[Haskell-beginners] Haskell triangular loop (correct use of (++))

[rohan sumant]

Suppose I have a list of distinct integers and I wish to generate all
possible unordered pairs (a,b) where a/=b.

Ex: [1,2,3,0] --> [(1,2),(1,3),(1,0),(2,3),(2,0),(3,0)]

 The approach I am following is this :-

mkpairs [] = []
mkpairs (x:xs) = (map (fn x) xs) ++ (mkpairs xs)

fn x y = (x,y)

It is generating the desired output but I am a bit unsure about the time
complexity of the function mkpairs. In an imperative language a nested
triangular for loop would do the trick in O(n^2) or more precisely
(n*(n-1)/2) operations. Does my code follow the same strategy? I am
particularly worried about the (++) operator. I think that (++) wouldn't
add to the time complexity since the initial code fragment (map (fn x) xs)
is to be computed anyway. Am I wrong here? Is this implementation running
O(n^2)? If not, could you please show me how to write a nested triangular
loop in Haskell?

Rohan Sumant
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