**Introduction:**

In the world of coding and data structures, the importance of efficiency cannot be understated. LeetCode challenges are a great way to practice these skills and improve your problem-solving abilities. In this article, we will explore a common coding interview challenge: *"**Product of Array Except Self.*** "** The twist is that we need to solve it in

`O(n)`

time without using division operations. We'll provide a step-by-step guide to tackling this problem efficiently.**Problem Description:**

The challenge is to write an algorithm that takes an array of integers, `nums`

, and returns another array, `answer`

, where `answer[i]`

is equal to the product of all the elements of `nums`

except `nums[i]`

. You need to solve this problem without using the division operation, and the product of any prefix or suffix of `nums`

is guaranteed to fit in a 32-bit integer.

**Solution Approach:**

To achieve an `O(n)`

time complexity without division, we'll break down the problem into a few key steps. We'll use two auxiliary arrays to store products of elements to the left and right of each element. Finally, we'll combine these two arrays to calculate the desired result.

Here's a step-by-step guide to solving this challenge:

**Step 1: Initialize Variables and Arrays**

```
function productExceptSelf(nums: number[]): number[] {
const n = nums.length;
let leftProducts = new Array(n);
let rightProducts = new Array(n);
let result = new Array(n).fill(1);
let leftProduct = 1;
```

We start by initializing three arrays: `leftProducts`

, `rightProducts`

, and `result`

. The `leftProducts`

array will store the product of elements to the left of each element, while `rightProducts`

will store the product of elements to the right of each element. We initialize `leftProduct`

to 1, which represents the product of elements to the left of the first element.

**Step 2: Calculate Left Products**

```
for (let i = 0; i < n; i++) {
leftProducts[i] = leftProduct;
leftProduct *= nums[i];
}
```

We traverse the `nums`

array from left to right, updating `leftProducts`

at each index. For each index `i`

, we set `leftProducts[i]`

to the `leftProduct`

value (which is initially 1) and then update `leftProduct`

by multiplying it by `nums[i]`

. This step efficiently computes the product of all elements to the left of each element in the `nums`

array.

**Step 3: Calculate Right Products**

```
let rightProduct = 1;
for (let i = n - 1; i >= 0; i--) {
rightProducts[i] = rightProduct;
rightProduct *= nums[i];
}
```

Similar to the left products, we now calculate the products of elements to the right of each element. We initialize `rightProduct`

to 1, representing the product of elements to the right of the last element. We traverse the `nums`

array from right to left and update `rightProducts`

at each index. This efficiently computes the product of all elements to the right of each element.

**Step 4: Calculate Final Result**

```
for (let i = 0; i < n; i++) {
result[i] = leftProducts[i] * rightProducts[i];
}
return result;
}
```

Finally, we calculate the desired result array. For each element at index `i`

, we multiply the corresponding values from the `leftProducts`

and `rightProducts`

arrays, effectively excluding `nums[i]`

. This step ensures that each element in the `result`

array is the product of all elements in `nums`

except itself.

**Conclusion:**

Solving the "** Product of Array Except Self**" challenge efficiently in

`O(n)`

time without division is an excellent example of how to approach array manipulation problems. By breaking the problem down into smaller subproblems and utilizing two auxiliary arrays, we achieve both a simple and efficient solution. This algorithm is particularly useful in interviews and competitive coding, where optimizing time and space complexity is essential.