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+---
+id: product-of-array-except-self
+title: Product of Array Except Self(LeetCode)
+sidebar_label: 0238-Product of Array Except Self
+tags:
+  - Array
+  - Prefix Sum
+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].
+---
+
+## Problem Statement
+
+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]`.
+
+The product of any prefix or suffix of `nums` is guaranteed to fit in a 32-bit integer.
+
+You must write an algorithm that runs in `O(n)` time and without using the division operation.
+
+### Examples
+
+**Example 1:**
+
+```plaintext
+Input: nums = [1,2,3,4]
+Output: [24,12,8,6]
+```
+
+**Example 2:**
+
+```plaintext
+Input: nums = [-1,1,0,-3,3]
+Output: [0,0,9,0,0]
+```
+
+### Constraints
+
+- `2 <= nums.length <= 105`
+- `30 <= nums[i] <= 30`
+- The product of any prefix or suffix of `nums` is guaranteed to fit in a 32-bit integer.
+
+## Solution
+
+We can solve this problem using multiple approaches. Here, I have explained all the possible solutions:
+
+1. Brute Force Approach: Using nested loops to calculate the product of elements except for the current index.
+2. Dynamic Programming (Tabulation): Using two arrays to store the left and right products.
+3. Dynamic Programming (Space Optimization): Optimizing space usage by combining the two product arrays into a single array.
+
+### Approach Brute Force (Two Nested Loops)
+
+### Algorithm
+1. Initialize an empty vector `output`.
+2. Iterate through each element `i` in the array `nums`:
+* Set `product` to 1.
+* Iterate through each element `j` in the array `nums`:
+  * If `i` equals `j`, skip this iteration.
+  * Multiply `product` by `nums[j]`.
+* Append `product` to the `output` vector.
+3. Return the `output` vector.
+
+#### Implementation
+
+```C++
+class Solution {
+public:
+    vector<int> productExceptSelf(vector<int>& nums) {
+        int n = nums.size();
+        vector<int> output;
+        for(int i = 0; i < n; i++) {
+            int product = 1;
+            for(int j = 0; j < n; j++) {
+                if(i == j) continue;
+                product *= nums[j];
+            }
+            output.push_back(product);
+        }
+        return output;
+    }
+};
+```
+
+### Complexity Analysis
+
+- **Time complexity**: O(N^2) - Two nested loops create a quadratic time complexity.
+- **Space complexity**: O(1) - No extra space except for the output array (which doesn't count towards space complexity).
+
+### Approach 2: Dynamic Programming (Tabulation)
+
+#### Algorithm
+
+1. Initialize vectors `left_Product` and `right_Product` of size `n` with all elements set to 1.
+2. Calculate `left_Product`:
+* Set `left_Product[0]` to 1.
+* For each element `i` from 1 to `n-1`, set `left_Product[i]` to `left_Product[i-1]` multiplied by `nums[i-1]`.
+3. Calculate `right_Product`:
+* Set `right_Product[n-1]` to 1.
+* For each element `i` from `n-2` to 0, set `right_Product[i]` to `right_Product[i+1]` multiplied by `nums[i+1]`.
+4. Initialize vector `ans` of size `n`.
+5. For each element `i` from 0 to `n-1`, set `ans[i]` to `left_Product[i]` multiplied by `right_Product[i]`.
+6. Return the `ans` vector.
+
+#### Implementation 
+
+```C++
+class Solution {
+public:
+    vector<int> productExceptSelf(vector<int>& nums) {
+        int n = nums.size();
+        vector<int> ans(n);
+        vector<int> left_Product(n);
+        vector<int> right_Product(n);
+
+        left_Product[0] = 1;
+        for(int i = 1; i < n; i++) {
+            left_Product[i] = left_Product[i-1] * nums[i-1];
+        }
+
+        right_Product[n-1] = 1;
+        for(int i = n-2; i >= 0; i--) {
+            right_Product[i] = right_Product[i+1] * nums[i+1];
+        }
+
+        for(int i = 0; i < n; i++) {
+            ans[i] = left_Product[i] * right_Product[i];
+        }
+        return ans;
+    }
+};
+```
+
+### Complexity Analysis
+
+- **Time complexity**: O(N) - We iterate through the array three times.
+- **Space complexity**: O(N) - Two additional arrays (left_Product and right_Product) are used.
+
+### Approach 3: Dynamic Programming (Space Optimization)
+
+#### Algorithm
+
+1. Initialize a vector `output` of size `n` with all elements set to 1.
+2. Calculate the left product:
+* Set `output[0]` to 1.
+* For each element `i` from 1 to `n-1`, set `output[i]` to `output[i-1]` multiplied by `nums[i-1]`.
+3. Calculate the right product:
+* Initialize `right` to 1.
+* For each element `i` from `n-1` to 0, set `output[i]` to `output[i]` multiplied by right, then set `right` to `right` multiplied by `nums[i]`.
+4. Return the `output` vector.
+
+#### Implementation 
+
+```C++
+class Solution {
+public:
+    vector<int> productExceptSelf(vector<int>& nums) {
+        int n = nums.size();
+        vector<int> output(n);
+
+        output[0] = 1;
+        for(int i = 1; i < n; i++) {
+            output[i] = output[i-1] * nums[i-1];
+        }
+
+        int right = 1;
+        for(int i = n-1; i >= 0; i--) {
+            output[i] *= right;
+            right *= nums[i];
+        }
+        return output;
+    }
+};
+```
+
+### Complexity Analysis
+
+- **Time complexity**: O(N) - We iterate through the array twice.
+- **Space complexity**: O(1) - Constant space is used, except for the output array.
+
+### Conclusion
+
+By exploring various approaches to solve this problem, we can see that the brute force method, while simple, is inefficient with a time complexity of O(N^2). The dynamic programming approach with tabulation optimizes the time complexity to O(N) but uses extra space. The most optimized approach uses space-efficient dynamic programming, maintaining O(N) time complexity while reducing space complexity to O(1). This final approach is the best in terms of both time and space efficiency.