[최원준] Day29

leetcode3/최원준/169. Majority Element.py

@@ -0,0 +1,11 @@
+class Solution:
+    def majorityElement(self, nums: list[int]) -> int:
+        n = len(nums)
+        majority = n/2
+
+        counter = Counter(nums)
+        for num, freq in counter.items():
+            if freq >= majority:
+                return num
+
+        return -1

leetcode3/최원준/242. Valid Anagram.py

@@ -0,0 +1,12 @@
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        if len(t) != len(s):
+            return False
+
+        counter1 = Counter(s)
+        counter2 = Counter(t)
+
+        for char1, freq in counter1.items():
+            if counter2[char1] != freq:
+                return False
+        return True

leetcode3/최원준/413. Arithmetic Slices.py

@@ -0,0 +1,40 @@
+#
+
+'''
+1. 아이디어 :
+diff가 다를때마다 자른다.
+자른 길이를 재귀함수를 통해 경우의 수를 구한다
+
+2. 시간복잡도 :
+    O(n + n)
+
+3. 자료구조/알고리즘 :
+dfs
+
+'''
+class Solution:
+    def numberOfArithmeticSlices(self, nums: list[int]) -> int:
+        if len(nums) <3:
+            return 0
+
+        @cache
+        def calc(n):
+            return n // 3 if n <= 3 else n-2 + calc(n-1)
+
+        ans = 0
+        prevs = [nums[0], nums[1]]
+        diff = nums[1] - nums[0]
+
+        for num in nums[2:]:
+            if num - prevs[-1] == diff:
+                prevs.append(num)
+
+            elif num - prevs[-1] != diff:
+                if len(prevs) >= 3:
+                    ans += calc(len(prevs))
+
+                diff = num - prevs[-1]
+                prevs = [prevs[-1], num]
+        ans += calc(len(prevs))
+        return ans            
+