From a93ca47fbdd40acd608c56cc8cc3bf896ce9b07a Mon Sep 17 00:00:00 2001
From: mt2324 <63892273+mt2324@users.noreply.github.com>
Date: Mon, 29 Jun 2026 19:01:40 +0900
Subject: [PATCH] Create 349.  Intersection of Two Arrays.md

https://leetcode.com/problems/intersection-of-two-arrays/
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+https://leetcode.com/problems/intersection-of-two-arrays/
+## STEP 1
+素直に両方setに変換してintersectionを取る。メモリはO(n)追加でいるけど速い。
+```python
+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        num1_set = set(nums1)
+        num2_set = set(nums2)
+        ans = []
+        for num in num1_set.intersection(num2_set):
+            ans.append(num)
+        return ans
+```
+
+ソートしてtwo pointersだけど複数回同じ数字が被った時のために結局setを作んないといけない。あと地味にソートも重い。intersectionが少なくてメモリをできるだけ節約したい時にはいいのか？
+
+```python
+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        nums1.sort()
+        nums2.sort()
+        i = 0
+        j = 0
+        intersection = set()
+        while i < len(nums1) and j < len(nums2):
+            if nums1[i] < nums2[j]:
+                i += 1
+            elif nums2[j] < nums1[i]:
+                j += 1
+            else:
+                intersection.add(nums1[i])
+                i += 1
+                j += 1
+        return list(intersection)
+```
+
+Counter で各要素を数えることもできるけどこれは本質的にsetに変換してるのと代わりない。
+```python 
+from collections import Counter
+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        counter1 = Counter(nums1)
+        counter2 = Counter(nums2)
+        intersection = []
+        for num1 in counter1.keys():
+            if num1 in counter2:
+                intersection.append(num1)
+        return intersection
+```
+
+## STEP2
+intersectionは`&`でも`.intersection()`でも取れるが`&`だとset同士でしか取れないが、`.intersection()` だと引数にset以外のiterable objectを取れるらしい。
+https://docs.python.org/3.14/library/stdtypes.html#set-types-set-frozenset
+
+あとsetを生成するはhashを計算する必要があって時間がかかるので小さい方をsetにした方がいいのはたしかにと思った。
+
+```python
+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        if len(nums1) < len(nums2):
+            return list(set(nums1).intersection(nums2))
+        else:
+            return list(set(nums2).intersection(nums1))
+```
+ついでにCounterを使った方も書きなおしたり、
+```python
+from collections import Counter
+class Solution:
+         
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        intersection = ()
+        if len(nums1) < len(nums2):
+            counter = Counter(nums1)
+            nums_to_iterate = nums2
+        else:
+            counter = Counter(nums2)
+            nums_to_iterate = nums1
+
+        intersection = []
+        for num in nums_to_iterate:
+            if num in counter and counter[num] > 0:
+                intersection.append(num)
+                counter[num] = 0
+        return intersection
+```
+分岐なしで書くならこうかな。
+```python
+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        return list(set(nums1) & set(nums2))
+```
+## STEP3
+練習するだけ
+```python
+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        if len(nums1) < len(nums2):
+            return list(set(nums1).intersection(nums2))
+        else:
+            return list(set(nums2).intersection(nums1))
+```