You are given two boxes: one box filled with mason jars of different sizes, and another box filled with corresponding screw top lids. You can test whether a given jar and lid are a match, from which1 you learn whether the jar is too large, too small, or an exact match for the lid. The differences insize between pairs of jars or lids are too small to see by eye, so you cannot compare the sizes oftwo jars or two lids directly. You are to match each jar to each lid, with the assumption that notwo jars or two lids are of the same size, and that every jar in one box has a corresponding lid inthe other box. Give a randomized O(nlogn) expected time algorithm for this problem.

For all machines-not just hammers- the user applies force force to the machine to the machine over a certain distance. a. input b. output c. duo d. none of the above

[a] create a class called “cycle” which has two instance integer variables as properties, “numberofwheels” and “weight.” create a constructor with two parameters, using the same variable names in the parameter list. assign each variable to numberofwheels” and “weight” respectively. write a separate application to test the class and display its properties. note: do not change the names of the instance variables or the variables listed in the constructor’s parameter list. [b] edit your class cycle by adding a default constructor which will assign the default values of 100 to represent the numberofwheels, and 1000 to represent the weight, by invoking a call to the other constructor. modify your application created in [a] to test the class.

What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i.e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }