[SunaDu] Week 1

contains-duplicate/dusunax.py

@@ -0,0 +1,23 @@
+'''
+# Leetcode 217. Contains Duplicate
+
+use set to store distinct elements 🗂️
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- iterating through the list just once to convert it to a set.
+
+### SC is O(n):
+- creating a set to store the distinct elements of the list.
+'''
+
+class Solution:
+    def containsDuplicate(self, nums: List[int]) -> bool:
+        return len(nums) != len(set(nums))
+

top-k-frequent-elements/dusunax.py

@@ -0,0 +1,33 @@
+'''
+# Leetcode 347. Top K Frequent Elements
+
+use **Counter** to count the frequency of each element in the list 🚀
+use **sorted()** to sort the elements by their frequency in descending order.
+
+## Time and Space Complexity
+
+```
+TC: O(n log n)
+SC: O(n)
+```
+
+### TC is O(n log n):
+- iterating through the list just once to count the frequency of each element. = O(n)
+- sorting the elements by their frequency in descending order. = O(n log n)
+
+### SC is O(n):
+- using a Counter to store the frequency of each element. = O(n)
+- sorted() creates a new list that holds the elements of frequency_map. = O(n)
+- result list that holds the top k frequent elements. = O(k)
+'''
+
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        frequency_map = Counter(nums) # TC: O(n), SC: O(n)
+        sorted_frequencies = sorted(frequency_map.items(), key=lambda x: x[1], reverse=True) # TC: O(n log n), SC: O(n)
+
+        result = [] # SC: O(k)
+        for e in sorted_frequencies: # TC: O(k)
+            result.append(e[0])
+
+        return result[0:k]

valid-palindrome/dusunax.py

@@ -0,0 +1,31 @@
+'''
+# Leetcode 125. Valid Palindrome
+
+use `isalnum()` to filter out non-alphanumeric characters 🔍
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- iterating through the string just once to filter out non-alphanumeric characters. O(n)
+
+### SC is O(n):
+- `s.lower()` creates a new string. O(n)
+- creating a new string `converted_s` to store the filtered characters. O(n)
+- `converted_s[::-1]` creates a new reversed string. O(n)
+'''
+
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+        if s == " ":
+            return True
+
+        s = s.lower()
+        converted_s = ''.join(c for c in s if c.isalnum())
+
+        return converted_s == converted_s[::-1]
+