From 4f4795bac9f25ae71612ab70b0a3a80869a6ea01 Mon Sep 17 00:00:00 2001
From: Taehun Lim <routiful@gmail.com>
Date: Sun, 8 Dec 2024 16:10:07 +0900
Subject: [PATCH 1/2] [First week][contains-duplicate] adds answer

---
 contains-duplicate/routiful.py | 37 ++++++++++++++++++++++++++++++++++
 1 file changed, 37 insertions(+)
 create mode 100644 contains-duplicate/routiful.py

diff --git a/contains-duplicate/routiful.py b/contains-duplicate/routiful.py
new file mode 100644
index 000000000..b708501c9
--- /dev/null
+++ b/contains-duplicate/routiful.py
@@ -0,0 +1,37 @@
+# FIRST WEEK
+
+# Question : 217. Contains Duplicate
+# Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.
+
+# Example 1:
+# Input: nums = [1,2,3,1]
+# Output: true
+# Explanation:
+#   The element 1 occurs at the indices 0 and 3.
+
+# Example 2:
+# Input: nums = [1,2,3,4]
+# Output: false
+# Explanation:
+#   All elements are distinct.
+
+# Example 3:
+# Input: nums = [1,1,1,3,3,4,3,2,4,2]
+# Output: true
+
+# Constraints:
+#   1 <= nums.length <= 105
+#   -109 <= nums[i] <= 109 
+
+# Notes:
+# Counts every elements(in the list) using 'count()' API. Luckily it is less than O(n) but maximum O(n)
+# Makes Dict; the key is elements of the list. If the length of them are different, the list has duplicate elements. It is O(n)
+
+class Solution:
+    def containsDuplicate(self, nums: List[int]) -> bool:
+        d = {}
+        for n in nums:
+            d[n] = 1
+        if len(d) != len(nums):
+            return True
+        return False

From 5d3df3d006e0a30fddf2eb9b7c2de433cff37063 Mon Sep 17 00:00:00 2001
From: Taehun Lim <routiful@gmail.com>
Date: Sun, 15 Dec 2024 20:24:15 +0900
Subject: [PATCH 2/2] [First week][valid-palindrome] adds answer

---
 valid-palindrome/routiful.py | 59 ++++++++++++++++++++++++++++++++++++
 1 file changed, 59 insertions(+)
 create mode 100644 valid-palindrome/routiful.py

diff --git a/valid-palindrome/routiful.py b/valid-palindrome/routiful.py
new file mode 100644
index 000000000..bff3b1a86
--- /dev/null
+++ b/valid-palindrome/routiful.py
@@ -0,0 +1,59 @@
+# FIRST WEEK
+
+# Question : 125. Valid Palindrome
+
+# A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and
+# removing all non-alphanumeric characters, it reads the same forward and backward.
+# Alphanumeric characters include letters and numbers.
+# Given a string s, return true if it is a palindrome, or false otherwise.
+
+# Example 1:
+# Input: s = "A man, a plan, a canal: Panama"
+# Output: true
+# Explanation: "amanaplanacanalpanama" is a palindrome.
+
+# Example 2:
+# Input: s = "race a car"
+# Output: false
+# Explanation: "raceacar" is not a palindrome.
+
+# Example 3:
+# Input: s = " "
+# Output: true
+# Explanation: s is an empty string "" after removing non-alphanumeric characters.
+# Since an empty string reads the same forward and backward, it is a palindrome.
+
+# Constraints:
+# 1 <= s.length <= 2 * 105
+# s consists only of printable ASCII characters.
+
+# Notes:
+# Using `reverse`(O(N)) api and matching all(max O(N)) look straightforward.
+# The two pointer method may useful to decrease time complexity.
+# If length of input is 0 or 1, return true.
+
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+        l = len(s)
+        if l < 2:
+            return True
+
+        f = 0
+        b = l - 1
+
+        while (f <= b):
+            if (s[f] == " " or (s[f].isalpha() == False and s[f].isnumeric() == False)):
+                f = f + 1
+                continue
+
+            if (s[b] == " " or (s[b].isalpha() == False and s[b].isnumeric() == False)):
+                b = b - 1
+                continue
+
+            if s[f].lower() != s[b].lower():
+                return False
+
+            f = f + 1
+            b = b - 1
+
+        return True