Add solutions for additional problems

Author: JoyJaewon

best-time-to-buy-and-sell-stock/JoyJaewon.py

@@ -0,0 +1,34 @@
+'''
+Brainstorm
+- Brute force: double for-loop O(N^2)
+- Efficient: single pass O(N)
+
+Plan
+1. Initialize the variables
+2. Iterate through each price
+    2-1. Update the min price
+    2-2. Calculate and update max profit
+3. Return the max profit
+'''
+
+
+def maxProfit(prices):
+    min_price = float('inf')
+    max_profit = 0
+    for price in prices:
+        if price < min_price:
+            min_price = price
+        elif price - min_price > max_profit:
+            max_profit = price - min_price
+    return max_profit
+
+#TC: O(N), SC: O(1)
+
+# Normal Case
+print(maxProfit([7, 1, 5, 3, 6, 4]) == 5)  
+print(maxProfit([7, 6, 4, 3, 1]) == 0)  
+
+# Edge Case
+print(maxProfit([]) == 0)  
+print(maxProfit([1]) == 0)  
+

contains-duplicate/JoyJaewon.py

@@ -27,4 +27,5 @@ def containsDuplicate(nums):
 
 #Edge case
 print(containsDuplicate([1]) == False)
-print(containsDuplicate([]) == False) 
+print(containsDuplicate([]) == False) 
+

two-sum/JoyJaewon.py

@@ -29,4 +29,5 @@ def twoSum(nums, target):
 
 #Edge case
 print(twoSum([1, 2], 3) == [0, 1])
-print(twoSum([1, 5, 10], 20) == None) 
+print(twoSum([1, 5, 10], 20) == None) 
+

valid-anagram/JoyJaewon.py

@@ -16,10 +16,7 @@ def isAnagram(s, t):
 
     count = {}
     for char in s:
-        if char in count:
-            count[char] += 1
-        else:
-            count[char] = 1
+        count[char] = count.get(char, 0) + 1  
 
     for char in t:
         if char in count:
@@ -40,4 +37,3 @@ def isAnagram(s, t):
 #Edge case
 print(isAnagram("", "") == True) 
 
-

valid-palindrome/JoyJaewon.py

@@ -0,0 +1,39 @@
+'''
+Brainstorm
+- Brute force: double for-loop O(N^2)
+- Efficient: two pointer O(N)
+
+Plan
+1. Preprocess the input data (lowercase, remove non-alphanumeric char)
+2. Use two pointers to check for palindrome
+'''
+#Version 1
+def isPalindrome(s):
+    filtered_chars = [c.lower() for c in s if c.isalnum()]
+    left, right = 0, len(filtered_chars) - 1
+
+    while left < right:
+        if filtered_chars[left] != filtered_chars[right]:
+            return False
+        left, right = left + 1, right - 1
+    
+    return True
+
+
+#Version 2
+def isPalindrome2(s):
+    filtered_chars = [c.lower() for c in s if c.isalnum()]
+    return filtered_chars == filtered_chars[::-1]
+
+
+#TC: O(N), SC: O(N)
+
+#Normal case
+print(isPalindrome("A man, a plan, a canal: Panama") == True) 
+print(isPalindrome("race a car") == False) 
+
+#Edge case
+print(isPalindrome("") == True) 
+print(isPalindrome(" ") == True) 
+print(isPalindrome("a.") == True) 
+