@Lucifer: a very good counter example involving the -ve numbers..[:)]
I never thought of negative numbers while looking for solution.
And I don't see a O(N) solution for this -->
1) Find the first pair (i,j) such that:
sum( A[0] .. till A[i]) = sum(B[0] ... B[i]) -- ESP when there are
-ve numbers, If No negative numbers its same as one provided before.
And, sorting the sums & comparing them like you suggested leads us to O(n^2
log n)
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