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Scala Daily Coding Problem #001

awwsmm profile image Andrew (he/him) Updated on ・3 min read

Daily Coding Problem is a website which will send you a programming challenge to your inbox every day. I want to show beginners how to solve some of these problems using Scala, so this will be an ongoing series of my solutions. Feel free to pick them apart in the comments!


Given a list of numbers and a number k, return whether any two numbers from the list add up to k.

For example, given [10, 15, 3, 7] and k of 17, return true since 10 + 7 is 17.

Bonus: Can you do this in one pass?


See my discussion of this problem (and my Java implementation) here!


Rather than using nested loops in an imperative style, as I did in my Java solution, I'd like to stick with a more functional approach for my Scala solution to this problem.

Scala provides a beautiful standard library, including Lists with methods like combinations(n: Int) (which gives all combinations of n elements in the List), and exists(p: List[T] => Boolean), which returns true if there exists at least one element in the list which satisfies the given predicate, p. Combined with Scala's implicit parameters and Numeric type class under the hood (neither of which exist in Java), we can solve this problem in a single line of Scala code (not including the definitions of the list and k themselves):

val ll: List[Int] = List(10, 15, 3, 7)

val k: Int = 17

ll.combinations(2).exists(_.sum == k) // returns true
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That's it!


ll.combinations(2) returns a List[List[Int]]. That is, a List which contains List[Int] (lists of ints). The inner lists are all of the possible distinct pairs of elements of the list ll.

Calling exists(...) on a List checks if any element of the list satisfies the function passed as the argument to exists(...). (Note that this requires Scala's support for higher-order functions.) The function I passed, _.sum = k, checks each inner list (each element of ll) using the underscore _ as a placeholder for each element in turn. Since each element is itself a List[Int], we can call the sum function on that inner List, and simply check if the sum is equal to k.

Removing all this syntactic sugar, you might end up with something like:

// with syntactic sugar
ll.combinations(2).exists(_.sum == k)

// de-sugared
  (innerList: List[Int]) => innerList.sum == k
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...but the Scala compiler knows that the type of innerList must be List[Int], since ll is of type List[List[Int]], so its elements are List[Int]s. So we can drop the explicit type annotation:

  (innerList) => innerList.sum == k
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...and when we declare an anonymous function with only a single argument, as above, we can drop the parentheses around that argument:

  innerList => innerList.sum == k
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Finally, we can replace any number of anonymous function arguments with underscores _, and the Scala compiler will replace them, one by one, in the function definition:

ll.combinations(2).exists(_.sum == k)
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The end result of all of this syntactic sugar is a compact bit of code that is easy to read and error check. You can see why I like Scala so much, especially compared to the verbosity of my (not purposefully obfuscated) Java solution!

All the code for my Daily Coding Problems solutions is available at

Suggestions? Let me know in the comments.

If you enjoyed this post, please consider supporting my work by buying me a coffee!

Discussion (3)

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rodiongork profile image
Rodion Gorkovenko • Edited

Andrew, Hi!

This problem is often at interviews, I dare say, so be careful :)

Your final solution with combinations and exists is definitely not one pass. It is N*N but wrapped in thin layer of deceiving FP style :)

Try filling list with 1 million random elements and you'll see what I mean...

awwsmm profile image
Andrew (he/him) Author

Thanks for pointing that out! How would you solve this in a single pass?

rodiongork profile image
Rodion Gorkovenko

Putting elements to HashSet and looking up for K-element... It will make O(N) solution.