.scripts/addnew

@@ -41,7 +41,7 @@ const $ch = cheerio.load(q.content);
 $ch("pre").wrapInner("<code></code>");
 
 const td = new Turndown({});
-const mdBody = td.turndown($ch("body").html()) + `\n\n${link}`;
+const mdBody = td.turndown($ch("body").html());
 
 await $`mkdir -p ${solutionDir}`;
 await $`touch ${solutionFilePath}`;

0042_trapping-rain-water/python3/extra_space.py

@@ -1,61 +0,0 @@
-# Time: O(N)
-# Space: O(N)
-
-class Solution:
-    def trap(self, heights: List[int]) -> int:
-        if len(heights) < 2: return 0
-        
-        # For every ith height, we need to calculate the max height
-        # on left and max on right of it
-        max_lefts, max_rights = [0] * len(heights), [0] * len(heights)
-        
-        maxl = 0
-        for i in range(len(heights)):
-            max_lefts[i] = maxl
-            maxl = max(maxl, heights[i])
-        
-        maxr = 0
-        for j in range(len(heights) - 1, -1, -1):
-            max_rights[j] = maxr
-            maxr = max(maxr, heights[j])
-            
-        print(max_lefts)
-        print(max_rights)
-        
-        # For every ith height, we can now compute the water output it can hold
-        # by doing the following:
-        #
-        # min(max_height_to_left, max_height_to_right) - height_of_bar
-        #
-        # If we ignore height_of_bar for a moment, the amount of water that can be
-        # held would be constrained by the least heights out of left/right.
-        #
-        #     3
-        # 2   |
-        # |   |
-        # | 0 |
-        #
-        # In the above case, would be min(2, 3) == 2. But, if we fill the space up with a bar
-        # then:
-        #
-        #     3
-        # 2   |
-        # | 1 |
-        # | | |
-        #
-        # We can't put 2 water units. We need to subtract the height of the bar from the prev
-        # min() which will give us 1.
-        #
-        # NOTE: We need to do this for every ith height in the array.
-        
-        output = 0
-        for i, h in enumerate(heights):
-            value = min(max_lefts[i], max_rights[i]) - heights[i]
-            
-            # It's possible that ith height is very large, causing
-            # the expression above to give us -ve value. In this case
-            # we just ignore it (i.e, 0 output for this ith height)
-            if value > 0:
-                output += value
-        
-        return output

0042_trapping-rain-water/python3/solution.py

@@ -1,48 +0,0 @@
-# Time: O(N)
-# Space: O(N)
-
-class Solution:
-    def trap(self, heights: List[int]) -> int:
-        # Two pointer approach
-        left, right = 0, len(heights) - 1
-        
-        # Since while scanning the array, at a given index i, we
-        # only need to consider the minimum of max_left_heights and
-        # max_right_heights, we just need to keep moving from both
-        # ends and keep track of the maximum height seen so far. We
-        # can then compare these two heights and decide from which
-        # side to move next.
-        max_left = max_right = 0
-        
-        output = 0
-        while left <= right:
-            # By the formula from the DP approach:
-            #
-            # min(max_left, max_right) - heights[i]
-            #
-            # If max_left <= max_right, formula becomes:
-            #
-            # max_left - heights[i]
-            #
-            if max_left <= max_right:
-                # Shorter form to avoid -ve values
-                output += max(max_left - heights[left], 0)
-                
-                max_left = max(max_left, heights[left])
-                left += 1
-            # By the formula from the DP approach:
-            #
-            # min(max_left, max_right) - heights[i]
-            #
-            # If max_left > max_right, formula becomes:
-            #
-            # max_right - heights[i]
-            #
-            else:
-                output += max(max_right - heights[right], 0)
-                
-                max_right = max(max_right, heights[right])
-                right -= 1
-            
-        
-        return output

