diff --git a/memo.md b/memo.md
index 4bd0397..546b61c 100644
--- a/memo.md
+++ b/memo.md
@@ -1 +1,349 @@
 # Step1
+
+## アプローチ
+
+* ノードとエッジが与えられて, 繋がっている塊の数を返せばいい
+    * グラフ理論で言うところの, 森？かな
+* Union-Findを使って, エッジごとに更新をしていけば, 各ノードのグループを表す親ノードを得られる
+    * エッジの個数 E
+    * ノードの個数 N
+    * １回のUnionにかかる最悪計算量
+        * O(logN)かな. 逆アッカーマンとかあったけど, あまり詳しく覚えていない.
+    * 最後に, すべてのノードをみて, 親番号の種類を確認する O(N)
+    * トータルで, O(ElogN + N)?
+    * step数としては, 5 * 10^3 * 10 + 2 * 10^3 = 10^4くらい
+    * 実行時間は, 10^7 step / secで見積もると, 0.001秒くらいか
+* DFSやBFSを使って全部見ることもできそう
+    * 最初のノードを適当にとって辿る
+    * 辿り切ったら, まだ見ていないとこから一個適当にとって同様の手順を繰り返す
+    * 全てのノードを見るので, O(N)
+    * だけど, まだ見ていないところから一個取るのがO(1)でできる？？
+        * visitedとかを使えば結局訪れるのはたかだか1回だからO(N)で良さそう
+
+
+
+## Code1-1 (DFS)
+
+* 繋がっているノードを確認するのにO(N)かかるから計算量の見積もりがアプローチで考えたものと違くなりそう
+* 各ノードに対して, 
+    * 繋がっているノードを確認 O(N)
+* O(N^2)かな
+* でも, 繋がっているノードをdictとして持つようにしたら, 繋がっているノードの確認はO(1)でよくなる
+    * ただ, 結局すべてのノードが辺で結ばれていたら, 繋がっているノードはN - 1個になるのでO(N^2)は免れないか？？
+* いや, 今回の場合は, 一度訪れられたノードに再び行くことはないから, O(N)
+* エッジの前処理も含めたらO(N + E)
+
+```python
+from typing import List
+
+class Solution:
+
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+
+        def create_adjacent_matrix():
+            adjacent_matrix = [[False] * n for _ in range(n)]
+            for edge in edges:
+                node1 = edge[0]
+                node2 = edge[1]
+                adjacent_matrix[node1][node2] = True
+                adjacent_matrix[node2][node1] = True
+            return adjacent_matrix
+        
+        def visit_all_connected_nodes(cur_node, adj_matrix, visited):
+            if visited[cur_node]:
+                return
+            visited[cur_node] = True
+            for next_node in range(n):
+                if cur_node == next_node:
+                    continue
+                if not adj_matrix[cur_node][next_node]:
+                    continue
+                visit_all_connected_nodes(next_node, adj_matrix, visited)
+            return
+
+        visited = [False] * n
+        adj_matrix = create_adjacent_matrix()
+        num_components = 0
+        for i in range(n):
+            if visited[i]:
+                continue
+            num_components += 1
+            visit_all_connected_nodes(i, adj_matrix, visited)
+        return num_components
+
+```
+
+## Code1-2 (Union Find)
+
+```python
+from typing import List
+
+
+class UnionFind:
+    def __init__(self, size):
+        self.parents = [i for i in range(size)]
+        self.rank = [0] * size
+    
+    def find(self, idx):
+        if self.parents[idx] != idx:
+            self.parents[idx] = self.find(self.parents[idx])
+        return self.parents[idx]
+    
+    def union(self, idx1, idx2):
+        parent1 = self.find(idx1)
+        parent2 = self.find(idx2)
+        if parent1 == parent2:
+            return
+        
+        if self.rank[parent1] < self.rank[parent2]:
+            self.parents[parent1] = parent2
+            return
+        elif self.rank[parent2] < self.rank[parent1]:
+            self.parents[parent2] = parent1
+            return
+        else:
+            self.parents[parent2] = parent1
+            self.rank[parent1] += 1
+            return
+        
+
+class Solution:
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+        uf = UnionFind(n)
+        for edge in edges:
+            uf.union(edge[0], edge[1])
+        
+        num_components = 0
+        seen_parents = set()
+        for i in range(n):
+            parent = uf.find(i)
+            if parent in seen_parents:
+                continue
+            seen_parents.add(parent)
+            num_components += 1
+        
+        return num_components
+            
+```
+
+# 他の人のコードを確認
+
+* 1 - https://github.com/mamo3gr/arai60/pull/56
+    * 言語: Python
+    * DFSとUnionFindの両方を実装
+    * DFS: DFSでスタックを使用し, 辺は隣接行列ではなくて辞書として用意
+    * UnionFind: 実装はほぼ自分のものと一致. 違いとしては`rank`じゃなくて`size`を使用している
+* 2 - https://github.com/Hiroto-Iizuka/coding_practice/pull/19
+    * 言語: Python
+    * 1とほぼ同様
+* 3 - https://github.com/TakayaShirai/leetcode_practice/pull/19
+    * 言語: Swift
+    * 手作業の様子を考えているのがいいと思った
+    * 実行時間が図られているのも良い
+    * 1, 2と同様
+* 4 - https://github.com/xbam326/leetcode/pull/21
+    * 言語: Python
+    * 1, 2, 3と同様
+* 5 - https://github.com/dxxsxsxkx/leetcode/pull/19
+    * 言語: C++
+    * 辞書ではなく, 二次元配列で辺のつながりを表現. adjacency_list[i] = (nodeiにつながるノードのリスト)
+    * あとは1,2,3,4と同様
+
+# 調べたこと
+
+* UnionFindの計算量
+    * 最適化なしだったら, O(N)
