**Postfix Notation Evaluation - Amazon Top Interview Questions**

### Problem Statement :

Postfix notation is a way to represent an expression where the operator comes after the operands. For example, ["2", "2", "+", "6", "*"] would be equal to 24, since we have (2 + 2) * 6 = 24. Given a list of strings exp, representing a postfix notation consisting of integers and operators ("+", "-", "*", "/"), evaluate the expression. "/" is integer division. Example 1 Input exp = ["9", "3", "+", "2", "/"] Output 6 Explanation (9 + 3) / 2 = 6 Example 2 Input exp = ["3", "9", "-", "4", "/"] Output -1 Explanation (3 - 9) / 4 = -1

### Solution :

` ````
Solution in C++ :
#define long long long
long cal(long first, string& op, long second) {
if (op == "*") return first * second;
if (op == "-") return first - second;
if (op == "+") return first + second;
return first / second;
}
int solve(vector<string>& exp) {
stack<long> st;
for (int i = 0; i < exp.size(); i++) {
if (exp[i] == "+" or exp[i] == "-" or exp[i] == "/" or exp[i] == "*") {
long a = st.top();
st.pop();
long b = st.top();
st.pop();
long res = cal(b, exp[i], a);
st.push(res);
continue;
}
long d = stol(exp[i]);
st.push(d);
}
return st.top();
}
```

` ````
Solution in Python :
OPERATIONS = {"+": add, "-": sub, "*": mul, "/": lambda a, b: int(a / b)}
class Solution:
def solve(self, exp):
stack = []
for x in exp:
if x in OPERATIONS:
b, a = stack.pop(), stack.pop()
stack.append(OPERATIONS[x](a, b))
else:
stack.append(int(x))
return stack[-1]
```

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