docs(docs): add MDX documentation pages for problems 150 (Evaluate Reverse Polish Notation) and 155 (Min Stack)

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Prad Nukala
2026-08-27 15:19:48 -04:00
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commit 95f10980d4
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---
title: '150. Evaluate Reverse Polish Notation'
description: You are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation
sidebar:
label: 'Evaluate Reverse Polish Notation'
badge: 'Medium'
---
<Badge variant="accent">Stack</Badge>
### Example 1:
- Input: `tokens = ["2","1","+","3","*"]`
- Output: `9`
- Explanation: `((2 + 1`) * `3`) = `9`
### Example 2:
- Input: `tokens = ["4","13","5","/","+"]`
- Output: `6`
- Explanation: `(4 + (13 / 5`)) = `6`
### Example 3:
- Input: `tokens = ["10","6","9","3","+","-11","*","/","*","17","+","5","+"]`
- Output: `22`
- 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`
### Constraints:
- `1 <= tokens.length <= 10^4`
- `tokens[i]` is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200].
## Solution
```py
class Solution:
def evalRPN(self, tokens: List[str]) -> int:
stack = []
operations = {
"+": lambda x, y: int(x + y),
"-": lambda x, y: int(x - y),
"*": lambda x, y: int(x * y),
"/": lambda x, y: int(x / y),
}
for c in tokens:
if c in operations:
y = stack.pop()
x = stack.pop()
calc_func = operations[c]
result = calc_func(x, y)
stack.append(result)
else:
stack.append(int(c))
return stack[0]
```
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---
title: '155. Min Stack'
description: Design a stack that supports push, pop, top, and retrieving the minimum element in constant time
sidebar:
label: 'Min Stack'
badge: 'Medium'
---
<Badge variant="accent">Stack</Badge>
::::warning
You must implement a solution with O(1) time complexity for each function.
::::
### Example 1:
- Input: ``
- Output: ``
- 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`
### Constraints:
- `-2^31 <= val <= 2^31 - 1`
- Methods pop, top and getMin operations will always be called on non-empty stacks.
- At most 3 * 10^4 calls will be made to push, pop, top, and getMin.
## Solution
```py
class MinStack:
def __init__(self):
self.stack = []
self.minStack = []
def push(self, value: int) -> None:
self.stack.append(value)
value = min(value, self.minStack[-1] if self.minStack else value)
self.minStack.append(value)
def pop(self) -> None:
self.stack.pop()
self.minStack.pop()
def top(self) -> int:
return self.stack[-1]
def getMin(self) -> int:
return self.minStack[-1]
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(value)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()
```