feat(algorithms): sliding window

DIRECTORY.md

@@ -133,6 +133,8 @@
     * Valid Path
       * [Test Valid Path](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/graphs/valid_path/test_valid_path.py)
   * Greedy
+    * Assign Cookies
+      * [Test Assign Cookies](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/greedy/assign_cookies/test_assign_cookies.py)
     * Boats
       * [Test Boats To Save People](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/greedy/boats/test_boats_to_save_people.py)
     * Gas Stations
@@ -231,6 +233,10 @@
       * [Test Longest Substring K Repeating Chars](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/sliding_window/longest_substring_with_k_repeating_chars/test_longest_substring_k_repeating_chars.py)
     * Longest Substring Without Repeating Characters
       * [Test Longest Substring Without Repeating Characters](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/sliding_window/longest_substring_without_repeating_characters/test_longest_substring_without_repeating_characters.py)
+    * Max Points From Cards
+      * [Test Max Points From Cards](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/sliding_window/max_points_from_cards/test_max_points_from_cards.py)
+    * Max Sum Of Subarray
+      * [Test Max Sum Sub Array](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/sliding_window/max_sum_of_subarray/test_max_sum_sub_array.py)
     * Repeated Dna Sequences
       * [Test Repeated Dna Sequences](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/sliding_window/repeated_dna_sequences/test_repeated_dna_sequences.py)
   * Sorting
@@ -482,9 +488,11 @@
         * [Binary Tree](https://github.com/BrianLusina/PythonSnips/blob/master/datastructures/trees/binary/tree/binary_tree.py)
         * [Test Binary Tree](https://github.com/BrianLusina/PythonSnips/blob/master/datastructures/trees/binary/tree/test_binary_tree.py)
         * [Test Binary Tree Deserialize](https://github.com/BrianLusina/PythonSnips/blob/master/datastructures/trees/binary/tree/test_binary_tree_deserialize.py)
+        * [Test Binary Tree Invert Tree](https://github.com/BrianLusina/PythonSnips/blob/master/datastructures/trees/binary/tree/test_binary_tree_invert_tree.py)
         * [Test Binary Tree Min Camera Cover](https://github.com/BrianLusina/PythonSnips/blob/master/datastructures/trees/binary/tree/test_binary_tree_min_camera_cover.py)
         * [Test Binary Tree Serialize](https://github.com/BrianLusina/PythonSnips/blob/master/datastructures/trees/binary/tree/test_binary_tree_serialize.py)
         * [Test Binary Tree Visible Nodes](https://github.com/BrianLusina/PythonSnips/blob/master/datastructures/trees/binary/tree/test_binary_tree_visible_nodes.py)
+        * [Tree Utils](https://github.com/BrianLusina/PythonSnips/blob/master/datastructures/trees/binary/tree/tree_utils.py)
       * [Utils](https://github.com/BrianLusina/PythonSnips/blob/master/datastructures/trees/binary/utils.py)
     * Btree
       * [Node](https://github.com/BrianLusina/PythonSnips/blob/master/datastructures/trees/btree/node.py)

algorithms/dynamic_programming/buy_sell_stock/README.md

@@ -126,5 +126,3 @@ Output: 6
 - Arrays
 - Dynamic Programming
 - Greedy
-
-

algorithms/dynamic_programming/buy_sell_stock/__init__.py

@@ -5,18 +5,23 @@
 def max_profit(prices: List[int]) -> int:
     """
     Find the maximum profit that can be made from buying and selling a stock once
+
+    Time complexity is O(n) as we iterate through each price to get the maximum profit
+    Space complexity is O(1) as no extra space is used
+    Args:
+        prices(list): list of prices
+    Returns:
+        int: maximum profit that can be made
     """
     if prices is None or len(prices) < 2:
         return 0
 
     start_price = prices[0]
     current_max_profit = 0
 
-    for price in prices:
-        if price < start_price:
-            start_price = price
-        elif price - start_price > current_max_profit:
-            current_max_profit = price - start_price
+    for price in prices[1:]:
+        start_price = min(start_price, price)
+        current_max_profit = max(current_max_profit, price - start_price)
 
     return current_max_profit
 
@@ -28,13 +33,14 @@ def max_profit_two_pointers(prices: List[int]) -> int:
     Space: O(1), no extra memory is used
     Time: O(n), where n is the size of the input list & we iterate through the list only once
     """
-    if prices is None or len(prices) < 2:
+    number_of_prices = len(prices)
+    if prices is None or number_of_prices < 2:
         return 0
 
     left, right = 0, 1
     current_max_profit = 0
 
-    while right < len(prices):
+    while right < number_of_prices:
         low = prices[left]
         high = prices[right]
 

