Write the routines with the following declarations: void permute( const string & str ); void permute( const string & str, int low, int high ); the first routine is a driver that calls the second and prints all the permutations of the characters in string str. if str is "abc", then the strings that are output are abc, acb, bac, bca, cab, and cba. use recursion for the second routine. 2 how many permutations for a string of length l?

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; }

In a personal computer, the system software

In many applications, only a small percentage of the points in a given set p of n points are extreme. in such a case, the convex hull of p has less than n vertices. this can actually make our algorithm convexhull run faster than θ(nlogn). assume, for instance, that the expected number of extreme points in a random sample of p of size r is o(rα ), for some constant α < 1. (this is true when the set p has been created by picking points uniformly at random in a ball.) prove that under this condition, the running time of the algorithm is o(n).