Evaluate an infix expression (with parentheses)?

Short answer: public int EvaluateInfix(string expression) { Stack<int> operands = new Stack<int>(); Stack<char> operators = new Stack<char>(); int i = 0; while (i < expression.Length) { if (char.IsWhiteSpace(expression[i])) { i++; continue; } if (char.IsDigit(expression[i])) { int val = 0; while (i < expression.Length && char.IsDigit(expression[i])) { val = val * 10 + (expression[i] - '0'); i++; } operands.Push(val); continue; }…

Explain a bit more

if (expression[i] == '(') { operators.Push(expression[i]); } else if (expression[i] == ')') { while (operators.Peek() != '(') { ApplyOp(operands, operators); } operators.Pop(); // remove '(' Follow on: } else if (IsOperator(expression[i])) { while (operators.Count > 0 && Precedence(operators.Peek()) >= Precedence(expression[i])) { ApplyOp(operands, operators); } operators.Push(expression[i]); } i++; } while (operators.Count > 0) { ApplyOp(operands, operators); } return operands.Pop(); } bool IsOperator(char c) { return c == '+' || c == '-' || c == '*' || c == '/'; } int Precedence(char op) { if (op == '+' || op == '-') return 1; if (op == '*' || op == '/') return 2; return 0; } void ApplyOp(Stack<int> operands, Stack<char> operators) { int b = operands.Pop(); int a = operands.Pop(); char op = operators.Pop(); int result = 0; switch (op) { case '+': result = a + b; break; case '-':…

Example code

public int EvaluateInfix(string expression) {
Stack<int> operands = new Stack<int>();
Stack<char> operators = new Stack<char>();
int i = 0; while (i < expression.Length) { if (char.IsWhiteSpace(expression[i])) { i++; continue; }
if (char.IsDigit(expression[i])) {
int val = 0; while (i < expression.Length && char.IsDigit(expression[i])) { val = val * 10 + (expression[i] - '0'); i++; } operands.Push(val); continue; }
if (expression[i] == '(') { operators.Push(expression[i]); } else if (expression[i] == ')') { while (operators.Peek() != '(') { ApplyOp(operands, operators); } operators.Pop(); // remove '(' Follow on: } else if (IsOperator(expression[i])) { while (operators.Count > 0 && Precedence(operators.Peek()) >= Precedence(expression[i])) { ApplyOp(operands, operators); } operators.Push(expression[i]); } i++; } while (operators.Count > 0) { ApplyOp(operands, operators); }
return operands.Pop();
} bool IsOperator(char c) { return c == '+' || c == '-' || c == '*' || c == '/';
}
int Precedence(char op) {
if (op == '+' || op == '-') return 1;
if (op == '*' || op == '/') return 2;
return 0;
} void ApplyOp(Stack<int> operands, Stack<char> operators) { int b = operands.Pop();
int a = operands.Pop();
char op = operators.Pop();
int result = 0; switch (op) { case '+': result = a + b; break;
case '-': result = a - b; break;
case '*': result = a * b; break; Follow on: case '/': result = a / b; break;
} operands.Push(result); } Explanation: Standard two-stack algorithm for evaluating infix expressions considering operator precedence and parentheses.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.