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Well, how about this? .... Let the sum of all elements in the array be S. First, forget about the wrap-around case and find the max sum of the 1D array using the Kadane approach. Say its M. Next, modify the Kadane algo a bit to find the max negative sum. Say its -N. Now check if (S+N) > M. If yes, the answer is S+N, otherwise its M. I think this should work. However please let me know if you find any case for which this logic might fail.
- sam8dec February 23, 2012