diff --git a/arai60/intersection-of-two-arrays/README.md b/arai60/intersection-of-two-arrays/README.md
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+## 考察
+- 過去に解いたことあり
+- 方針
+    - 積集合をとる(Pythonだと、setに変換して集合演算)
+        - time: O(n + m), space: O(n + m)
+        - n: len(nums1), m: len(nums2)
+    - 片方をHashSetにしてループ処理
+        - 結果をユニークにして返す
+        - time: O(n + m), space: O(n + m)
+- 直感的な積集合をとる方針でやる
+- あとは実装
+
+## Step1
+- 集合演算で実装
+- time: O(n + m), space: O(n + m)
+
+## Step2
+- ほかの人のPRを検索してみた
+- sortしてtwo pointersを使う方法もあるみたい
+    - Ref. https://github.com/Mike0121/LeetCode/pull/30#discussion_r1641596790
+    - そういえば昨日ちょうどソート済みの2つの配列のintersectionを活用する問題をやった
+        - https://leetcode.com/problems/get-the-maximum-score/description/
+    - もともとソート済みならO(min(n, m))なので速い
+    - この手法でもやってみる
+- この問題の場合は、ソートがボトルネックになる
+- time: O(n log n + m log m), space: O(min(n, m))
+
+## Step3
+- 1回目: 22s
+- 2回目: 21s
+- 3回目: 22s
+
+## Step4
+- レビューを元に修正
+- ワンライナーでも良さそうなので、一行で記述した
+- two pointersの方も修正
+    - `answer` -> `intersection`
+    - `list.sort()` を使ってもメモリO(n)のため、引数に破壊的変更を加えない形に修正
diff --git a/arai60/intersection-of-two-arrays/step1.py b/arai60/intersection-of-two-arrays/step1.py
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+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        nums1_set = set(nums1)
+        nums2_set = set(nums2)
+        return list(nums1_set & nums2_set)
diff --git a/arai60/intersection-of-two-arrays/step2.py b/arai60/intersection-of-two-arrays/step2.py
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+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        nums1.sort()
+        nums2.sort()
+
+        index_nums1 = 0
+        index_nums2 = 0
+        answer = []
+        while index_nums1 < len(nums1) and index_nums2 < len(nums2):
+            if nums1[index_nums1] < nums2[index_nums2]:
+                index_nums1 += 1
+                continue
+            if nums1[index_nums1] > nums2[index_nums2]:
+                index_nums2 += 1
+                continue
+
+            common = nums1[index_nums1]
+            answer.append(common)
+            while index_nums1 < len(nums1) and nums1[index_nums1] == common:
+                index_nums1 += 1
+            while index_nums2 < len(nums2) and nums2[index_nums2] == common:
+                index_nums2 += 1
+
+        return answer
diff --git a/arai60/intersection-of-two-arrays/step3.py b/arai60/intersection-of-two-arrays/step3.py
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+++ b/arai60/intersection-of-two-arrays/step3.py
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+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        nums1_set = set(nums1)
+        nums2_set = set(nums2)
+        return list(nums1_set & nums2_set)
diff --git a/arai60/intersection-of-two-arrays/step4.py b/arai60/intersection-of-two-arrays/step4.py
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+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        return list(set(nums1) & set(nums2))
diff --git a/arai60/intersection-of-two-arrays/step4_two_pointers.py b/arai60/intersection-of-two-arrays/step4_two_pointers.py
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+++ b/arai60/intersection-of-two-arrays/step4_two_pointers.py
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+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        nums1_sorted = sorted(nums1)
+        nums2_sorted = sorted(nums2)
+
+        index_nums1 = 0
+        index_nums2 = 0
+        intersection = []
+        while index_nums1 < len(nums1_sorted) and index_nums2 < len(nums2_sorted):
+            if nums1_sorted[index_nums1] < nums2_sorted[index_nums2]:
+                index_nums1 += 1
+                continue
+            if nums1_sorted[index_nums1] > nums2_sorted[index_nums2]:
+                index_nums2 += 1
+                continue
+
+            common = nums1_sorted[index_nums1]
+            intersection.append(common)
+            while index_nums1 < len(nums1_sorted) and nums1_sorted[index_nums1] == common:
+                index_nums1 += 1
+            while index_nums2 < len(nums2_sorted) and nums2_sorted[index_nums2] == common:
+                index_nums2 += 1
+
+        return intersection