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apps/docs/content/prefix-sum/238-product-of-array-except-self.mdx

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+---
+title: '238. Product of Array Except Self'
+description: Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i]
+sidebar:
+  label: 'Product of Array Except Self'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Array</Badge>
+
+::::warning
+You must write an algorithm that runs in O(n) time and without using the division operation.
+Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)
+::::
+
+### Example 1:
+- Input: `nums = [1,2,3,4]`
+- Output: `[24,12,8,6]`
+
+### Example 2:
+- Input: `nums = [-1,1,0,-3,3]`
+- Output: `[0,0,9,0,0]`
+
+### Constraints:
+
+- `2 <= nums.length <= 10^5`
+- `-30 <= nums[i] <= 30`
+- The input is generated such that `answer[i]` is guaranteed to fit in a 32-bit integer.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["productExceptSelf(nums)"]) --> I["answer = ones(n), rightArr = ones(n)"]
+  I --> P1["Pass 1 — left to right: answer[i] = nums[i-1] * answer[i-1]"]
+  P1 --> P2["Pass 2 — right to left: rightArr[i] = nums[i+1] * rightArr[i+1]"]
+  P2 --> P3["Pass 3: answer[i] *= rightArr[i]"]
+  P3 --> E(["return answer — no division needed"])
+```
+
+## Solution
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var productExceptSelf = function(nums) {
+  const n = nums.length;
+  const answer = new Array(n).fill(1);
+  const rightArr = new Array(n).fill(1);
+
+  for (let i = 1; i < n; i++) {
+    answer[i] = nums[i - 1] * answer[i - 1];
+  }
+  for (let i = n - 2; i >= 0; i--) {
+    rightArr[i] = nums[i + 1] * rightArr[i + 1];
+  }
+  for (let i = 0; i < n; i++) {
+    answer[i] *= rightArr[i];
+  }
+  return answer;
+};
+```
+
+## Explanation
+
+<YouTube id="bNvIQI2wAjk" title="Product of Array Except Self - Leetcode 238 - Python" />