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Find the Quadruplets - Hard


Problem - Given 4 arrays A,B,C,D. Find out if there exists an instance where A[i] + B[j] + C[k] + D[l] = 0

Like the Find the Triple problem, we're going to develop 4 algorithms to solve this. Starting with the naive O(n^4) solution.

Then we proceed to eliminate the inner-most loop with a Binary Search, reducing the complexity to O(n^3 logn)

Now, we replace the last 2 loops with the left-right traversal we did in the previous 3 posts.

Now the complexity is O(n^3).

Finally, we reduce the complexity to O(n^2 logn) at the cost of O(n^2) Space Complexity. We store every combination of A[i] + B[j] and store it in AB[]. Similarly we make CD[] out of C[i] + D[j].


So,

AB = A x B
CD = C x D

We then sort AB and CD (which costs O(n^2 log(n^2)) ~ O(n^2 logn) ) and then run a left-right linear Algorithm on AB and CD. (Note : Their size is of the order O(n^2))

So the overall complexity is due to sorting the large array of size n^2. which is O(n^2 logn).
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