problems do you think is harderit is probably not at all obvious to you that one problem is harder or easier than the other there is intuition that argues for either case on the one hand intuition might argue that the process of finding the maximum should tell you something about the second biggest valuemore than that process should tell you about the minimum value on the other handany given comparison tells you something about which of two can be candidate for maximum valueand which can be candidate for minimum valuethus making progress in both directions we will start by considering simple divide-and-conquer approach to finding the minimum and maximum split the list into two parts and find the minimum and maximum elements in each part then compare the two minimums and maximums to each other with further two comparisons to get the final result the algorithm is as follows /*return the minimum and maximum values in between positions and *static void minmax(int []int lint rint out[]if ( = / = out[ [ ]out[ [ ]else if ( + = / = out[ math min( [ ] [ ])out[ math max( [ ] [ ])else / > int[out new int[ ]int[out new int[ ]int mid ( )/ minmax(almidout )minmax(amid+ rout )out[ math min(out [ ]out [ ])out[ math max(out [ ]out [ ])the cost of this algorithm can be modeled by the following recurrence = = (nt(bn/ ct(dn/ this is rather interesting recurrenceand its solution ranges between / (when or + and / (when we can infer from this behavior that how we divide the list affects the performance of the algorithm