further improve our run time by choosing the shorter array for the outer loop since if is much smaller than mthen log(mis much smaller than log(nthis is the best solution if one set is much smaller than the other howeverit is not the best when the array lengths are similar because we are not exploiting the fact that both arrays are sorted we can achieve linear runtime by simultaneously advancing through the two input arrays in increasing order at each iterationif the array elements differthe smaller one can be eliminated if they are equalwe add that value to the intersection and advance both (we handle duplicates by comparing the current element with the previous one for exampleif the arrays are and ithen we know by inspecting the first element of each that cannot belong to the intersectionso we advance to the second element of now we have common element which we add to the resultand then we advance in both arrays now we are at in both arraysbut we know has already been added to the result since the previous element in is also we advance in both again without adding to the intersection comparing to we can eliminate and advance to the fourth element in awhich is and equal to the element that ' iterator holdsso it is added to the result we then eliminate and since no elements remain in awe return def intersect_two_sorted_arrays (ab)ijintersection_a_b [while len(aand len( )if [ = [ ]if = or [ ! [ ]intersection_a_b append ( [ ]ij elif [ib[ ] + elsea[ib[jj + return intersection_a_b since we spend ( time per input array elementthe time complexity for the entire algorithm is ( render calendar consider the problem of designing an online calendaring application one component of the design is to render the calendari display it visually suppose each day consists of number of eventswhere an event is specified as start time and finish time individual events for day are to be rendered as nonoverlapping rectangular regions whose sides are parallel to the xand -axes let the -axis correspond to time if an event starts at time and ends at time ethe upper and lower sides of its corresponding rectangle must be at and erespectively figure on the next page represents set of events suppose the -coordinates for each day' events must lie between and ( pre-specified constant)and each event' rectangle must have the same "height(distance between the sides parallel to the -axisyour task is to compute the maximum height an event rectangle can have in essencethis is equivalent to the following problem write program that takes set of eventsand determines the maximum number of events that take place concurrently