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Find the Triple - Hard

Problem : Given 3 Integer Arrays A,B,C. Find out if an instance exists where A[i] + B[j] + C[k] == 0.

Very deceptive question but very similar to the 3SUM Problem. The Naive solution is O(n^3). If you're a little clever, you can do a O(n^2 logn). At this point, you might give up assuming that this is pretty efficient. But then there is also a O(n^2) solution as well!

Let's go from worst to best.

The Naive solution is trivial.
FIND-TRIPLET-1(A,B,C)
FOR i = 1 to len(A)
FOR j = 1 to len(B)
FOR k = 1 to len(C)
if A[i] + B[j] + C[k] == 0
RETURN {A[i],B[j],C[k]}
RETURN {}
END-FUNCTION
view raw gistfile1.txt hosted with ❤ by GitHub
Try every combination until we hit a solution, if we don't there is no solution.

So let's try to speed things up a bit, since the solution is already O(n^3), I think we can afford to sort any of the  arrays since nlogn = o(n^3). Therefore, we cleverly sort the Array C. Now in the third loop we replace the linear scan with a binary search to find (-A[i]-B[j]). The complexity is now reduced to O(n^2 logn)

FIND-TRIPLET-2(A,B,C)
SORT(C)
FOR i = 1 to len(A)
FOR j = 1 to len(B)
k = BINARY-SEARCH(C,-A[i]-B[j])
if k > 0
RETURN {A[i],B[j],C[k]}
RETURN {}
END FUNCTION
view raw gistfile1.txt hosted with ❤ by GitHub
At this point, you've leveled up. The next algorithm might not be immediately apparent but works. In fact the method is similar to the O(n^2) solution to 3SUM.

FIND-TRIPLET-3(A,B,C) // A[1..len(A)]
SORT(A)
SORT(B)
FOR i = 1 to len(C)
left = 1
right = len(B)
WHILE left <= len(A) AND right > 0
SUM = A[left] + B[right] + C[i]
IF SUM == 0
return {A[left] , B[right] , C[i]}
ELSE IF SUM < 0
left = left + 1
ELSE
right = right - 1
END-WHILE
RETURN {}
END FUNCTION
view raw gistfile1.txt hosted with ❤ by GitHub
Java Code for all these solutions can be found here.

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