Word Machine - Microsoft Top Interview Questions


Problem Statement :


You are given a list of strings ops where each element is either:

A non-negative integer that should be pushed into a stack

"POP" meaning pop the top element in the stack

"DUP" meaning duplicate the top element in the stack

"+" meaning pop the top two and push the sum

"-" meaning pop the top two and push top - second

Return the top element in the stack after applying all operations. If there are any invalid operations, 
return -1.



Constraints



1 ≤ n ≤ 100,000 where n is the length of ops

Example 1

Input

ops = ["1", "5", "DUP", "3", "-"]

Output

-2

Explanation

Following the operations:



We push 1 into the stack: [1]

We push 5 into the stack: [1, 5]
We duplicate the top element: [1, 5, 5]

We push 3 into the stack: [1, 5, 5, 3]

We pop 3 and 5 and push their difference 3 - 5: [1, 5, -2]

We return the top element which is -2



Example 2

Input

ops = ["+"]

Output
-1

Explanation

There's no elements in the stack so this is invalid.



Solution :



title-img




                        Solution in C++ :

int solve(vector<string>& ops) {
    stack<int> stk;
    for (auto s : ops) {
        if (s == "POP") {
            if (stk.size() < 1) {
                return -1;
            } else {
                stk.pop();
            }
        } else if (s == "DUP") {
            if (stk.size() < 1) {
                return -1;
            } else {
                stk.push(stk.top());
            }
        } else if (s == "+") {
            if (stk.size() < 2) {
                return -1;
            }
            int x = stk.top();
            stk.pop();
            int y = stk.top();
            stk.pop();
            stk.push(x + y);
        } else if (s == "-") {
            if (stk.size() < 2) {
                return -1;
            }
            int x = stk.top();
            stk.pop();
            int y = stk.top();
            stk.pop();
            stk.push(x - y);
        } else {
            stk.push(stoi(s));
        }
    }
    return stk.top();
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(String[] ops) {
        if (ops == null)
            return 0;
        Stack<Integer> element = new Stack<>();
        int totalElement = 0;
        for (String str : ops) {
            switch (str) {
                case "POP":
                    if (element.empty())
                        return -1;
                    element.pop();
                    break;
                case "DUP":
                    if (element.empty())
                        return -1;
                    element.push(element.peek());
                    break;
                case "+":
                    if (element.size() < 2)
                        return -1;
                    element.push(element.pop() + element.pop());
                    break;
                case "-":
                    if (element.size() < 2)
                        return -1;
                    element.push(element.pop() - element.pop());
                    break;
                default:
                    element.push(Integer.parseInt(str));
            }
        }
        return element.peek();
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, ops):
        stack = []
        try:
            for n in ops:
                if n == "DUP":
                    stack.append(stack[-1])
                elif n == "POP":
                    stack.pop()
                elif n == "-":
                    stack.append(stack.pop() - stack.pop())
                elif n == "+":
                    stack.append(stack.pop() + stack.pop())
                else:
                    stack.append(int(n))
            return stack[-1]
        except:
            return -1
                    


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