Got the exact same solution! It's a nice feeling haha.

regex is the best thing

yes, get a new way :)

same =(

Nice solution ;)

feel dumb

Because the second j starts from i+1 so it will never be equal to i and it skips unnecessary steps.
If your loop has j=0 replace it with j=i+1

I get the same solution lol

If there's no solution, better return an empty array than null.

I think by "different items" means different items of the array, not different values of those items. Hence values can be equal but items (indexes) must be different.

How can O(N^2) be "Best practice"?

The problem statement is "The input will always be valid (numbers will be an array of length 2 or greater, and all of the items will be numbers; target will always be the sum of two different items from that array)."

The input can be repeated numbers, i.e if target is 4 , input can be [2,2,1,3] . The above solution gives (0,1) , but actual ans is(2,3) as target will always be the sum of two different items from that array.

check() takes no input and each boggle board is constructed with a char[][] board and a word to search for.

I considered it! But I thought it was a bit pedantic. Glad someone is paying attention to the niceties... :-)