+    * rankやsizeによって木の高さを制限したら, O(logN)
+    * Path Compression + rankにしたら, O(α(N))で, これはほぼ定数時間と見ていい
+
+
+# Step2
+
+## Code2-1 (DFS)
+
+* 再帰をスタックで行うようにした
+* 辞書として辺を登録した
+
+```python
+from typing import List
+from collections import defaultdict
+
+class Solution:
+
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+
+        node_to_neighbors = defaultdict(set)
+        for edge in edges:
+            node_to_neighbors[edge[0]].add(edge[1])
+            node_to_neighbors[edge[1]].add(edge[0])
+
+        visited = [False] * n
+
+        def visit_connected_nodes(init_node):
+            frontier = [init_node]
+            while frontier:
+                node = frontier.pop()
+                if visited[node]:
+                    continue
+                visited[node] = True
+                for neighbor_node in node_to_neighbors[node]:
+                    if not visited[neighbor_node]:
+                        frontier.append(neighbor_node)
+            return
+
+        num_components = 0
+        for i in range(n):
+            if visited[i]:
+                continue
+            num_components += 1
+            visit_connected_nodes(i)
+        return num_components
+
+```
+
+## Code2-2 (Union Find)
+
+```python
+from typing import List
+
+
+class UnionFind:
+    def __init__(self, size):
+        self.parents = [i for i in range(size)]
+        self.rank = [0] * size
+    
+    def find(self, idx):
+        if self.parents[idx] != idx:
+            self.parents[idx] = self.find(self.parents[idx])
+        return self.parents[idx]
+    
+    def union(self, idx1, idx2):
+        parent1 = self.find(idx1)
+        parent2 = self.find(idx2)
+        if parent1 == parent2:
+            return
+        
+        if self.rank[parent1] < self.rank[parent2]:
+            self.parents[parent1] = parent2
+            return
+        elif self.rank[parent2] < self.rank[parent1]:
+            self.parents[parent2] = parent1
+            return
+        else:
+            self.parents[parent2] = parent1
+            self.rank[parent1] += 1
+            return
+        
+
+class Solution:
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+        uf = UnionFind(n)
+        for edge in edges:
+            uf.union(edge[0], edge[1])
+        
+        num_components = 0
+        seen_parents = set()
+        for i in range(n):
+            parent = uf.find(i)
+            if parent in seen_parents:
+                continue
+            seen_parents.add(parent)
+            num_components += 1
+        
+        return num_components
+
+``` 
+
+# Step3
+
+## Code3-1 (DFS)
+
+```python
+from typing import List
+from collections import defaultdict
+
+
+class Solution:
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+
+        node_to_neighbors = defaultdict(set)
+        for edge in edges:
+            node_to_neighbors[edge[0]].add(edge[1])
+            node_to_neighbors[edge[1]].add(edge[0])
+        
+        visited = [False] * n
+
+        def visit_connected(init_node):
+            if visited[init_node]:
+                return
+            frontier = [init_node]
+            while frontier:
+                node = frontier.pop()
+                if visited[node]:
+                    continue
+                visited[node] = True
+                for neighbor_node in node_to_neighbors[node]:
+                    if visited[neighbor_node]:
+                        continue
+                    frontier.append(neighbor_node)
+            return
+        
+        num_components = 0
+        for node in range(n):
+            if visited[node]:
+                continue
+            num_components += 1
+            visit_connected(node)
+        return num_components
+
+```
+
+## Code3-2 (Union Find)
+
+```python
+from typing import List
+
+class UnionFind:
+    def __init__(self, size):
+        self.parents = [i for i in range(size)]
+        self.rank = [0] * size
+    
+    def find(self, idx):
+        if self.parents[idx] != idx:
+            self.parents[idx] = self.find(self.parents[idx])
+        return self.parents[idx]
+    
+    def union(self, idx1, idx2):
+        parent1 = self.find(idx1)
+        parent2 = self.find(idx2)
+        if parent1 == parent2:
+            return
+        if self.rank[parent1] < self.rank[parent2]:
+            self.parents[parent1] = parent2
+            return
+        if self.rank[parent2] < self.rank[parent2]:
+            self.parents[parent2] = parent1
+            return
