diff --git a/memo.md b/memo.md
index 4bd0397..6834a6d 100644
--- a/memo.md
+++ b/memo.md
@@ -1 +1,248 @@
 # Step1
+
+## アプローチ
+
+* `len(arr1) = N` , `len(arr2) = M`, `len(arr1[i]) = L`とする
+* `arr1`からTrie木を作っておいて, `arr2`の各要素に対してprefixとなっている文字数を測る
+* 計算量: O(NL + ML)
+    * 実行時間: 10^4 * 10^8 / 10^6 ~= 10^6 sec
+* 時間はかかりそうだけど, 他に方法が思いつかないので実装する
+
+## Code1-1 (Trie)
+
+* TLEになると思ったがAC.
+* 10^8は桁数としては8桁だからだ！！！
+
+```python
+class Node:
+    def __init__(self):
+        self.children = [None] * 10
+
+class Trie:
+    def __init__(self):
+        self.root = Node()
+
+    def _get_digits(self, num: int) -> list[int]:
+        digits = []
+        if num == 0:
+            digits.append(0)
+        else:
+            while num > 0:
+                digits.append(num % 10)
+                num = num // 10
+
+        digits.reverse()
+        return digits
+
+
+    def add(self, num: int) -> None:
+        digits = self._get_digits(num)
+        
+        node = self.root
+        for d in digits:
+            if node.children[d] is None:
+                node.children[d] = Node()
+            node = node.children[d]
+
+        return
+
+
+    def get_common_prefix_length(self, num: int) -> int:
+        digits = self._get_digits(num)
+
+        node = self.root
+        length = 0
+        for d in digits:
+            if node.children[d] is None:
+                return length
+            length += 1
+            node = node.children[d]
+        
+        return length
+
+class Solution:
+    def longestCommonPrefix(self, arr1: List[int], arr2: List[int]) -> int:
+        trie = Trie()
+        for num1 in arr1:
+            trie.add(num1)
+
+        longest_length = float("-inf")
+        for num2 in arr2:
+            prefix_length = trie.get_common_prefix_length(num2)
+            longest_length = max(longest_length, prefix_length)
+        
+        return longest_length
+
+```
+
+# Step2
+
+## Code2-1 (Trie)
+
+```python
+class IntTrieNode:
+    def __init__(self):
+        self.children = [None] * 10
+
+class IntTrieTree:
+    def __init__(self):
+        self.root = IntTrieNode()
+
+    def _get_digits(self, num: int) -> list[int]:
+        digits = []
+        if num == 0:
+            digits.append(0)
+        else:
+            while num > 0:
+                digits.append(num % 10)
+                num = num // 10
+
+        digits.reverse()
+        return digits
+
+
+    def add(self, num: int) -> None:
+        digits = self._get_digits(num)
+        
+        node = self.root
+        for d in digits:
+            if node.children[d] is None:
+                node.children[d] = IntTrieNode()
+            node = node.children[d]
+
+        return
+
+
+    def get_longest_common_prefix_length(self, num: int) -> int:
+        digits = self._get_digits(num)
+
+        node = self.root
+        length = 0
+        for d in digits:
+            if node.children[d] is None:
+                return length
+            length += 1
+            node = node.children[d]
+        
+        return length
+
+class Solution:
+    def longestCommonPrefix(self, arr1: List[int], arr2: List[int]) -> int:
+        trie_tree = IntTrieTree()
+
+        for num1 in arr1:
+            trie_tree.add(num1)
+
+        max_prefix_length = float("-inf")
+        for num2 in arr2:
+            prefix_length = trie_tree.get_longest_common_prefix_length(num2)
+            max_prefix_length = max(max_prefix_length, prefix_length)
+        
+        return max_prefix_length
+
+```
+
+## Code2-2 (Hash)
+
+* Hashを使う方法もあるらしい
+* `arr1`から存在しうる`prefix`をすべて`set`に追加する
+    * O(Nd^2) => O(N)
+* `arr2`の存在しうる`prefix`が`set`にあるかどうかを探る
+    * O(Md^2) => O(M)
+
+```python
+class Solution:
+    def longestCommonPrefix(self, arr1: List[int], arr2: List[int]) -> int:
+        prefixes = set()
+
+        for num1 in arr1:
+            num1_str = str(num1)
+            for i in range(1, len(num1_str) + 1):
+                prefixes.add(num1_str[:i])
+        
+        longest_common_prefix = 0
+        for num2 in arr2:
+            num2_str = str(num2)
+            for i in range(len(num2_str), 0, -1):
+                if num2_str[:i] not in prefixes:
+                    continue
+                longest_common_prefix = max(longest_common_prefix, i)
+                break
+        
+        return longest_common_prefix
+
+```
+
+# Step3
+
+## Code3-1 (Trie)
+
+* 1st: 5:26
+* 2nd: 4:22
+* 3rd: 3:29
+
+```python
+
+class Node:
+    def __init__(self):
+        self.children = [None] * 10
+
+
+class Trie:
+    def __init__(self):
+        self.root = Node()
+
+    def _get_digits(self, num: int) -> list[int]:
+        if num == 0:
+            return [0]
+        
+        digits_reversed = []
+        while num > 0:
+            digits_reversed.append(num % 10)
+            num //= 10
+        
+        return reversed(digits_reversed)
+
+
+    def add(self, num: int) -> None:
+        digits = self._get_digits(num)
+
+        node = self.root
+        for d in digits:
+            if node.children[d] is None:
+                node.children[d] = Node()
+            node = node.children[d]
+
+        return
+
+    
+    def get_longest_common_prefix_length(self, num: int) -> int:
+        digits = self._get_digits(num)
+
+        node = self.root
+        prefix_length = 0
+        for d in digits:
+            if node.children[d] is None:
+                return prefix_length
+            node = node.children[d]
+            prefix_length += 1
+        
+        return prefix_length
+
+
+class Solution:
+    def longestCommonPrefix(self, arr1: List[int], arr2: List[int]) -> int:
+        trie = Trie()
+
+        for num1 in arr1:
+            trie.add(num1)
+        
+        max_prefix_length = 0 
+        for num2 in arr2:
+            prefix_length = trie.get_longest_common_prefix_length(num2)
+            max_prefix_length = max(max_prefix_length, prefix_length)
+        
+        return max_prefix_length
+
+
+```