From bb6378c4eb17fb2c6384d6cbbab466fab77f3c4e Mon Sep 17 00:00:00 2001
From: Dusuna <94776135+dusunax@users.noreply.github.com>
Date: Wed, 26 Feb 2025 23:50:29 +0900
Subject: [PATCH 1/4] add solution: same-tree

---
 same-tree/dusunax.py | 37 +++++++++++++++++++++++++++++++++++++
 1 file changed, 37 insertions(+)
 create mode 100644 same-tree/dusunax.py

diff --git a/same-tree/dusunax.py b/same-tree/dusunax.py
new file mode 100644
index 000000000..d1e4bd7ee
--- /dev/null
+++ b/same-tree/dusunax.py
@@ -0,0 +1,37 @@
+'''
+# 100. Same Tree
+
+## Base case
+- if both nodes are None, return True
+- if one of the nodes is None, return False
+- if the values of the nodes are different, return False
+
+## Recursive case
+- check if the left subtrees are the same
+- check if the right subtrees are the same
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+#### TC is O(n):
+- visiting each node once, balanced or skewed, it's O(n)
+
+#### SC is O(n):
+- for h is tree's height.
+  - in best case(balanced): h is logN, so SC is O(logN)
+  - in worst case(skewed): h is N, so SC is O(N)
+'''
+class Solution:
+    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+        if not p and not q:
+            return True
+        if not p or not q:
+            return False
+        if p.val != q.val:
+            return False
+
+        return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)

From 94106fc467c76a0ee178d55ad0ad584cf57e7a48 Mon Sep 17 00:00:00 2001
From: Dusuna <94776135+dusunax@users.noreply.github.com>
Date: Thu, 27 Feb 2025 23:54:13 +0900
Subject: [PATCH 2/4] add solution: remove-nth-node-from-end-of-list

---
 remove-nth-node-from-end-of-list/dusunax.py | 70 +++++++++++++++++++++
 1 file changed, 70 insertions(+)
 create mode 100644 remove-nth-node-from-end-of-list/dusunax.py

diff --git a/remove-nth-node-from-end-of-list/dusunax.py b/remove-nth-node-from-end-of-list/dusunax.py
new file mode 100644
index 000000000..805d34d5f
--- /dev/null
+++ b/remove-nth-node-from-end-of-list/dusunax.py
@@ -0,0 +1,70 @@
+'''
+# 19. Remove Nth Node From End of List
+
+## 풀이 접근방식 레퍼런스
+- https://www.algodale.com/problems/remove-nth-node-from-end-of-list/
+'''
+class Solution:
+    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        '''
+        # 1. 노드 리스트
+        - 음수 인덱스 원소 접근
+        - nodes 배열을 사용하여 마지막 n번째 노드를 찾는다.
+        - 메모리 낭비
+        TC: O(N), SC: O(N)
+        '''
+        # nodes = [] # 배열
+        # temp = ListNode(None, head)
+        # node = temp
+
+        # while node:
+        #     nodes.append(node)
+        #     node = node.next
+        
+        # nodes[-n - 1].next = nodes[-n - 1].next.next
+
+        # return temp.next
+
+        '''
+        # 2. 큐
+        - 큐를 사용하여 마지막 n번째 노드를 찾는다.
+        - 최대 n+1개의 노드(필요한 범위)만 유지, 슬라이딩 윈도우
+        TC: O(N), SC: O(N)
+        '''
+        # queue = deque() #큐
+        # temp = ListNode(None, head)
+        # node = temp
+
+        # for _ in range(n + 1):
+        #     queue.append(node)
+        #     node = node.next
+        
+        # while node:
+        #     queue.popleft()
+        #     queue.append(node)
+        #     node = node.next
+
+        # queue[0].next = queue[0].next.next
+        # return temp.next
+
+        '''
+        # 3. 포인터
+        - 연결 리스트의 특성 이용
+        - 첫번째 포인터는 n번 이동하여 마지막 노드를 찾는다.
+        - 두번째 포인터는 첫번째 포인터가 마지막 노드를 찾을 때까지 이동한다.
+        - 두번째 포인터는 첫번째 포인터가 마지막 노드를 찾으면 첫번째 포인터의 이전 노드를 삭제한다.
+        TC: O(N), SC: O(1)
+        '''
+        first = head
+        for _ in range(n):
+            first = first.next
+
+        temp = ListNode(None, head)
+        second = temp
+        
+        while first:
+            first = first.next
+            second = second.next
+
+        second.next = second.next.next
+        return temp.next 