+        self.parents[parent2] = parent1
+        self.rank[parent1] += 1
+        return
+
+class Solution:
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+        uf = UnionFind(n)
+        for edge in edges:
+            uf.union(edge[0], edge[1])
+
+        num_components = 0
+        seen_parent = set()
+        for node in range(n):
+            parent = uf.find(node)
+            if parent in seen_parent:
+                continue
+            seen_parent.add(parent)
+            num_components += 1
+        
+        return num_components
+
+
+```
\ No newline at end of file
diff --git a/step1-1_dfs.py b/step1-1_dfs.py
new file mode 100644
index 0000000..852eb31
--- /dev/null
+++ b/step1-1_dfs.py
@@ -0,0 +1,36 @@
+from typing import List
+
+class Solution:
+
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+
+        def create_adjacent_matrix():
+            adjacent_matrix = [[False] * n for _ in range(n)]
+            for edge in edges:
+                node1 = edge[0]
+                node2 = edge[1]
+                adjacent_matrix[node1][node2] = True
+                adjacent_matrix[node2][node1] = True
+            return adjacent_matrix
+        
+        def visit_all_connected_nodes(cur_node, adj_matrix, visited):
+            if visited[cur_node]:
+                return
+            visited[cur_node] = True
+            for next_node in range(n):
+                if cur_node == next_node:
+                    continue
+                if not adj_matrix[cur_node][next_node]:
+                    continue
+                visit_all_connected_nodes(next_node, adj_matrix, visited)
+            return
+
+        visited = [False] * n
+        adj_matrix = create_adjacent_matrix()
+        num_components = 0
+        for i in range(n):
+            if visited[i]:
+                continue
+            num_components += 1
+            visit_all_connected_nodes(i, adj_matrix, visited)
+        return num_components
\ No newline at end of file
diff --git a/step1-2_union_find.py b/step1-2_union_find.py
new file mode 100644
index 0000000..c71c95f
--- /dev/null
+++ b/step1-2_union_find.py
@@ -0,0 +1,48 @@
+from typing import List
+
+
+class UnionFind:
+    def __init__(self, size):
+        self.parents = [i for i in range(size)]
+        self.rank = [0] * size
+    
+    def find(self, idx):
+        if self.parents[idx] != idx:
+            self.parents[idx] = self.find(self.parents[idx])
+        return self.parents[idx]
+    
+    def union(self, idx1, idx2):
+        parent1 = self.find(idx1)
+        parent2 = self.find(idx2)
+        if parent1 == parent2:
+            return
+        
+        if self.rank[parent1] < self.rank[parent2]:
+            self.parents[parent1] = parent2
+            return
+        elif self.rank[parent2] < self.rank[parent1]:
+            self.parents[parent2] = parent1
+            return
+        else:
+            self.parents[parent2] = parent1
+            self.rank[parent1] += 1
+            return
+        
+
+class Solution:
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+        uf = UnionFind(n)
+        for edge in edges:
+            uf.union(edge[0], edge[1])
+        
+        num_components = 0
+        seen_parents = set()
+        for i in range(n):
+            parent = uf.find(i)
+            if parent in seen_parents:
+                continue
+            seen_parents.add(parent)
+            num_components += 1
+        
+        return num_components
+            
\ No newline at end of file
diff --git a/step2-1_dfs.py b/step2-1_dfs.py
new file mode 100644
index 0000000..063cc63
--- /dev/null
+++ b/step2-1_dfs.py
@@ -0,0 +1,33 @@
+from typing import List
+from collections import defaultdict
+
+class Solution:
+
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+
+        node_to_neighbors = defaultdict(set)
+        for edge in edges:
+            node_to_neighbors[edge[0]].add(edge[1])
+            node_to_neighbors[edge[1]].add(edge[0])
+
+        visited = [False] * n
+
+        def visit_connected_nodes(init_node):
+            frontier = [init_node]
+            while frontier:
+                node = frontier.pop()
+                if visited[node]:
+                    continue
+                visited[node] = True
+                for neighbor_node in node_to_neighbors[node]:
+                    if not visited[neighbor_node]:
+                        frontier.append(neighbor_node)
+            return
+
