Question

In: Computer Science

Write a program to check given expression is valid or not.The expression consists of paranthsis like  [{(...

Write a program to check given expression is valid or not.The expression consists of paranthsis like  [{( if valid then convert into postfix expression and after conversion then evaluate postfix expession(using stack) and do not use build in stack. Use the c++ laguage.

Solutions

Expert Solution

#include<iostream>
#include<stack>
#include<string>

using namespace std;

// Function to convert Infix expression to postfix
string InfixToPostfix(string expression);

// Function to verify whether an operator has higher precedence over other
int HasHigherPrecedence(char operator1, char operator2);

// Function to verify whether a character is operator symbol or not.
bool IsOperator(char C);

// Function to verify whether a character is alphanumeric chanaracter
//(letter or numeric digit) or not.
bool IsOperand(char C);

// Function to check whether a char is an opening parenthesis i.e '(' or '{' or '['
bool IsOpeningParentheses(char C);

// Function to check whether a char is an closing parenthesis i.e ')' or '}' or ']'
bool IsClosingParentheses(char C);

int main()
{
        string expression;
        cout<<"Enter Infix Expression \n";
        getline(cin,expression);
        string postfix = InfixToPostfix(expression);
        cout<<"Output = "<<postfix<<"\n";
}

// Function to evaluate Postfix expression and return output
string InfixToPostfix(string expression)
{
        // Declaring a Stack from Standard template library in C++.
        stack<char> S;
        string postfix = ""; // Initialize postfix as empty string.
        for(int i = 0;i< expression.length();i++) {

                // Scanning each character from left.
                // If character is a delimiter, move on.
                if(expression[i] == ' ' || expression[i] == ',') continue;

                // Else if character is an operand then just append it in res string
                else if(IsOperand(expression[i]))
                {
                        postfix +=expression[i];
                }

                //If character is operator, check for Higher precedence operator in Stack top
                //till first opening bracket and pop all such operators
                //Finally push the current operator on the Stack
                else if(IsOperator(expression[i]))
                {
                        while(!S.empty() && !IsOpeningParentheses(S.top())
                        && HasHigherPrecedence(S.top(),expression[i]))
                        {
                                postfix+= S.top();
                                S.pop();
                        }
                        S.push(expression[i]);
                }

                //If opening Parentheses simply push it on the Stack
                else if(IsOpeningParentheses(expression[i]))
                {
                        S.push(expression[i]);
                }

        //If Closing Parentheses then pop all the operators and append to res string
        // until Stack top is opening Parentheses and finally one extra pop for Opening Parentheses
                else if(IsClosingParentheses(expression[i]))
                {
                        while(!S.empty() && !IsOpeningParentheses(S.top())) {
                                postfix += S.top();
                                S.pop();
                        }
                        S.pop();
                }
        }

    //Finally pop and append all operators till Stack is not empty
        while(!S.empty()) {
                postfix += S.top();
                S.pop();
        }

        return postfix;
}

// Function to verify whether a character is english letter or numeric digit.
// We are assuming in this solution that operand will be a single character
bool IsOperand(char C)
{
        if(C >= '0' && C <= '9') return true;
        if(C >= 'a' && C <= 'z') return true;
        if(C >= 'A' && C <= 'Z') return true;
        return false;
}

// Function to verify whether a character is operator symbol or not.
bool IsOperator(char C)
{
        if(C == '+' || C == '-' || C == '*' || C == '/' || C== '^')
                return true;

        return false;
}

bool IsOpeningParentheses(char C)
{
    if(C == '(' || C == '{' || C=='[')
        return true ;
    return false;
}

bool IsClosingParentheses(char C)
{
    if(C == ')' || C == '}' || C==']')
        return true ;
    return false;
}
// Function to verify whether an operator is right associative or not.
int IsRightAssociative(char op)
{
        if(op == '^') return true;
        return false;
}

// Function to get weight of an operator. An operator with higher weight will have higher precedence.
int GetOperatorWeight(char op)
{
        int weight = -1;
        switch(op)
        {
        case '+':
        case '-':
                weight = 1;
                break ;
        case '*':
        case '/':
                weight = 2;
                break ;
        case '^':
                weight = 3;
                break;
        }
        return weight;
}

// Function to perform an operation and return output.
int HasHigherPrecedence(char op1, char op2)
{
        int op1Weight = GetOperatorWeight(op1);
        int op2Weight = GetOperatorWeight(op2);

        // If operators have equal precedence, return true if they are left associative.
        // return false, if right associative.
        // if operator is left-associative, left one should be given priority.
        if(op1Weight == op2Weight)
        {
                if(IsRightAssociative(op1)) return false;
                else return true;
        }
        return op1Weight > op2Weight ?  true: false;
}

Note: I hope the above source code will help you with your question and this is what you wanted. If in case you find any error or wrong implementation then please do let me know through comments. I'll try to resolve your issue and post the updated source code as you want in the comments section. Thank You!!!!1

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