From 50791a818ce5b7fd80ee349c1049a62c9d2619d4 Mon Sep 17 00:00:00 2001
From: Dusuna <94776135+dusunax@users.noreply.github.com>
Date: Fri, 28 Feb 2025 15:01:18 +0900
Subject: [PATCH 3/4] add solution:
 number-of-connected-components-in-an-undirected-graph

---
 .../dusunax.py                                | 107 ++++++++++++++++++
 1 file changed, 107 insertions(+)
 create mode 100644 number-of-connected-components-in-an-undirected-graph/dusunax.py

diff --git a/number-of-connected-components-in-an-undirected-graph/dusunax.py b/number-of-connected-components-in-an-undirected-graph/dusunax.py
new file mode 100644
index 000000000..33d471546
--- /dev/null
+++ b/number-of-connected-components-in-an-undirected-graph/dusunax.py
@@ -0,0 +1,107 @@
+'''
+# 323. Number of Connected Components in an Undirected Graph
+
+- 무방향 그래프에서 연결된 컴포넌트(connected component)의 개수 구하기
+- 노드 개수 n: 0 ~ n-1
+- 간선 리스트 edges: 노드 쌍 [u, v]
+- 연결된 컴포넌트의 개수를 반환한다.
+
+## 그래프
+- edges를 통해 인접 리스트(adjacency list)를 생성한다.
+- 무방향 그래프이므로, u ↔ v 양쪽 방향으로 연결
+
+## 풀이 알고리즘
+- DFS(스택)/BFS(큐)
+  - 각 컴포넌트를 탐색하며 방문 처리
+  - 방문하지 않은 노드를 찾으면 새로운 컴포넌트이므로 카운트 증가
+- Union-Find
+  - 각 노드가 어떤 그룹에 속하는 지 parent 배열로 표현
+  - find로 부모를 찾고, union으로 노드를 합친다.
+  - 최종적으로 독립된 그룹의 개수를 반환한다.
+'''
+
+'''
+1. DFS(스택)
+'''
+def countComponentsDFS(self, n: int, edges: List[List[int]]) -> int:
+  graph = defaultdict(list)
+  for u, v in edges: # 인접 리스트 생성
+    graph[u].append(v)
+    graph[v].append(u)
+  
+  visited = set() # 방문처리할 set
+  count = 0
+  
+  def dfs(start):
+    stack = [start]
+    visited.add(start)
+
+    while stack:
+      curr = stack.pop()
+      for neighbor in graph[curr]:
+        if neighbor not in visited:
+          visited.add(neighbor)
+          stack.append(neighbor)
+          
+  for node in range(n):
+    if node not in visited:
+      dfs(node)
+      count += 1
+  
+  return count
+
+'''
+2. BFS(큐)
+'''
+def countComponentsBFS(self, n: int, edges: List[List[int]]) -> int:
+  graph = defaultdict(list)
+  for u, v in edges: # 인접 리스트 생성
+    graph[u].append(v)
+    graph[v].append(u)
+  
+  visited = set() # 방문처리할 set
+  count = 0
+  
+  def bfs(start):
+    queue = deque([start])
+    visited.add(start)
+
+    while queue:
+      curr = queue.popleft()
+      for neighbor in graph[curr]:
+        if neighbor not in visited:
+          visited.add(neighbor)
+          queue.append(neighbor)
+        
+  for node in range(n):
+    if node not in visited:
+      bfs(node)
+      count += 1
+  
+  return count
+
+'''
+3. Union-Find
+- 간선을 순회하며 두 노드를 union으로 합친다.(parent 업데이트)
+- find는 경로 압축을 통해 대표(root)를 찾는다.(parent에서 재귀로 찾기)
+- 최종 root를 조사하고, 서로 다른 root의 개수를 반환한다.
+'''
+def countComponentsUnionFind(self, n: int, edges: List[List[int]]) -> int:
+  parent = list(range(n)) # 각 노드의 부모를 자신으로 초기화
+
+  def find(x):
+    if parent[x] != x:
+      parent[x] = find(parent[x]) # 경로 압축 path compression
+    return parent[x]
+  
+  def union(x, y):
+    rootX = find(x)
+    rootY = find(y)
+    if rootX != rootY:
+      parent[rootX] = rootY
+  
+  for u, v in edges:
+    union(u, v)
+  
+  # 모든 노드의 최종 부모의, 유일한 root의 개수를 반환
+  return len(set(find(i) for i in range(n)))

