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//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;

Given the Following Fucntions, Give the Space and time analysis for each function

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Answer #1

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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