Pretty! :) Channa Bankapur wrote: > Elegant.. I think it can't be better than this. Identifying that each > of them are on different sides of S/2 was the key! > > > On Tue, Aug 4, 2009 at 10:05 AM, Prunthaban Kanthakumar > <pruntha...@gmail.com <mailto:pruntha...@gmail.com>> wrote: > > Here is the right answer: > > Find the sum of missing numbers. Call it S (this is a easy to do). > Now the two missing numbers are such that one is <=S/2 and the > other is > S/2 > Have two variables S1 and S2, traverse the array and add > everything <= S/2 to S1 and > S/2 to S2. > Now > First number = (Sum of numbers from 1 to S/2) - S1 > Second Number = (Sum of numbers from [S/2 + 1] to n+2) - S2 > > O(n) time and O(1) space. > > > On Tue, Aug 4, 2009 at 3:28 AM, Karthik Singaram Lakshmanan > <karthiksinga...@gmail.com <mailto:karthiksinga...@gmail.com>> wrote: > > > well..will this work? > > x + y = SUM(1:N+2) - SUM(array) = a > x^2 + y^2 = SUM(1^2:(N+2)^2) - SUM(array.^2) = b > so (a^2 - b) = 2xy > > so xy = (a^2-b)/2 = k (say) > > now, > > x + (k/x) = a > > x^2 + k = ax > (x, y) = (a +/- sqrt(a^2-4k))/2 > > I may not have written the equations correctly (need coffee !!!) > but you get the general idea... > solve a quadratic equation to solve for (x+y) = a and (x^2 + > y^2) = b > > - Karthik > > > > > > > > >
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