![]() The last value in the value stack will be the result.Once we have iterated the entire expression, we pop one operator from the operator stack and two values from the value operator and apply the operator to the values, store the result in the value stack, and keep on repeating this, until we have just a single value left in the value stack.And, push the current iterated operator in the operator stack.If operator stack is not empty and the operator present in the stack has higher precedence than the current operator, then we pop the operator with high precedence, and two values from the value stack, apply the operator to the values and store the result in the value stack.If we encounter an operator, we will push it in the operator stack.If we encounter a closing parenthesis ), we pop an operator from the operator stack and two data elements from the value stack and apply the operator to the numeric values and store the result in the value stack.For example: If we have expression (22+2)*4, in this case our expression is made of following tokens, hence on encountering the first 2, we should look for consecutive tokens, if its a number, which in this case is, we append it to existing number to make it 22.We will do this, until we encounter token other than a number. ![]() Your code would hit that branch when the character is not recognized, meaning that the infix expression is invalid. So when we encounter a numeric value, we look for consecutive members of the expression, if we find another number, we append it to existing number. There are several things below that you should check as you do the conversion to decide if the infix expression is valid: Add the final else to the chain determining the character type, i.e. But, what if the number is a two digit or three digit number, we first have to form the complete number.If we encounter any numeric value, we have to push it in the values stack.If we encounter an opening parenthesis (, we will push it in the operator stack.We will start iterating the expression from left to right.Here is the algorithm for solving an arithmetic expression using Stacks. We will be using two stacks, one for operators and parenthesis, and the other one to store the numerical values.Although it will be easier for us as we are writing the algorithm for only 4 operators, specified in point 1. Operator precedence is another tricky factor here.So we will have to deal with opening and closing parenthesis carefully. It divides a simple linear expression into sections to be solved separately. We will keep the program simple and will only evaluate expressions with +.There are a few important points to note: ![]() Here we will be writing a simple algorithm to solve a given arithmetic expression in infix form using Stack. Now that we already know how to implement a Stack in python, it's time to use it. Arithmetic Expression(Infix) Evaluation using Stack
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