Chegg repeat increment t if there is at least one active process choose from the active processes a process Pᵢ to run next, according to the scheduling algorithm. decrement Rᵢ (Pᵢ has accumulated 1 CPU time unit) if Rᵢ == 0 (process i has terminated) set active flag of Pᵢ = 0 TTᵢ = t - Aᵢ end if until Rᵢ == 0 for all n processes (all processes have terminated) compute the average turnaround time, ATT, by averaging all values TTᵢ

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The editing of digital photos us about the same level of difficulty as editing an analog photo

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; }