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 :
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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