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docs(docs): add MDX documentation pages for problems 150 (Evaluate Reverse Polish Notation) and 155 (Min Stack)
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---
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title: '150. Evaluate Reverse Polish Notation'
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description: You are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation
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sidebar:
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label: 'Evaluate Reverse Polish Notation'
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badge: 'Medium'
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---
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<Badge variant="accent">Stack</Badge>
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### Example 1:
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- Input: `tokens = ["2","1","+","3","*"]`
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- Output: `9`
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- Explanation: `((2 + 1`) * `3`) = `9`
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### Example 2:
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- Input: `tokens = ["4","13","5","/","+"]`
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- Output: `6`
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- Explanation: `(4 + (13 / 5`)) = `6`
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### Example 3:
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- Input: `tokens = ["10","6","9","3","+","-11","*","/","*","17","+","5","+"]`
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- Output: `22`
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- Explanation: `((10 * (6 / ((9 + 3`) * `-11`))) + `17`) + `5 = ((10 * (6 / (12 * -11`))) + `17`) + `5 = ((10 * (6 / -132`)) + `17`) + `5 = ((10 * 0`) + `17`) + `5 = (0 + 17`) + `5 = 17 + 5 = 22`
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### Constraints:
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- `1 <= tokens.length <= 10^4`
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- `tokens[i]` is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200].
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## Solution
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```py
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class Solution:
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def evalRPN(self, tokens: List[str]) -> int:
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stack = []
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operations = {
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"+": lambda x, y: int(x + y),
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"-": lambda x, y: int(x - y),
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"*": lambda x, y: int(x * y),
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"/": lambda x, y: int(x / y),
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}
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for c in tokens:
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if c in operations:
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y = stack.pop()
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x = stack.pop()
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calc_func = operations[c]
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result = calc_func(x, y)
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stack.append(result)
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else:
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stack.append(int(c))
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return stack[0]
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```
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---
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title: '155. Min Stack'
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description: Design a stack that supports push, pop, top, and retrieving the minimum element in constant time
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sidebar:
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label: 'Min Stack'
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badge: 'Medium'
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---
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<Badge variant="accent">Stack</Badge>
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::::warning
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You must implement a solution with O(1) time complexity for each function.
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::::
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### Example 1:
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- Input: ``
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- Output: ``
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- Explanation: Input `["MinStack","push","push","push","getMin","pop","top","getMin"]` [[],[-2],[0],[-3],[],[],[],[]] Output `[null,null,null,null,-3,null,0,-2]` Explanation MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); // return `-3` minStack.pop(); minStack.top(); // return `0` minStack.getMin(); // return `-2`
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### Constraints:
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- `-2^31 <= val <= 2^31 - 1`
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- Methods pop, top and getMin operations will always be called on non-empty stacks.
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- At most 3 * 10^4 calls will be made to push, pop, top, and getMin.
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## Solution
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```py
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class MinStack:
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def __init__(self):
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self.stack = []
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self.minStack = []
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def push(self, value: int) -> None:
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self.stack.append(value)
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value = min(value, self.minStack[-1] if self.minStack else value)
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self.minStack.append(value)
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def pop(self) -> None:
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self.stack.pop()
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self.minStack.pop()
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def top(self) -> int:
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return self.stack[-1]
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def getMin(self) -> int:
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return self.minStack[-1]
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# Your MinStack object will be instantiated and called as such:
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# obj = MinStack()
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# obj.push(value)
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# obj.pop()
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# param_3 = obj.top()
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# param_4 = obj.getMin()
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```
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