Fast checking if array has repeated elements or not

I need to check whether a given array, A has repeated elements or not and return in true or false. The function runs near about 5*10^6 times. Initially I tried this

function flag = hasNoRepeats(A)
    flag = numel(A)==numel(unique(A)) ;
end

This is very slow, it takes about 90-100 seconds. It is given that array has all positive integers and they are in between 1 and K.

function flag = hasNoRepeats(A,K)
    tmp = zeros(1,K);
    flag = true;
    for i=1:numel(A)
        if tmp(A(i))
            flag = false;
            break
        end
        tmp(A(i)) = 1;
    end
end

This version is faster (~20 secs) but still slower than my requirements. Is there any way to do it faster?

[Mehmed Saad]

flag = sum(diff(sort(A)))>1;

Try this and tell me what time it got?

[Dhritishmam Sarmah]

Sorry, your code doesn't work. Try taking a random array, A=randi(20,1,6). But still I tried using sort and got improved time (~12sec).

function flag = hasNoRepeats(A,K)
    flag = true;
    A = sort(A);
    for i=1:(numel(A)-1)
        if A(i)==A(i+1)
            flag = false;
            break;
        end
    end
end

[Alex Mcaulley]

Using sort and diff you can try with:

flag = ~sum(~diff(sort(A)));

[Dhritishmam Sarmah]

Thanks Alex. But it seems like my implementation using sort gives the best timing yet so I will stick with it for the moment.

Answer 1

[KSSV]

You can simply use unique to find out whether a array has repeated elements or not.

Let A be your array.

if length(A)==length(unique(A))
    fprintf("A has no repeated elements") ; 
else
    fprintf("A has repeated elements") ; 
end