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+# 238. Product of Array Except Self
+
+## STEP1
+
+### 発想
+
+* 全ての要素に対して、自分以外の積を求める。
+時間計算量O(N^2)
+空間計算量はO(N)
+
+* 最初に全ての要素の積を求めて、各要素に対して割り算を行う。
+問題の条件で割り算をしてはいけないという記載があるため、不可。
+また、要素には0が含まれてうるため、解法としても不可。
+
+* 配列の各要素に対して、
+その要素の直前までの積とその要素の後ろ以降の積を掛け合わせる。
+
+### 想定されるユースケース
+
+### 何が分からなかったか?
+
+* 空間計算量をO(1)で行う方法が、思いつかなかった。
+
+### コード
+
+```javascript
+const productExceptSelf = function(nums) {
+    const prefixProductForward = new Array(nums.length).fill(0)
+    const prefixProductReversed = new Array(nums.length).fill(0)
+
+    let currentProduct = 1
+    for (let i = 0; i < nums.length; i++) {
+        currentProduct *= nums[i]
+        prefixProductForward[i] = currentProduct
+    }
+    
+    currentProduct = 1
+    for (let i = nums.length - 1; i >=0; i--) {
+        currentProduct *= nums[i]
+        prefixProductReversed[i] = currentProduct
+    }
+
+    const result = []
+    for (let i = 0; i < nums.length; i++) {
+        if (i === 0) {
+            result.push(prefixProductReversed[i + 1])
+            continue
+        }
+        if (i === nums.length - 1) {
+            result.push(prefixProductForward[i - 1])
+            continue
+        }
+        result.push(prefixProductForward[i - 1] * prefixProductReversed[i + 1])
+    }
+    return result
+};
+```
+
+## STEP2
+
+```javascript
+const productExceptSelf = function(nums) {
+    const prefixProduct = new Array(nums.length)
+    const prefixProductReversed = new Array(nums.length)
+
+    let currentProduct = 1
+    for (let i = 0; i < nums.length; i++) {
+        currentProduct *= nums[i]
+        prefixProduct[i] = currentProduct
+    }
+    
+    let currentProductReversed = 1
+    for (let i = nums.length - 1; 0 <= i; i--) {
+        currentProductReversed *= nums[i]
+        prefixProductReversed[i] = currentProductReversed
+    }
+
+    const result = []
+    result.push(prefixProductReversed[1])
+    for (let i = 1; i < nums.length - 1; i++) {
+        result.push(prefixProduct[i - 1] * prefixProductReversed[i + 1])
+    }
+    result.push(prefixProduct[nums.length - 2])
+    return result
+};
+```
+
+## STEP3
+
+* `*1` 累積積を用いた方法
+
+```javascript
+const productExceptSelf = function(nums) {
+    if (nums.length === 1) {
+        throw new Error("invalid nums length. length should be larger than 1.")
+    }
+    if (nums.length === 2) {
+        return [nums[1], nums[0]]
+    }
+    const prefixProduct = new Array(nums.length)
+    const prefixProductReversed = new Array(nums.length)
+
+    let currentProduct = 1
+    for (let i = 0; i < nums.length; i++) {
+        currentProduct *= nums[i]
+        prefixProduct[i] = currentProduct
+    }
+    let currentProductReversed = 1
+    for (let i = nums.length - 1; 0 <= i; i--) {
+        currentProductReversed *= nums[i]
+        prefixProductReversed[i] = currentProductReversed
+    }
+
+    const result = []
+    result.push(prefixProductReversed[1])
+    for (let i = 1; i < nums.length - 1; i++) {
+        const productExceptSelf = prefixProduct[i - 1] * prefixProductReversed[i + 1]
+        result.push(productExceptSelf)
+    }
+    result.push(prefixProduct[nums.length - 2])
+    return result
+};
+```
+
+## 感想
+
+### コメント集を読んで
+
+## 他の人のPRを読んで
+
+- https://github.com/Ryotaro25/leetcode_first60/pull/67
+- https://github.com/rihib/leetcode/pull/21
+- https://github.com/Exzrgs/LeetCode/pull/31
+- https://github.com/t-ooka/leetcode/pull/5
+- https://hayapenguin.com/notes/LeetCode/238/ProductOfArrayExceptSelf
+
+
+## その他の方法
+
+* `*2` 累積積で番兵を用いた方法
+
+```javascript
+const productExceptSelf = function(nums) {
+    const prefixProductForward = new Array(nums.length + 2)
+    const prefixProductReversed = new Array(nums.length + 2)
+
+    prefixProductForward[0] = 1
+    prefixProductForward[nums.length + 1] = 1
+    prefixProductReversed[0] = 1
+    prefixProductReversed[nums.length + 1] = 1
+
+    for (let i = 0; i < nums.length; i++) {
+        prefixProductForward[i + 1] = prefixProductForward[i] * nums[i]
+    }
+    
+    for (let i = nums.length - 1; i >=0; i--) {
+        prefixProductReversed[i + 1] = prefixProductReversed[i + 2] * nums[i]
+    }
+
+    const result = []
+    for (let i = 0; i < nums.length; i++) {
+        result.push(prefixProductForward[i] * prefixProductReversed[i + 2])
+    }
+    return result
+};
+```
+
+* `*3` 空間計算量をO(1)で行う方法
+
+```javascript
+const productExceptSelf = function(nums) {
+    if (nums.length <= 1) {
+        throw new Error("array lengths is invalid. nums size should be larger than 1.")
+    }
+    const result = new Array(nums.length).fill(1)
+    
+    let currentProduct = 1
+    for (let i = 1; i < nums.length; i++) {
+        currentProduct *= nums[i - 1]
+        result[i] = currentProduct
+    }
+    let currentProductReversed = 1
+    for (let i = nums.length - 2; 0 <= i; i--) {
+        currentProductReversed *= nums[i + 1]
+        result[i] *= currentProductReversed
+    }
+    return result
+};
+```
+
+- `*4` ナイーブに求める自分以外の席をFor文で計算する方法
+  - Time Limmit Exceeded エラーが発生する。
+
+```javascript
+const productExceptSelf = function(nums) {
+    const result = []
+    for (let i = 0; i < nums.length; i++) {
+        let product = 1
+        for (let j = 0; j < nums.length; j++) {
+            if (i === j) {
+                continue
+            }
+            product *= nums[j]
+        }
+        result.push(product)
+    }
+    return result
+};
+```
+
+### コードの良し悪し
+
+numsの配列の長さをNとする。
+
+* `*1`
+  * 時間計算量: O(N)
+  * 空間計算量: O(N)
+
+* `*2`
+  * 時間計算量: O(N)
+  * 空間計算量: O(N)
+
+* `*3`
+  * 時間計算量: O(N)
+  * 空間計算量: O(1)
+
+* `*4`
+  * 時間計算量: O(N^2)
+  * 空間計算量: O(1)
+
+## 調べたこと
+