+        num_components = 0
+        for i in range(n):
+            if visited[i]:
+                continue
+            num_components += 1
+            visit_connected_nodes(i)
+        return num_components
\ No newline at end of file
diff --git a/step2-2_union_find.py b/step2-2_union_find.py
new file mode 100644
index 0000000..c71c95f
--- /dev/null
+++ b/step2-2_union_find.py
@@ -0,0 +1,48 @@
+from typing import List
+
+
+class UnionFind:
+    def __init__(self, size):
+        self.parents = [i for i in range(size)]
+        self.rank = [0] * size
+    
+    def find(self, idx):
+        if self.parents[idx] != idx:
+            self.parents[idx] = self.find(self.parents[idx])
+        return self.parents[idx]
+    
+    def union(self, idx1, idx2):
+        parent1 = self.find(idx1)
+        parent2 = self.find(idx2)
+        if parent1 == parent2:
+            return
+        
+        if self.rank[parent1] < self.rank[parent2]:
+            self.parents[parent1] = parent2
+            return
+        elif self.rank[parent2] < self.rank[parent1]:
+            self.parents[parent2] = parent1
+            return
+        else:
+            self.parents[parent2] = parent1
+            self.rank[parent1] += 1
+            return
+        
+
+class Solution:
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+        uf = UnionFind(n)
+        for edge in edges:
+            uf.union(edge[0], edge[1])
+        
+        num_components = 0
+        seen_parents = set()
+        for i in range(n):
+            parent = uf.find(i)
+            if parent in seen_parents:
+                continue
+            seen_parents.add(parent)
+            num_components += 1
+        
+        return num_components
+            
\ No newline at end of file
diff --git a/step3-1_dfs.py b/step3-1_dfs.py
new file mode 100644
index 0000000..d3a5612
--- /dev/null
+++ b/step3-1_dfs.py
@@ -0,0 +1,36 @@
+from typing import List
+from collections import defaultdict
+
+
+class Solution:
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+
+        node_to_neighbors = defaultdict(set)
+        for edge in edges:
+            node_to_neighbors[edge[0]].add(edge[1])
+            node_to_neighbors[edge[1]].add(edge[0])
+        
+        visited = [False] * n
+
+        def visit_connected(init_node):
+            if visited[init_node]:
+                return
+            frontier = [init_node]
+            while frontier:
+                node = frontier.pop()
+                if visited[node]:
+                    continue
+                visited[node] = True
+                for neighbor_node in node_to_neighbors[node]:
+                    if visited[neighbor_node]:
+                        continue
+                    frontier.append(neighbor_node)
+            return
+        
+        num_components = 0
+        for node in range(n):
+            if visited[node]:
+                continue
+            num_components += 1
+            visit_connected(node)
+        return num_components
diff --git a/step3-2_union_find.py b/step3-2_union_find.py
new file mode 100644
index 0000000..f73d18b
--- /dev/null
+++ b/step3-2_union_find.py
@@ -0,0 +1,43 @@
+from typing import List
+
+class UnionFind:
+    def __init__(self, size):
+        self.parents = [i for i in range(size)]
+        self.rank = [0] * size
+    
+    def find(self, idx):
+        if self.parents[idx] != idx:
+            self.parents[idx] = self.find(self.parents[idx])
+        return self.parents[idx]
+    
+    def union(self, idx1, idx2):
+        parent1 = self.find(idx1)
+        parent2 = self.find(idx2)
+        if parent1 == parent2:
+            return
+        if self.rank[parent1] < self.rank[parent2]:
+            self.parents[parent1] = parent2
+            return
+        if self.rank[parent2] < self.rank[parent2]:
+            self.parents[parent2] = parent1
+            return
+        self.parents[parent2] = parent1
+        self.rank[parent1] += 1
+        return
+
+class Solution:
+    def countComponents(self, n: int, edges: List[List[int]]) -> int:
+        uf = UnionFind(n)
+        for edge in edges:
+            uf.union(edge[0], edge[1])
+
+        num_components = 0
+        seen_parent = set()
+        for node in range(n):
+            parent = uf.find(node)
+            if parent in seen_parent:
+                continue
+            seen_parent.add(parent)
+            num_components += 1
+        
+        return num_components