![//deletes the specified member from the set //compares the specified member to each member in the set 曰void DELETE(vector char» myArray[90], string myLine) 曰 for (int i = e; i < myArray[myLine[9]].size(); i++) // if the member is found it is erased if (myLine[7] myArrayImyLine[911[i]) myArray[myLine[9]].erase(myArray[myLine[9]].begin) (i)); 曰void PRINTSET(vectorc char> myArray[90], string myLine, ofstream &outs) //prints the members of the specified set //if the set is empty it prints EMPTY SET if (ImyArray [myLine [91].emptyO) for (int i e; i< myArray[myLine[911.size); i++) if (myArray [myLine[9] ] [1] != NULL) cout < myArray[myLine[9]]li]< outs <<myArray[myLine [911i]<< 白else cout < EMPTY SET outs << EMPTY SET cout<endl; outs <endl;](http://img.homeworklib.com/questions/c747e570-0288-11ec-963e-ff494be41eb3.png?x-oss-process=image/resize,w_560)
Given the Following Fucntions, Give the Space and time analysis for each function
Homework Answers
In the DELETE() for loop will be repeted based on the length of myLine[9] size.
so its time complexity is Order(n) where n is length of the line.
As we are using one variable myArray[90] which is not depened on any thing as 90 is a constant
its space complexity is order(1)
In PRINTSET() again for loop depends on length of myline[9] .
so its time complexity is Order(n) where n is the length of the line
and it is not using any extra space so its space complexity is Order(1)
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