No of comparisons to find 2nd smallest number? you have 32 numbers. What is the least number of comparison needed to find the 2nd smallest out of them.
As per me it 61 comparsisons are required. Compare first two number in list. Assign largest and second largest to them  based on comparison. Now compare rest 30 numbers with two of them and assign largest and second largest based on comparison. So for rest of 30 numbers we have to do 2 comparison for each number . this answer is 30*2 + 1 = 61. Can 61 be reduced?

35 comparisons are optimal. Pair off the numbers and test them against each other. Take the winners, the winners' winners, and so forth to find the maximal element; this takes 31 comparisons in all. Now the maximal element was compared against five other elements, and the second-largest element is precisely the largest of these five. You can find it in four comparisons, for a total of 35 comparisons.
