feat(work): add Minimum Window Substring & Basic Calculator solutions and implement StockSpanner algorithm

work/1/Hard/Hash Table/76.minimum-window-substring.py

@@ -0,0 +1,56 @@
+"""
+76. Minimum Window Substring
+Difficulty: Hard
+https://leetcode.com/problems/minimum-window-substring/
+
+──────────────────────────────────────────────────
+
+Given two strings s and t of lengths m and n respectively, return the
+minimum window substring of s such that every character in t
+(including duplicates) is included in the window. If there is no such
+substring, return the empty string "".
+
+The testcases will be generated such that the answer is unique.
+
+ 
+
+Example 1:
+
+Input: s = "ADOBECODEBANC", t = "ABC"
+Output: "BANC"
+Explanation: The minimum window substring "BANC" includes 'A', 'B',
+and 'C' from string t.
+
+Example 2:
+
+Input: s = "a", t = "a"
+Output: "a"
+Explanation: The entire string s is the minimum window.
+
+Example 3:
+
+Input: s = "a", t = "aa"
+Output: ""
+Explanation: Both 'a's from t must be included in the window.
+Since the largest window of s only has one 'a', return empty string.
+
+ 
+
+Constraints:
+
+	• m == s.length
+
+	• n == t.length
+
+	• 1 <= m, n <= 10^5
+
+	• s and t consist of uppercase and lowercase English letters.
+
+ 
+
+Follow up: Could you find an algorithm that runs in O(m + n) time?
+"""
+
+class Solution:
+    def minWindow(self, s: str, t: str) -> str:
+        

work/1/Hard/Math/224.basic-calculator.py

@@ -0,0 +1,55 @@
+"""
+224. Basic Calculator
+Difficulty: Hard
+https://leetcode.com/problems/basic-calculator/
+
+──────────────────────────────────────────────────
+
+Given a string s representing a valid expression, implement a basic
+calculator to evaluate it, and return the result of the evaluation.
+
+Note: You are not allowed to use any built-in function which
+evaluates strings as mathematical expressions, such as eval().
+
+ 
+
+Example 1:
+
+Input: s = "1 + 1"
+Output: 2
+
+Example 2:
+
+Input: s = " 2-1 + 2 "
+Output: 3
+
+Example 3:
+
+Input: s = "(1+(4+5+2)-3)+(6+8)"
+Output: 23
+
+ 
+
+Constraints:
+
+	• 1 <= s.length <= 3 * 10^5
+
+	• s consists of digits, '+', '-', '(', ')', and ' '.
+
+	• s represents a valid expression.
+
+• '+' is not used as a unary operation (i.e., "+1" and "+(2 + 3)" is
+invalid).
+
+• '-' could be used as a unary operation (i.e., "-1" and "-(2 + 3)"
+is valid).
+
+	• There will be no two consecutive operators in the input.
+
+• Every number and running calculation will fit in a signed 32-bit
+integer.
+"""
+
+class Solution:
+    def calculate(self, s: str) -> int:
+        

work/1/Medium/Stack/901.online-stock-span.py

@@ -24,12 +24,12 @@ less than or equal 8 for 3 consecutive days.
 
 Implement the StockSpanner class:
 
-	• StockSpanner() Initializes the object of the class.
+        • StockSpanner() Initializes the object of the class.
 
 • int next(int price) Returns the span of the stock's price given
 that today's price is price.
 
- 
+
 
 Example 1:
 
@@ -52,23 +52,27 @@ stockSpanner.next(75);  // return 4, because the last 4 prices
 price.
 stockSpanner.next(85);  // return 6
 
- 
+
 
 Constraints:
 
-	• 1 <= price <= 10^5
+        • 1 <= price <= 10^5
 
-	• At most 10^4 calls will be made to next.
+        • At most 10^4 calls will be made to next.
 """
 
-class StockSpanner:
 
+class StockSpanner:
     def __init__(self):
-        prices = []
-        
+        self.stack = []
 
     def next(self, price: int) -> int:
-        
+        ans = 1
+        while self.stack and self.stack[-1][0] <= price:
+            ans += self.stack.pop()[1]
+
+        self.stack.append([price, ans])
+        return ans
 
 
 # Your StockSpanner object will be instantiated and called as such: