From 315d3654737e71dc2de7e4d8bf966cd359fa0ce2 Mon Sep 17 00:00:00 2001
From: SEONG JUN SHIN <shinsj4653@gmail.com>
Date: Sun, 20 Apr 2025 14:55:30 +0900
Subject: [PATCH 1/6] feat: add week04 problems

---
 coin-change/shinsj4653.py                         | 15 +++++++++++++++
 .../shinsj4653.py                                 | 15 +++++++++++++++
 maximum-depth-of-binary-tree/shinsj4653.py        | 15 +++++++++++++++
 merge-two-sorted-lists/shinsj4653.py              | 15 +++++++++++++++
 word-search/shinsj4653.py                         | 15 +++++++++++++++
 5 files changed, 75 insertions(+)
 create mode 100644 coin-change/shinsj4653.py
 create mode 100644 find-minimum-in-rotated-sorted-array/shinsj4653.py
 create mode 100644 maximum-depth-of-binary-tree/shinsj4653.py
 create mode 100644 merge-two-sorted-lists/shinsj4653.py
 create mode 100644 word-search/shinsj4653.py

diff --git a/coin-change/shinsj4653.py b/coin-change/shinsj4653.py
new file mode 100644
index 000000000..f2dd9f373
--- /dev/null
+++ b/coin-change/shinsj4653.py
@@ -0,0 +1,15 @@
+"""
+[문제풀이]
+# Inputs
+
+# Outputs
+
+# Constraints
+
+# Ideas
+
+[회고]
+
+"""
+
+
diff --git a/find-minimum-in-rotated-sorted-array/shinsj4653.py b/find-minimum-in-rotated-sorted-array/shinsj4653.py
new file mode 100644
index 000000000..f2dd9f373
--- /dev/null
+++ b/find-minimum-in-rotated-sorted-array/shinsj4653.py
@@ -0,0 +1,15 @@
+"""
+[문제풀이]
+# Inputs
+
+# Outputs
+
+# Constraints
+
+# Ideas
+
+[회고]
+
+"""
+
+
diff --git a/maximum-depth-of-binary-tree/shinsj4653.py b/maximum-depth-of-binary-tree/shinsj4653.py
new file mode 100644
index 000000000..f2dd9f373
--- /dev/null
+++ b/maximum-depth-of-binary-tree/shinsj4653.py
@@ -0,0 +1,15 @@
+"""
+[문제풀이]
+# Inputs
+
+# Outputs
+
+# Constraints
+
+# Ideas
+
+[회고]
+
+"""
+
+
diff --git a/merge-two-sorted-lists/shinsj4653.py b/merge-two-sorted-lists/shinsj4653.py
new file mode 100644
index 000000000..f2dd9f373
--- /dev/null
+++ b/merge-two-sorted-lists/shinsj4653.py
@@ -0,0 +1,15 @@
+"""
+[문제풀이]
+# Inputs
+
+# Outputs
+
+# Constraints
+
+# Ideas
+
+[회고]
+
+"""
+
+
diff --git a/word-search/shinsj4653.py b/word-search/shinsj4653.py
new file mode 100644
index 000000000..f2dd9f373
--- /dev/null
+++ b/word-search/shinsj4653.py
@@ -0,0 +1,15 @@
+"""
+[문제풀이]
+# Inputs
+
+# Outputs
+
+# Constraints
+
+# Ideas
+
+[회고]
+
+"""
+
+

From 6d98663b51f49179ad792d9256e235e1bf101619 Mon Sep 17 00:00:00 2001
From: SEONG JUN SHIN <shinsj4653@gmail.com>
Date: Mon, 21 Apr 2025 22:44:56 +0900
Subject: [PATCH 2/6] feat: add word-search solution

