural when there exists an element x in the array if more than half of the array is the same as x.
I am wondering if there is a more efficient divide and conquer algorithm that runs better than the current one i have now.
Assume you have the array
aaabbcac
i will recursively split the array up until i get subarrays of size 2, as follows.
aa ab bc ac
from here, i will compare each element in the SUBARRAY to see if they are equal: If EQUAL, return the element, Else, return false.
aa ab bc ac
a f f f
so now i have an array of the element A and 3 false.
i was thinking of combining them like so:
a f f f
a f <----- combining a and f will give a
a <----- returns a
But, if we look at the array, we have 4 A's, which does not fulfill the criteria of a plural array. It should return false, should the array not be a plural array.
I believe my algorithm will run in O(n lgn), should it be a correct algorithm, which unfortunately is not.
Can anyon开发者_Go百科e point me in the right direction for this?
This is a problem of counting number of occurrences of x
. Split the array into sub arrays and send the x
along with sub arrays. Return count, sum counts and check if it's greater then the size of the array.
Since it's homework, here's clues - you should be able to prove these easily and finish the question.
- A single element array trivially has a plural element
- An array has at most one plural element, suppose it exists and call it x.
- If you partition the array into two halves, x will be a plural element of at least one of the halves.
You can also sort array by some sorting algorithm (such as quicksort), after that loop until (N+1)/2 element by checking if n+1 element is the same as n.
When using quicksort such approach would be with complexity O(n*log n + n/2)
. So basically my algorithm will be bound by sorting speed.
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