From a9ced98a3a9ed0b03ea6fd018a768b5d8474232c Mon Sep 17 00:00:00 2001
From: Nayeon <nayeonkn0330@gmail.com>
Date: Sun, 23 Feb 2025 12:18:42 +0900
Subject: [PATCH 1/5] chore: add placeholder for "Same Tree" problem

---
 same-tree/KwonNayeon.py | 13 +++++++++++++
 1 file changed, 13 insertions(+)
 create mode 100644 same-tree/KwonNayeon.py

diff --git a/same-tree/KwonNayeon.py b/same-tree/KwonNayeon.py
new file mode 100644
index 000000000..75ef8d540
--- /dev/null
+++ b/same-tree/KwonNayeon.py
@@ -0,0 +1,13 @@
+"""
+Constraints:
+- 
+
+Time Complexity: 
+- 
+
+Space Complexity: 
+- 
+
+풀이방법:
+1. 
+"""

From 9e611b5cce5235ea0817576ec3e517a1d8ca50c2 Mon Sep 17 00:00:00 2001
From: Nayeon <nayeonkn0330@gmail.com>
Date: Tue, 25 Feb 2025 04:58:19 +0900
Subject: [PATCH 2/5] solve: Same Tree

---
 same-tree/KwonNayeon.py | 33 +++++++++++++++++++++++++++------
 1 file changed, 27 insertions(+), 6 deletions(-)

diff --git a/same-tree/KwonNayeon.py b/same-tree/KwonNayeon.py
index 75ef8d540..c7c7adee7 100644
--- a/same-tree/KwonNayeon.py
+++ b/same-tree/KwonNayeon.py
@@ -1,13 +1,34 @@
 """
 Constraints:
-- 
+- The number of nodes in both trees is in the range [0, 100].
+- -10^4 <= Node.val <= 10^4
 
-Time Complexity: 
-- 
+Time Complexity: O(n)
+- 각 노드를 한 번씩 방문
 
-Space Complexity: 
-- 
+Space Complexity: O(h)
+- 재귀 호출 스택의 깊이는 트리의 높이(h)에 비례함
 
 풀이방법:
-1. 
+1. DFS와 재귀를 활용하여 두 트리를 동시에 탐색
+2. base case:
+   - p와 q가 모두 None이면 → 같은 트리
+   - 둘 중 하나만 None이거나 노드의 값이 다르면 → 다른 트리
+3. 재귀로 왼쪽과 오른쪽 서브트리가 모두 같은지 확인
 """
+# 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 isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+        if p is None and q is None:
+            return True
+        
+        if p is None or q is None or p.val != q.val:
+            return False
+
+        return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
+

From 3f7d0f8afd4100e4366e97d7c35b5cf71c8466d1 Mon Sep 17 00:00:00 2001
From: Nayeon <nayeonkn0330@gmail.com>
Date: Tue, 25 Feb 2025 09:25:11 +0900
Subject: [PATCH 3/5] solve: Remove Nth Node From End of List

---
 .../KwonNayeon.py                             | 42 +++++++++++++++++++
 1 file changed, 42 insertions(+)
 create mode 100644 remove-nth-node-from-end-of-list/KwonNayeon.py

diff --git a/remove-nth-node-from-end-of-list/KwonNayeon.py b/remove-nth-node-from-end-of-list/KwonNayeon.py
new file mode 100644
index 000000000..24914e107
--- /dev/null
+++ b/remove-nth-node-from-end-of-list/KwonNayeon.py
@@ -0,0 +1,42 @@
+"""
+Constraints:
+- The number of nodes in the list is sz.
+- 1 <= sz <= 30
+- 0 <= Node.val <= 100
+- 1 <= n <= sz
+
+Time Complexity: O(n)
+- 여기서 n은 리스트의 길이, 리스트를 한 번만 순회함
+
+Space Complexity: O(1)
+- 추가 공간을 사용하지 않음, 정해진 개수의 변수만 사용
+
+풀이방법:
+1. fast 포인터를 n번 앞으로 이동
+2. base case: 만약 fast가 None이라면, 첫 번째 노드를 제거함
+3. fast.next가 None일 때까지 두 포인터를 함께 이동
+4. slow의 다음 노드를 제거함 (slow.next = slow.next.next로 연결을 끊어냄)
+"""
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        slow = head
+        fast = head
+
+        for i in range(n):
+            fast = fast.next
+
+        if not fast:
+            return head.next
+
+        while fast.next:
+            slow = slow.next
+            fast = fast.next
+
+        slow.next = slow.next.next
+
+        return head

From 2c5eb230343e095f8ac823afdab0d85a94944b6e Mon Sep 17 00:00:00 2001
From: Nayeon <nayeonkn0330@gmail.com>
Date: Thu, 27 Feb 2025 22:21:56 +0900
Subject: [PATCH 4/5] solve: Non-overlapping Intervals

---
 non-overlapping-intervals/NayeonKwon.py | 36 +++++++++++++++++++++++++
 1 file changed, 36 insertions(+)
 create mode 100644 non-overlapping-intervals/NayeonKwon.py

diff --git a/non-overlapping-intervals/NayeonKwon.py b/non-overlapping-intervals/NayeonKwon.py
new file mode 100644
index 000000000..a6d83fd79
--- /dev/null
+++ b/non-overlapping-intervals/NayeonKwon.py
@@ -0,0 +1,36 @@
+"""
+Constraints:
+- 1 <= intervals.length <= 10^5
+- intervals[i].length == 2
+- -5 * 10^4 <= starti < endi <= 5 * 10^4
+
+Time Complexity: O(n * log n)
+- 구간 정렬 시 O(n * log n) 사용, 정렬된 구간을 순회할 때 O(n)
+
+Space Complexity: O(1)
+- 정해진 변수 외에는 공간 사용하지 않음
+
+풀이방법:
+1. 끝점을 기준으로 오름차순 정렬 <- 제거할 구간의 수를 최소화하기 위해
+2. 첫 번째 구간을 선택하고 그 구간의 끝점을 기록함
+3. 두 번째 구간부터 end보다 시작점이 크거나 같은(겹치지 않는) 구간을 찾아서 카운트
+4. 전체 구간의 수에서 count를 빼서 제거해야 할 구간의 수를 반환
+"""
+class Solution:
+    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
+        if not intervals:
+            return 0
+
+        intervals.sort(key=lambda x: x[1])
+
+        count = 1
+        end = intervals[0][1]
+
+        for interval in intervals[1:]:
+
+            if interval[0] >= end:
+                count += 1
+                end = interval[1]
+                
+        return len(intervals) - count
+

From a0a82eba9ecf6e8d38f349e396d5eb6d7693a22e Mon Sep 17 00:00:00 2001
From: Nayeon <nayeonkn0330@gmail.com>
Date: Thu, 27 Feb 2025 22:23:56 +0900
Subject: [PATCH 5/5] Edit file name to fix error

---
 non-overlapping-intervals/{NayeonKwon.py => KwonNayeon.py} | 0
 1 file changed, 0 insertions(+), 0 deletions(-)
 rename non-overlapping-intervals/{NayeonKwon.py => KwonNayeon.py} (100%)

diff --git a/non-overlapping-intervals/NayeonKwon.py b/non-overlapping-intervals/KwonNayeon.py
similarity index 100%
rename from non-overlapping-intervals/NayeonKwon.py
rename to non-overlapping-intervals/KwonNayeon.py