---
 word-search/shinsj4653.py | 141 +++++++++++++++++++++++++++++++++++++-
 1 file changed, 138 insertions(+), 3 deletions(-)

diff --git a/word-search/shinsj4653.py b/word-search/shinsj4653.py
index f2dd9f373..ceccff14c 100644
--- a/word-search/shinsj4653.py
+++ b/word-search/shinsj4653.py
@@ -1,15 +1,150 @@
 """
 [문제풀이]
 # Inputs
-
+- board: m*n grid of characters
+- word: str
 # Outputs
-
+- word 가 grid 안에 있으면 true, 아니면 false
 # Constraints
+- 탐색: 상하좌우
+- m == board.length
+- n = board[i].length
+- 1 <= m, n <= 6
+- 1 <= word.length <= 15
 
+- board, word => 무조건 대소문자로 구성
 # Ideas
+1. 퍼지면서 탐색 -> bfs?
+q = cur.append(i, j, "")
+q.pop() 해서
+if 현재 w 가 word랑 같다면 그 즉시 bfs stop하고 return True
+만약 bfs 끝까지 했는데도 안나오면 return False
 
-[회고]
+bfs: q에 다음 원소 넣을 때 현재 값에 이어붙인문자가 w랑 어떤식으로 비교해야 맞는 조건절이 되는걸까?
+=> cur_w의 len 값을 i로 사용하려했지만 w가 더 짧으면 idxError 뜰까봐 보류
+=> 메모리 초과 오류 뜸..
+=> 아..같은 자리를 방문하면 안됨!! 즉, 원복이 필요함!
+-> 백트래킹으로 풀이 change
+
+2. dfs로도 꼭 풀어보기!
+백트래킹 복습 용으로 굿
+
+board안의 숫자는 오직 대소문자 -> isalpha() 로 체킹 가능
+=> 방문하면 0으로 바꾸기
+
+풀기는 풀었지만..7090ms : Beats 7.34%
+
+Follow up : Could you use search pruning to make your solution faster with a larger board
+가지치기 기법 어떻게 도입할까??
+=> cur_idx 하나 둬서, AB.. ABC -> 진행 도중, 하나라도 word랑 다르면 바로 그 경로 컷
+=> 그래도 4587ms.. Beats 41.00%
+=> 먼가 더 정석 성능 개선이 필요
+
+해설 참고
+=> word_idx를 이용, any문을 이용한 리팩토링
+=> 해설 풀이 : 4385ms
 
+[회고]
+idx를 이용한 가지치기 방법 잘 알아두자
 """
 
+# from collections import deque
+# class Solution:
+#     def exist(board, word):
+#
+#         q = deque([])
+#         n, m = len(board), len(board[0])
+#
+#         q.append((0, 0, board[0][0]))
+#
+#         dy = [-1, 0, 1, 0]
+#         dx = [0, 1, 0, -1]
+#
+#         while q:
+#             cur_y, cur_x, cur_w = q.popleft()
+#
+#             if cur_w == word:
+#                 return True
+#
+#             for i in range(3):
+#                 ny = cur_y + dy[i]
+#                 nx = cur_x + dx[i]
+#
+#                 if 0 <= ny < n and 0 <= nx < m:
+#                     q.append((ny, nx, cur_w + board[ny][nx]))
+#
+#         return False
+#
+#     exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], "ABCCDE")
+
+
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+
+        n, m = len(board), len(board[0])
+
+        dy = [-1, 0, 1, 0]
+        dx = [0, 1, 0, -1]
+
+        v = [[False for _ in range(m)] for _ in range(n)]
+
+        def dfs(y, x, w):
+            #print('y, x : ', y, x)
+            #print('w: ', w)
+            if w == word:
+                return True
+
+            if len(w) >= len(word):
+                return False
+
+            for k in range(4):
+                ny = y + dy[k]
+                nx = x + dx[k]
+
+                if 0 <= ny < n and 0 <= nx < m and not v[ny][nx]:
+                    v[ny][nx] = True
+                    if dfs(ny, nx, w + board[ny][nx]):
+                        return True
+                    v[ny][nx] = False
+
+        for i in range(n):
+            for j in range(m):
+                if not v[i][j] and board[i][j] == word[0]:
+                    v[i][j] = True
+                    if dfs(i, j, board[i][j]):
+                        return True
+                    v[i][j] = False
+
+        return False
+
+# 해설
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+
+        n, m = len(board), len(board[0])
+
+        dy = [-1, 0, 1, 0]
+        dx = [0, 1, 0, -1]
+
+        def dfs(y, x, idx):
+            if idx == len(word):
+                return True
+
+            if not (0 <= y < n and 0 <= x < m):
+                return False
+
+            if board[y][x] != word[idx]:
+                return False
+
+            temp = board[y][x]
+            board[y][x] = ""
+
+            for i in range(4):
+                if dfs(y + dy[i], x + dx[i], idx + 1):
+                    return True
+
+            board[y][x] = temp
+            return False
+
+        return any(dfs(i, j, 0) for i in range(n) for j in range(m))
 

