[Week 4] Add KyrieJ95's solution for 3 / 5 problems

maximum-product-subarray/joon.py

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+from typing import List
+
+
+class Solution:
+    # Time: O(n)
+    # Space: O(n)
+    def maxProduct(self, nums: List[int]) -> int:
+        maxProducts = [nums[0]]
+        minProducts = [nums[0]]
+        # Store all max/minProducts[i]: the max/min product of all subarrays that have nums[i] as the last element.
+        # Time: O(n)
+        # Space: O(n)
+        for num in nums[1:]:
+            newMaxProduct = max(maxProducts[-1] * num, minProducts[-1] * num, num)
+            newMinProduct = min(maxProducts[-1] * num, minProducts[-1] * num, num)
+            maxProducts.append(newMaxProduct)
+            minProducts.append(newMinProduct)
+        return max(maxProducts)

missing-number/joon.py

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+from typing import List
+
+
+class Solution:
+    # Time: O(n)
+    # Space: O(1)
+    def missingNumber(self, nums: List[int]) -> int:
+        n = len(nums)
+        # MissingNumber = (Sum of 1, 2, ..., n) - Sum of nums)
+        # Time: O(n)
+        return n * (n + 1) // 2 - sum(nums)

valid-palindrome/joon.py

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+import re
+
+
+class Solution:
+    # Time: O(n)
+    # Space: O(n)
+    def isPalindrome(self, s: str) -> bool:
+        # 1. Convert string
+        # Time: O(n)
+        # Space: O(n) since re.sub() will internally use a new string of length n.
+        s = re.sub('[^a-z0-9]', '', s.lower())
+        length = len(s)
+        # 2. Check if the string reads the same forward and backward, one by one.
+        # Time: O(n)
+        # Space: O(1)
+        for i in range(length):
+            if (s[i] != s[length - 1 - i]):
+                return False
+        return True