Convert into 24 hours clock time. (a).1.40pm​

Requirement types discussed during software development include functional and color scheme nonfunctional and code style constraint and nonfunctional fashionable and functional.

We as humans write math expression in infix notation, e. g. 5 + 2 (the operators are written in-between the operands). in a computer’s language, however, it is preferred to have the operators on the right side of the operands, i. e. 5 2 +. for more complex expressions that include parenthesis and multiple operators, a compiler has to convert the expression into postfix first and then evaluate the resulting postfix. write a program that takes an “infix” expression as input, uses stacks to convert it into postfix expression, and finally evaluates it. it must support the following operations: + - / * ^ % (example infix expression: (7 - 3) / (2 + 2)postfix expression: 7 3 - 2 2 + /result: 1guidelines: 1. you will need to use stacks in three placesa. one for the parenthesis check [char stack]b. one during infix to postfix [char stack]c. one during evaluation [int stack]for a and b above, you can use same array and same push, pop method as both ofthem are char. but for evaluation you have int stack and you might consider to createanother push pop method to handle it. maybe push_int, pop_int, etc. or find otherstrategy to utilize existing push pop method2. you can create a function for obtaining operator priority. that function should take anoperator as input and return its priority as an integer. this function will you a lot andreduce repeated code3. during evaluation you will need to convert char into integer. example for single digit: char c = '5'; int x = c - '0';

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; }