From acf37c0cff16e645622b1279f2c98283c7787800 Mon Sep 17 00:00:00 2001
From: SEONG JUN SHIN <shinsj4653@gmail.com>
Date: Sat, 26 Apr 2025 20:43:03 +0900
Subject: [PATCH 3/6] feat: add coin-change solution

---
 coin-change/shinsj4653.py | 38 ++++++++++++++++++++++++++++++++++----
 1 file changed, 34 insertions(+), 4 deletions(-)

diff --git a/coin-change/shinsj4653.py b/coin-change/shinsj4653.py
index f2dd9f373..246706c18 100644
--- a/coin-change/shinsj4653.py
+++ b/coin-change/shinsj4653.py
@@ -1,15 +1,45 @@
 """
 [문제풀이]
 # Inputs
-
+- 정수 배열 coins
+- 총 돈의 값 amount
 # Outputs
-
+- amount 를 채울 수 있는 동전의 가장 적은 수
+- 만약 만들 수 없다면 -1 return
 # Constraints
-
+- 1 <= coins.length <= 12
+- 1 <= coins[i] <= 2^31 - 1
+- 0 <= amount <= 10^4
 # Ideas
+- 우선 이전 풀이 본 거 연습해보기
+- DP 탑다운 방식으로! 이 문제 풀이를 내가 100% 이해하고 있는게 맞는지 손으로 도식화 해보기로함
 
-[회고]
 
+[회고]
+재귀랑 dp는 다양한 유형을 많이 풀어보는 수 밖에..
 """
 
 
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        memo = {}
+
+        def dp(amount):
+
+            if amount == 0:
+                return 0
+
+            retList = []
+            for coin in coins:
+                if amount - coin >= 0:
+                    if amount - coin not in memo:
+                        memo[amount - coin] = dp(amount - coin)
+
+                    if memo[amount - coin] != -1:
+                        retList.append(memo[amount - coin])
+
+            return min(retList) + 1 if retList else -1
+
+        return dp(amount)
+
+

From caca32904a8b88c68160a5486bb20052c139cd9f Mon Sep 17 00:00:00 2001
From: SEONG JUN SHIN <shinsj4653@gmail.com>
Date: Sat, 26 Apr 2025 23:17:56 +0900
Subject: [PATCH 4/6] feat: add merge-two-sorted-lists solution

