## LuanJunYi

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I thought of a O(k*(logm + logn) solution.

The idea is to find the 1th minimum sum(which is a1+b1), the 2nd minimum sum, the 3rd..., until the kth minumum sum.

Denote the ith minimum sum is min(i). The key observation is that if min(i) = a(p) + b(q), min(i+1) is either a(p+1) + b(x) or a(y) + b(q+1), where b(x) is the minimum element in array b that hold b(x) + a(p+1) >= min(i), a(y) is the minimum element in array a that hold a(y) + b(q+1) >= min(i). min(i+1) is the smaller of the two.

Obviously, m(1) = a(1) + b(1), from where we can use iteration to compute m(k). Use binary search to locate a(x) need log(m) time, and locate b(y) need log(n) time. So k iteration need O(k*(logm + logn)) time.

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Correct solution. Besides, I think the same algorithm can be extended easily for input with negative elements in it.

- LuanJunYi January 07, 2013