[chordpli] WEEK 02 solutions (#1776)

Author: chordpli

3sum/chordpli.py

@@ -0,0 +1,31 @@
+from typing import List
+
+
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        answer = []
+        nums.sort()
+        nums_len = len(nums)
+
+        for i in range(nums_len - 2):
+            if i > 0 and nums[i] == nums[i - 1]:
+                continue
+            left, right = i + 1, nums_len - 1
+
+            while left < right:
+                current_sum = nums[i] + nums[left] + nums[right]
+                if current_sum < 0:
+                    left += 1
+                elif current_sum > 0:
+                    right -= 1
+
+                else:
+                    answer.append([nums[i], nums[left], nums[right]])
+                    while left < right and nums[left] == nums[left + 1]:
+                        left += 1
+                    while left < right and nums[right] == nums[right - 1]:
+                        right -= 1
+                    left += 1
+                    right -= 1
+
+        return answer

climbing-stairs/chordpli.py

@@ -0,0 +1,13 @@
+class Solution:
+    def climbStairs(self, n: int) -> int:
+        if n <= 2:
+            return n
+
+        dp = [0] * (n + 1)
+        dp[1] = 1
+        dp[2] = 2
+
+        for i in range(3, n + 1):
+            dp[i] = dp[i - 1] + dp[i - 2]
+
+        return dp[n]

valid-anagram/chordpli.py

@@ -0,0 +1,19 @@
+class Solution:
+    # dictionary 사용 11ms
+    def isAnagram(self, s: str, t: str) -> bool:
+        check = {}
+        for char in s:
+            if char not in check:
+                check[char] = 0
+            check[char] += 1
+        for char in t:
+            if char not in check:
+                return False
+            check[char] -= 1
+            if check[char] == 0:
+                del check[char]
+        return not check
+
+    # 정렬 사용 15 - 20ms
+    # def isAnagram(self, s: str, t: str) -> bool:
+    #     return sorted(s) == sorted(t)