---
 merge-two-sorted-lists/shinsj4653.py | 51 ++++++++++++++++++++++++++--
 1 file changed, 48 insertions(+), 3 deletions(-)

diff --git a/merge-two-sorted-lists/shinsj4653.py b/merge-two-sorted-lists/shinsj4653.py
index f2dd9f373..2b7a9f575 100644
--- a/merge-two-sorted-lists/shinsj4653.py
+++ b/merge-two-sorted-lists/shinsj4653.py
@@ -1,15 +1,60 @@
 """
 [문제풀이]
 # Inputs
-
+- two sorted linked lists
 # Outputs
-
+- 두 링크드 리스트를 합쳐서 하나의 정렬된 (오름차순) 리스트 반환
 # Constraints
-
+- The number of nodes in both lists is in the range [0, 50].
+- -100 <= Node.val <= 100
+- Both list1 and list2 are sorted in non-decreasing order.
 # Ideas
+두 링크드 리스트는 모두 오름차순으로 정렬되어 있음
+-> ListNode는 val, next 로 되어있음
+
+우선 두 리스트 이어붙이고,
+먼가 리스트 정렬 알고리즘 써야할듯?
+리스트라서, 두 노드를 바꾸는 swap() 함수 쓰면서,
+왼쪽보다 오른쪽이 작으면 바꾸는 식으로 하면 될듯?
+
+즉, 붙어있는 두 노드 비교하면서 쭉 돌기
+하지만 인덱스가 없어서 어떻게 순회할지?
+-> point라는 기준을 둬서 이 point를 기준으로 prev, next를 만들어서 순회
+-> 근데 prev, next만으로는 swap이 어려워서 2개가 아닌 3개를 둬야할 것 같은데,
+구현 방법이 떠오르지 않아서 해설 참고
+
+해설 참고
+- 기존 리스트 변경이 아닌, 아예 새로운 리스트를 재창조 하는 방식으로!
+
 
 [회고]
+기존 리스트를 swap하려고 하다 보니, 어렵게 생각하게됨..
 
 """
 
 
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
+
+        retList = ListNode(None)
+        head = retList
+
+        while list1 and list2:
+            if list1.val < list2.val:
+                head.next = list1
+                list1 = list1.next
+
+            else:
+                head.next = list2
+                list2 = list2.next
+
+            head = head.next
+
+        head.next = list1 or list2
+        return retList.next
+    
\ No newline at end of file

From 94686144d3ad089a0e9e5d2c65718f896ac2527d Mon Sep 17 00:00:00 2001
From: SEONG JUN SHIN <shinsj4653@gmail.com>
Date: Sat, 26 Apr 2025 23:34:08 +0900
Subject: [PATCH 5/6] feat: add maximum-depth-of-binary-tree sol

---
 maximum-depth-of-binary-tree/shinsj4653.py | 47 ++++++++++++++++++++--
 merge-two-sorted-lists/shinsj4653.py       |  2 +-
 2 files changed, 45 insertions(+), 4 deletions(-)

diff --git a/maximum-depth-of-binary-tree/shinsj4653.py b/maximum-depth-of-binary-tree/shinsj4653.py
index f2dd9f373..7d09b54ac 100644
--- a/maximum-depth-of-binary-tree/shinsj4653.py
+++ b/maximum-depth-of-binary-tree/shinsj4653.py
@@ -1,15 +1,56 @@
 """
 [문제풀이]
 # Inputs
-
+- 이진 트리의 root
 # Outputs
-
+- 가장 최고 깊이
 # Constraints
-
+The number of nodes in the tree is in the range [0, 104].
+-100 <= Node.val <= 100
 # Ideas
+dfs, 즉 재귀 돌면서
+depth 를 max로 업데이트 하면 될 것 같은 느낌?
 
 [회고]
+내 풀이는 정답
+해설 방법도 궁금하여 참고
+-> TC, SC도 파악하는 연습!
+
+TC: o(n)
+SC: o(n)
 
 """
 