algorithms/dynamic_programming/buy_sell_stock/test_max_profit.py

@@ -1,6 +1,9 @@
 import unittest
 
-from . import max_profit, max_profit_two_pointers
+from algorithms.dynamic_programming.buy_sell_stock import (
+    max_profit,
+    max_profit_two_pointers,
+)
 
 
 class MaxProfitTestCases(unittest.TestCase):

algorithms/greedy/assign_cookies/README.md

@@ -0,0 +1,84 @@
+# Assign Cookies
+
+Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most
+one cookie.
+
+Each child i has a greed factor g[i], which is the minimum size of a cookie that the child will be content with; and
+each cookie j has a size s[j]. If s[j] >= g[i], we can assign the cookie j to the child i, and the child i will be content.
+Your goal is to maximize the number of your content children and output the maximum number.
+
+## Constraints
+
+- 1 <= greed.length <= 3 * 104
+- 0 <= cookies.length <= 3 * 104
+- 1 <= greed[i], cookies[j] <= 231 - 1
+
+## Examples
+
+Example 1:
+```text
+Input: g = [1,2,3], s = [1,1]
+Output: 1
+Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3. 
+And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1
+content.
+You need to output 1.
+```
+
+Example 2:
+```text
+Input: g = [1,2], s = [1,2,3]
+Output: 2
+Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. 
+You have 3 cookies and their sizes are big enough to gratify all of the children, 
+You need to output 2.
+```
+
+## Topics
+
+- Array
+- Two Pointers
+- Greedy
+- Sorting
+
+## Solution
+
+Intiutively, we want to give each child the smallest cookie that satisfies them. This allows us to save the larger
+cookies for the greedier children and allows us to maximize the number of satisfied children.
+
+The greedy algorithm sorts both the greeds and cookies arrays in ascending order. This places the child with the smallest
+greed and the smallest cookie at the front of each array. 
+
+For example:
+
+```text
+greeds = [1, 3, 3, 4]
+cookies = [2, 2, 3, 4]
+```
+
+We then initialize two pointers `i` and `j` to the start of the `greeds` and `cookies` arrays, respectively. `i` 
+represents the current child and `j` represents the current cookie.
+
+If `cookies[j] >= greeds[i]`, that means the current cookie can satisfy the current child. We increment the number of
+satisfied children and move to the next child and cookie.
+
+![Solution 1](./images/solutions/assign_cookies_solution_1.png)
+
+If `cookies[j] < greeds[i]`, that means the current cookie cannot satisfy the current child, so we move to the next cookie
+to see if it can.
+
+![Solution 2](./images/solutions/assign_cookies_solution_2.png)
+
+We can continue this process until we reach the end of either the greeds or cookies arrays, and return the number of
+satisfied children as the result.
+
+### Complexity Analysis
+
+#### Time Complexity 
+
+O(n log n + m log m) where n is the number of children and m is the number of cookies. We sort the
+greeds and cookies arrays in O(n log n + m log m) time, and then iterate through the arrays in O(n + m) time.
+
+#### Space Complexity
+
+O(1) We only use a constant amount of space for variables.

algorithms/greedy/assign_cookies/__init__.py

@@ -0,0 +1,21 @@
+from typing import List
+
+
+def find_content_children(greeds: List[int], cookies: List[int]) -> int:
+    # This in-place sorting of both g and s results in a time complexity of O(n log(n) + m log(m))
+    greeds.sort()
+    cookies.sort()
+
+    cookie, greed = 0, 0
+    count = 0
+
+    # We iterate through each greed factor and cookie
+    while greed < len(greeds) and cookie < len(cookies):
+        # When we get a cookie that satisfies kid i, we assign that cookie to the child
+        # and move along, increasing the count as well
+        if cookies[cookie] >= greeds[greed]:
+            count += 1
+            greed += 1
+        cookie += 1
+
+    return count

algorithms/greedy/assign_cookies/test_assign_cookies.py

@@ -0,0 +1,24 @@
+import unittest
+from typing import List
+from parameterized import parameterized
+from algorithms.greedy.assign_cookies import find_content_children
+
+
+ASSIGN_COOKIES_TEST_CASES = [
+    ([1, 2, 3], [1, 1], 1),
+    ([1, 2], [1, 2, 3], 2),
+    ([10, 9, 8, 7], [5, 6, 7, 8], 2),
+]
+
+
+class AssignCookiesTestCase(unittest.TestCase):
+    @parameterized.expand(ASSIGN_COOKIES_TEST_CASES)
+    def test_find_content_children(
+        self, greed: List[int], cookies: List[int], expected: int
+    ):
+        actual = find_content_children(greed, cookies)
+        self.assertEqual(expected, actual)
+
+
+if __name__ == "__main__":
+    unittest.main()

algorithms/greedy/gas_stations/README.md

@@ -42,3 +42,54 @@ Therefore, you can't travel around the circuit once no matter where you start.
 