From 6079c79ed5b35827d27900e8715965ccb812a0cb Mon Sep 17 00:00:00 2001
From: Dusuna <94776135+dusunax@users.noreply.github.com>
Date: Fri, 28 Feb 2025 21:01:01 +0900
Subject: [PATCH 4/4] add solution:
 number-of-connected-components-in-an-undirected-graph

---
 non-overlapping-intervals/dusunax.py | 53 ++++++++++++++++++++++++++++
 1 file changed, 53 insertions(+)
 create mode 100644 non-overlapping-intervals/dusunax.py

diff --git a/non-overlapping-intervals/dusunax.py b/non-overlapping-intervals/dusunax.py
new file mode 100644
index 000000000..182744ee3
--- /dev/null
+++ b/non-overlapping-intervals/dusunax.py
@@ -0,0 +1,53 @@
+'''
+# 435. Non-overlapping Intervals
+
+## understanding the problem
+- 겹치지 않는 인터벌을 최대한 많이 남기기 위해, 지워야하는 인터벌의 최소값 반환
+- not it: 그래프에 순환이 있는지 여부를 알아본다❌ overkill
+- core approach: Greedy 
+
+## Greedy
+- Whenever you detect an overlap, you should remove one.
+    - not physically. we'll going to return counts, so just keep counts is enough.
+- Goal: keep many non-overlapping intervals as possible to minimize removal.
+
+### how to detect overlap?
+1. sort intervals by ends
+2. iterate through intervals while tracking down the prev_end(last vaild end time)
+  - if `current_start < prev_end`, it's overlap.
+    - Example: 
+      - [2, 4] is overlap with [1, 3]
+      - start (2) is smaller than end (3)
+
+### which one should removed?
+- when overlap happens, remove the interval that ends later 
+  - it will restrict more future intervals.
+    - the longer an interval lasts, the more it blocks others(leaving less room for non-overlapping intervals)
+
+## complexity
+
+### TC is O(n log n)
+- sorting: O(n log n)
+- iterating: O(n)
+- total: O(n log n)
+
+### SC is O(1)
+- no extra space is used
+'''
+class Solution:
+    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
+        intervals.sort(key=lambda x: x[1])
+
+        count = 0 # number of removals
+        prev_end = intervals[0][1] # last valid end time
+        
+        for start, end in intervals[1:]: # 1 ~ n-1
+            if start < prev_end: # overlap detected
+                count += 1 
+                # <do NOT move the prev_end pointer>
+                # prev_end is still pointing at the previous interval 
+                # so it's keeping the interval ends earlier. (removing the longer one)
+            else: 
+                prev_end = end
+        
+        return count