0121_best-time-to-buy-and-sell-stock/README.md

@@ -1,25 +0,0 @@
-You are given an array `prices` where `prices[i]` is the price of a given stock on the `ith` day.
-
-You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
-
-Return _the maximum profit you can achieve from this transaction_. If you cannot achieve any profit, return `0`.
-
-**Example 1:**
-
-    Input: prices = [7,1,5,3,6,4]
-    Output: 5
-    Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
-    Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
-    
-
-**Example 2:**
-
-    Input: prices = [7,6,4,3,1]
-    Output: 0
-    Explanation: In this case, no transactions are done and the max profit = 0.
-    
-
-**Constraints:**
-
-*   `1 <= prices.length <= 105`
-*   `0 <= prices[i] <= 104`

0121_best-time-to-buy-and-sell-stock/python3/solution.py

@@ -1,20 +0,0 @@
-# Time: O(N)
-# Space: O(1)
-class Solution:
-    def maxProfit(self, prices: List[int]) -> int:
-        if len(prices) < 1: return 0
-        
-        i, j = 0, 1
-        max_profit = 0
-        
-        while i <= j < len(prices):
-            max_profit = max(prices[j] - prices[i], max_profit)
-            
-            # Buy low, sell high — so we need to keep moving i (like
-            # a buy pointer) if jth price is smaller than what ith is.
-            if prices[j] < prices[i]:
-                i = j
-            else:
-                j += 1
-        
-        return max_profit

0535_encode-and-decode-tinyurl/README.md

@@ -1,29 +0,0 @@
-> Note: This is a companion problem to the [System Design](https://leetcode.com/discuss/interview-question/system-design/) problem: [Design TinyURL](https://leetcode.com/discuss/interview-question/124658/Design-a-URL-Shortener-(-TinyURL-)-System/).
-
-TinyURL is a URL shortening service where you enter a URL such as `https://leetcode.com/problems/design-tinyurl` and it returns a short URL such as `http://tinyurl.com/4e9iAk`. Design a class to encode a URL and decode a tiny URL.
-
-There is no restriction on how your encode/decode algorithm should work. You just need to ensure that a URL can be encoded to a tiny URL and the tiny URL can be decoded to the original URL.
-
-Implement the `Solution` class:
-
-*   `Solution()` Initializes the object of the system.
-*   `String encode(String longUrl)` Returns a tiny URL for the given `longUrl`.
-*   `String decode(String shortUrl)` Returns the original long URL for the given `shortUrl`. It is guaranteed that the given `shortUrl` was encoded by the same object.
-
-**Example 1:**
-
-    Input: url = "https://leetcode.com/problems/design-tinyurl"
-    Output: "https://leetcode.com/problems/design-tinyurl"
-    
-    Explanation:
-    Solution obj = new Solution();
-    string tiny = obj.encode(url); // returns the encoded tiny url.
-    string ans = obj.decode(tiny); // returns the original url after deconding it.
-    
-
-**Constraints:**
-
-*   `1 <= url.length <= 104`
-*   `url` is guranteed to be a valid URL.
-
-https://leetcode.com/problems/encode-and-decode-tinyurl

0535_encode-and-decode-tinyurl/python3/solution.py

@@ -1,37 +0,0 @@
-# NOTE: This one is an open-ended question, and relevant to system design. Refer
-# the notes.
-
-import string
-import random
-
-class Codec:
-    store = {}
-    base = 'https://tiny.url/'
-
-    def encode(self, longUrl: str) -> str:
-        """Encodes a URL to a shortened URL.
-        """
-        tiny = None
-        
-        while True:
-            tiny = ''.join(random.choices(string.ascii_letters + string.digits, k = 5))
-            
-            if tiny not in self.store:
-                break
-        
-        self.store[tiny] = longUrl
-        
-        return f'{self.base}{tiny}'
-        
-
-    def decode(self, shortUrl: str) -> str:
-        """Decodes a shortened URL to its original URL.
-        """
-        _, tiny = shortUrl.split(self.base)
-        
-        return self.store[tiny]
-        
-
-# Your Codec object will be instantiated and called as such:
-# codec = Codec()
-# codec.decode(codec.encode(url))