+# 내 풀이
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def maxDepth(self, root: Optional[TreeNode]) -> int:
+        def dfs(node, depth):
+            if node is None:
+                return depth - 1
+
+            return max(dfs(node.left, depth + 1), dfs(node.right, depth + 1))
+
+        return dfs(root, 1)
+# 해설
+# bfs 풀이도 존재
+class Solution:
+    def maxDepth(self, root: Optional[TreeNode]) -> int:
+        if not root:
+            return 0
+        max_depth = 0
+        stack = [(root, 1)]
+        while stack:
+            node, depth = stack.pop()
+            max_depth = max(depth, max_depth)
+            if node.left:
+                stack.append((node.left, depth + 1))
+            if node.right:
+                stack.append((node.right, depth + 1))
+        return max_depth
 
diff --git a/merge-two-sorted-lists/shinsj4653.py b/merge-two-sorted-lists/shinsj4653.py
index 2b7a9f575..c85a4ed69 100644
--- a/merge-two-sorted-lists/shinsj4653.py
+++ b/merge-two-sorted-lists/shinsj4653.py
@@ -57,4 +57,4 @@ def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) ->
 
         head.next = list1 or list2
         return retList.next
-    
\ No newline at end of file
+

From afe0d4a35998401a585d551ee3b4a022fd37c7f9 Mon Sep 17 00:00:00 2001
From: SEONG JUN SHIN <shinsj4653@gmail.com>
Date: Sun, 27 Apr 2025 00:24:07 +0900
Subject: [PATCH 6/6] feat: add find-minimum-in-rotated-sorted-array sol

---
 .../shinsj4653.py                             | 74 ++++++++++++++++++-
 1 file changed, 71 insertions(+), 3 deletions(-)

diff --git a/find-minimum-in-rotated-sorted-array/shinsj4653.py b/find-minimum-in-rotated-sorted-array/shinsj4653.py
index f2dd9f373..ff858b4a9 100644
--- a/find-minimum-in-rotated-sorted-array/shinsj4653.py
+++ b/find-minimum-in-rotated-sorted-array/shinsj4653.py
@@ -1,15 +1,83 @@
 """
 [문제풀이]
 # Inputs
-
+- sorted rotated array : nums
+- nums안의 요소는 unique
 # Outputs
-
+- 배열의 가장 작은 값 Return
 # Constraints
-
+- O(logN) 시간대로 풀어야함
+- n == nums.length
+- 1 <= n <= 5000
+- -5000 <= nums[i] <= 5000
+- All the integers of nums are unique.
+- nums is sorted and rotated between 1 and n times.
 # Ideas
+그냥 정렬은 O(NlogN) 이어서 불가능
+rotate : 맨 끝의 요소가 맨 처음으로 옴
+이진탐색이 O(logN) 이긴한데..->하지만, 이진탐색은 정렬된 배열에서만 사용가능
+그리고, 이 문제에선 아닌 것 같다
+
+끝의 원소를 빼서 stack에 append -> 모두 o(1)
+
+회전되어 있는 배열의 특성을 활용하여
+-> 끝의 원소 빼고, 그걸 stack 에 넣기
+-> 그리고 stack의 맨 끝 원소보다 nums의 끝 원소가 더 클때
+stack[-1] 이 정답
+-> 근데 이건 o(n) 아닌가??
+
+우선 내 풀이 정답
+해설 참고
+-> 내 풀이는 O(n) 이다.. O(logN) 풀이가 필요
+이분탐색 응용문제이다
 
 [회고]
 
 """
 
+# 내 풀이
+from collections import deque
+
+
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+
+        if len(nums) == 1:
+            return nums[0]
+
+        nums = deque(nums)
+
+        st = []
+        ret = 0
+
+        while nums:
+            num = nums.pop()
+
+            if not nums:
+                ret = num
+                break
+
+            if num < nums[-1]:
+                ret = num
+                break
+
+        return ret
+
+# 해설
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+
+        left, right = 1, len(nums) - 1
+
+        while left <= right:
+            mid = (left + right) // 2
+            if nums[mid - 1] > nums[mid]:
+                return nums[mid]
+            if nums[0] < nums[mid]:
+                left = mid + 1
+            else:
+                right = mid - 1
+
+        return nums[0]
+