Merge pull request DaleStudy#2513 from gcount85/main

SamTheKorean · web-flow · commit 42d1f47db3d4 · 2026-04-13T19:41:28.000-04:00
[gcount85] WEEK 06 Solutions
diff --git a/container-with-most-water/gcount85.py b/container-with-most-water/gcount85.py
@@ -0,0 +1,24 @@
+"""
+# Approach
+투포인터로 양 끝에서 물의 양을 계산해나갑니다.
+더 낮은 height인 쪽의 포인터를 줄여나가는데, 높이가 같은 경우에는 왼쪽 포인터만 움직입니다.
+
+# Complexity
+height의 길이가 N일 때
+- Time complexity: O(N)
+- Space complexity: O(1)
+"""
+
+
+class Solution:
+    def maxArea(self, height: list[int]) -> int:
+        n = len(height)
+        left, right = 0, n - 1
+        best = 0
+        while left < right:
+            best = max(best, (right - left) * min(height[left], height[right]))
+            if height[left] > height[right]:
+                right -= 1
+            else:
+                left += 1
+        return best
diff --git a/longest-increasing-subsequence/gcount85.py b/longest-increasing-subsequence/gcount85.py
@@ -0,0 +1,22 @@
+"""
+# Approach
+dp[n]은 nums[n] 위치까지 lis 길이입니다.
+
+# Complexity
+nums의 길이를 n이라고 할 때,
+- Time complexity: 이중 반복문 O(N^2)
+- Space complexity: dp 배열 O(N)
+"""
+
+
+class Solution:
+    def lengthOfLIS(self, nums: list[int]) -> int:
+        n = len(nums)
+        dp = [1] * n
+        answer = 1
+        for i in range(1, n):
+            for j in range(i):
+                if nums[j] < nums[i]:
+                    dp[i] = max(dp[j] + 1, dp[i])
+            answer = max(answer, dp[i])
+        return answer
diff --git a/spiral-matrix/gcount85.py b/spiral-matrix/gcount85.py
@@ -0,0 +1,41 @@
+"""
+# Approach
+상하좌우 경계값을 설정하고 matrix을 순회하며 경계값을 줄여나갑니다.
+
+# Complexity
+matrix 크기의 가로를 M, 세로를 N이라고 할 때
+- Time complexity: O(M*N)
+- Space complexity: O(M*N)
+"""
+
+
+class Solution:
+    def spiralOrder(self, matrix: list[list[int]]) -> list[int]:
+        start_row, start_col = 0, 0
+        end_row, end_col = len(matrix) - 1, len(matrix[0]) - 1
+        output = []
+
+        while start_row <= end_row and start_col <= end_col:
+            # 좌->우
+            for col in range(start_col, end_col + 1):
+                output.append(matrix[start_row][col])
+            start_row += 1
+
+            # 상->하
+            for row in range(start_row, end_row + 1):
+                output.append(matrix[row][end_col])
+            end_col -= 1
+
+            # 우->좌
+            if start_row <= end_row:
+                for col in range(end_col, start_col - 1, -1):
+                    output.append(matrix[end_row][col])
+                end_row -= 1
+
+            # 하->상
+            if start_col <= end_col:
+                for row in range(end_row, start_row - 1, -1):
+                    output.append(matrix[row][start_col])
+                start_col += 1
+
+        return output
diff --git a/valid-parentheses/gcount85.py b/valid-parentheses/gcount85.py
@@ -0,0 +1,23 @@
+"""
+# Approach
+스택을 이용하여 닫는 괄호가 나올 때 스택 top과 비교하여 짝이 맞으면 pop합니다.
+짝이 맞지 않으면 invalid, 최종적으로 스택이 비어있지 않으면 invalid 합니다.
+
+# Complexity
+s의 길이를 N이라고 할 때
+- Time complexity: O(N)
+- Space complexity: O(N)
+"""
+
+
+class Solution:
+    def isValid(self, s: str) -> bool:
+        stack = []
+        brackets = {"(": ")", "{": "}", "[": "]"}
+        for b in s:
+            if b in brackets:
+                stack.append(b)
+                continue
+            if not stack or brackets[stack.pop()] != b:
+                return False
+        return not stack
