diff --git a/035-search-insert-position/memo.md b/035-search-insert-position/memo.md
new file mode 100644
index 0000000..dca3186
--- /dev/null
+++ b/035-search-insert-position/memo.md
@@ -0,0 +1,88 @@
+## 問題文
+Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
+
+You must write an algorithm with O(log n) runtime complexity.
+
+ 
+
+Example 1:
+
+Input: nums = [1,3,5,6], target = 5
+Output: 2
+Example 2:
+
+Input: nums = [1,3,5,6], target = 2
+Output: 1
+Example 3:
+
+Input: nums = [1,3,5,6], target = 7
+Output: 4
+
+
+Constraints:
+
+1 <= nums.length <= 104
+-104 <= nums[i] <= 104
+nums contains distinct values sorted in ascending order.
+-104 <= target <= 104
+
+## Step1で考えたこと
+O(log n)という制約がある +　問題の特徴から -> Binary Search
+Listを半分に分割し、右側の先頭の数字がTargetよりも大きければそのまま右側を走査、
+小さければ左側を走査する
+そのまま見つかればIndexをreturn
+
+見つからない場合、
+
+## Step2での気づき
+解答解説を見た時、left, rightポインタを使えば良いのかと学んだ。
+
+Let's look at the case where the target number is not present.
+
+Input: nums = [1,3,5,6], target = 2
+[1,3,5,6]
+ L M   R
+Now middle number is greater than the target, so move R to M - 1.
+
+[1,3,5,6]
+ L
+ R
+ M
+Now middle number is less than the target, so move L to M + 1.
+
+[1,3,5,6]
+ R L
+Now L is greater than R. We stop binary search. As a result, we don't find the target.
+
+The description says "return the index where it would be if it were inserted in order."
+
+In this case, we should insert the target into index 1. As you see, L is now at index 1, so
+
+return 1 (= left index)
+
+### Points
+Return left index if you don't find the target.
+見つからない場合のreturnするIndexの考え方も吸収できた。
+
+### 間違えていたところとしては、
+1. left <= rightとしていなかった。
+2. right, leftの値更新の際の1ずらすところができていなかった。
+
+right = mid - 1, left = mid + 1としないと、left < rightという制約を超えることがないため無限ループを引き起こしてしまう。
+
+## GPTからもらった解答
+```
+class Solution:
+    def searchInsert(self, nums: List[int], target: int) -> int:
+        left, right = 0, len(nums)  # 右端は“外”
+        while left < right:
+            mid = (left + right) // 2
+            if nums[mid] < target:
+                left = mid + 1
+            else:
+                right = mid
+        return left  # 最初に target 以上が出る位置
+```
+target が配列内にあればその index、なければ挿入位置を返す  
+target > 最後の要素 の場合でも left == len(nums) になって正しく 末尾 を返す.  
+distinct でなくても“先頭の位置”が取れる（一般に強い）
diff --git a/035-search-insert-position/step1.py b/035-search-insert-position/step1.py
new file mode 100644
index 0000000..765a72a
--- /dev/null
+++ b/035-search-insert-position/step1.py
@@ -0,0 +1,7 @@
+from typing import List
+
+class Solution:
+    def searchInsert(self, nums: List[int], target: int) -> int:
+        mid = nums.len() / 2
+        if target > List[mid]:
+            
\ No newline at end of file
diff --git a/035-search-insert-position/step2.py b/035-search-insert-position/step2.py
new file mode 100644
index 0000000..3022131
--- /dev/null
+++ b/035-search-insert-position/step2.py
@@ -0,0 +1,17 @@
+# 答えを見てから自力で書いてみたコード
+class Solution:
+    def searchInsert(self, nums: List[int], target: int) -> int:
+        left = 0
+        right = len(nums) - 1
+
+        while left < right:
+            mid = (left + right) // 2
+
+            if target == nums[mid]:
+                return mid
+            elif target < nums[mid]:
+                right = mid
+            else:
+                left = mid
+        
+        return left
diff --git a/035-search-insert-position/step3.py b/035-search-insert-position/step3.py
new file mode 100644
index 0000000..5e51ea6
--- /dev/null
+++ b/035-search-insert-position/step3.py
@@ -0,0 +1,15 @@
+class Solution:
+    def searchInsert(self, nums: List[int], target: int) -> int:
+        left = 0
+        right = len(nums) - 1
+        
+        while left <= right:
+            mid = (left + right) // 2
+            
+            if nums[mid] == target:
+                return target
+            elif nums[mid] > target:
+                right = mid - 1
+            else:
+                left = mid + 1
+        return left