Finding ah...isnt that simple... you have to do a binary search for it... that means that if you're on step j in the for above...you will have to do (n-j)lg(n-j)computations well...I think it would bring the algorithm to a total of O(sum from j=0 to n of (n-j)lg(n-j)< O(n^2lgn) if anyone can improve or find a better way...
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