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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; }
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The value of sin(x) (in radians) can be approximated by the alternating infinite series create a function (prob3_2) that takes inputs of a scalar angle measure (in radians) and the number of approximation terms, n, and estimates sin(x). do not use the sin function in your solution. you may use the factorial function. though this can be done without a loop (more efficiently), your program must use (at least) one. you may find the mod() function useful in solving the problem.
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