Very Nice Simple approach @Dave On Sep 24, 12:56 am, Dave <[email protected]> wrote: > Do a single-elimination tournament of the numbers, where the larger > wins each "game". This will take n/2 + n/4 + ... + 1 <= n-1 > comparisons. The second largest will be among the numbers that lost to > the largest in one of the "games". As you conduct the tournament, keep > track of the losers. Since there will be at most ceiling(log_2(n)) > stages in the tournament, you can find the largest of the losers to > the winner in ceiling(log_2(n))-1 comparisons. > > Dave > > On Sep 23, 1:36 pm, Divesh Dixit <[email protected]> > wrote: > > > finding largest and second largest elements from a set of n elements > > by means of > > Minimum comparison of n+celling(log n) +2....
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