 - Array
 - Greedy
+
+## Solution
+
+If there is more gas along the route than the cost of the route, then there is guaranteed to be a solution to the problem.
+So the first step is to check if the sum of the gas is greater than or equal to the sum of the cost. If it is not, then
+we return -1.
+
+Next, we iterate through the gas station to find the starting index of our circuit using a greedy approach: whenever we
+don't have enough gas to reach the next station, we move our starting gas station to the next station and reset our gas
+tank.
+
+We start at the first station, and fill our tank with gas[0] = 5 units of gas. From there, it takes cost[0] = 1 units of
+gas to travel to the next station, so we arrive at station 2 (index 1) with 4 units of gas.
+
+![Solution 1](./images/solutions/gas_stations_solution_1.png)
+![Solution 2](./images/solutions/gas_stations_solution_2.png)
+![Solution 3](./images/solutions/gas_stations_solution_3.png)
+
+At station 2, we fill our tank with gas[1] = 2 units of gas, for a total of 6 units of gas. It takes cost[1] = 5 units
+of gas to travel to the next station, so we arrive at station 3 with 1 unit of gas
+
+![Solution 4](./images/solutions/gas_stations_solution_4.png)
+![Solution 5](./images/solutions/gas_stations_solution_5.png)
+
+Now at station 3, we fill our tank with gas[2] = 0 units of gas, for a total of 1 unit of gas. It takes cost[2] = 5 units
+of gas to travel to the next station, which we don't have.
+
+This is where our greedy approach comes in. We reset our starting station to the next station i + 1 and reset our gas
+tank to 0. We can do this because all other start indexes between 0 and 2 will also run into the same problem of not
+having enough gas to reach the next station, so we can rule them out.
+
+![Solution 6](./images/solutions/gas_stations_solution_6.png)
+![Solution 7](./images/solutions/gas_stations_solution_7.png)
+
+If we follow this approach of resetting the start index and gas tank whenever we don't have enough gas to reach the next
+station, then when we finish iterating, the last start index will be the solution to the problem.
+
+![Solution 8](./images/solutions/gas_stations_solution_8.png)
+![Solution 9](./images/solutions/gas_stations_solution_9.png)
+![Solution 10](./images/solutions/gas_stations_solution_10.png)
+![Solution 11](./images/solutions/gas_stations_solution_11.png)
+
+### Complexity Analysis
+
+#### Time Complexity
+
+O(n) where n is the number of gas stations. We only iterate through the gas stations once.
+
+#### Space Complexity
+
+O(1) We only use constant extra space for variables.

algorithms/greedy/gas_stations/__init__.py

@@ -22,10 +22,14 @@ def can_complete_circuit(gas: List[int], cost: List[int]) -> int:
 
     gas_tank, start_index = 0, 0
 
-    for i in range(len(gas)):
-        gas_tank += gas[i] - cost[i]
+    for gas_station in range(len(gas)):
+        # can reach next station:
+        # update remaining fuel
+        gas_tank += gas[gas_station] - cost[gas_station]
 
         if gas_tank < 0:
-            start_index = i + 1
+            # can't reach next station:
+            # try starting from next station
+            start_index = gas_station + 1
             gas_tank = 0
     return start_index

algorithms/greedy/gas_stations/test_gas_stations.py

@@ -3,15 +3,16 @@
 from parameterized import parameterized
 from algorithms.greedy.gas_stations import can_complete_circuit
 
+CAN_COMPLETE_CIRCUIT_TEST_CASES = [
+    ([1, 2, 3, 4, 5], [3, 4, 5, 1, 2], 3),
+    ([2, 3, 4], [3, 4, 3], -1),
+    ([1, 2], [2, 1], 1),
+    ([5, 2, 0, 3, 3], [1, 5, 5, 1, 1], 3),
+]
+
 
 class CanCompleteCircuitTestCase(unittest.TestCase):
-    @parameterized.expand(
-        [
-            ([1, 2, 3, 4, 5], [3, 4, 5, 1, 2], 3),
-            ([2, 3, 4], [3, 4, 3], -1),
-            ([1, 2], [2, 1], 1),
-        ]
-    )
+    @parameterized.expand(CAN_COMPLETE_CIRCUIT_TEST_CASES)
     def test_can_complete_circuit(self, gas: List[int], cost: List[int], expected: int):
         actual = can_complete_circuit(gas, cost)
         self.assertEqual(expected, actual)