prolog: maximally repeated element in a list - prolog

Question

Any ideas how to retrieve the maximally repeated element in a list.
i.e. something like below,
?- maxRepeated([1,2,7,3,6,1,2,2,3],M).
M = 2.

Answer 1

I like so much the relational Prolog power:
maxRepeated(L, M) :-
sort(L, S),
maplist(count(L), S, C),
keysort(C, [_-M|_Ms]).
count(L, S, I-S) :-
aggregate(count, member(S, L), C), I is -C.
test:
?- maxRepeated([1,2,7,3,6,1,2,2,3],M).
M = 2.
edit and now, still more compact!
maxRepeated(L, M) :-
setof(I-E, C^(aggregate(count, member(E, L), C), I is -C), [_-M|_]).

Answer 2

This solution sorts the list, granting elements to appear sequentially -- there's no need to maintain all elements, once they're not repeating later.
Your prolog interpreter must have the function msort(), which sorts a list maintaining duplicated entries.
maxRepeated([], []).
maxRepeated(L, E) :-
msort(L, [H|T]),
maxRepeated(T, H, H, 1, 0, E).
maxRepeated([], H, _, C1, C2, H) :- C1 >= C2.
maxRepeated([], _, X, C1, C2, X) :- C1 < C2.
maxRepeated([H|T], H, LastF, C1, C2, E) :-
maxRepeated(T, H, LastF, C1 + 1, C2, E).
maxRepeated([X|T], H, LastF, C1, C2, E) :-
(
C1 > C2
-> maxRepeated(T, X, H, 1, C1, E)
; maxRepeated(T, X, LastF, 1, C2, E)
).
The complexity is given by the sort used, usually O(n log n), once, after the sort, the list is traversed only once, aggregating the elements and keeping track of the most frequent one.
Regards!

Answer 3

If you know the max value that Ai can take and if this Amax is such that you can create an array as large as Amax then there is a method by which you can find the most repeated element in O(n) time.
int A[max+1]; // set all elements to 0
int S[n]; // Set S
for (i=0;i<n;i++) A[ S[i] ]++;
int m=0, num; // num is the number to be found
for (i=1;i<=max;i++)
if (A[i] > m)
{
m = A[i];
num = i;
}
print (num)

Answer 4

Here is a quick and dirty answer. I constrained the problem to a set of allowed elements. Works but needs elaboration.
maxRepeated([],_,Current,_,Current).
maxRepeated([H|T],L,Current,MaxCount,X) :-
(
count(L,H,N),
N > MaxCount,
maxRepeated(T,L,H,N,X)
)
;
maxRepeated(T,L,Current,MaxCount,X).
count([],X,0).
count([X|T],X,Y):- count(T,X,Z), Y is 1+Z.
count([X1|T],X,Z):- X1\=X,count(T,X,Z).