Learn how to solve the Product of Array Except Self problem with our comprehensive guide. Follow 4 easy steps and get solutions in Python, Java, JavaScript, and C++. Perfect for coding interviews and mastering algorithmic challenges.

Welcome back, visitors! In this article, we will delve into the Product of Array Except Self problem and Product of Array Except Self problem is a part of blind 75 leetcode list, a popular intermediate-level interview question often found on platforms like LeetCode.

We will explore various approaches to solve this problem using different programming languages such as Python, Java, JavaScript, and C++. This article is designed to help you understand the concept thoroughly, making it easier for you to explain and implement it during your interviews.

**FREE** : If you’re a new visitor preparing for an interview and need help or **want to request a new question**, you can use **the form on the right side to ask your question**.** It’s free and doesn’t require signing in**. Alternatively, you can ask in our **Quora space**, which is also **free**, and you’ll receive **a response within 24** hours. Go ahead and check it out! do check **Master the Stock Span Problem by 3 Easy Steps in Python**, **Master Abstract Factory Design Pattern for Programming Interviews with 5 easy steps** ,**Mastering Object-Oriented Programming in C++** , **Method Overloading vs. Method Overriding in C#**

## Table of Contents

- What Is the Product of Array Except Self Problem?
- Explanation of Product of Array Except Self Problem:
- Approaches to Solve the Problem
- Implementations in Various Languages
- Summarizing Product of Array Except Self Problem Solution
- Key Points:
- FAQs:
- What is the “Product of Array Except Self” problem?
- Why is the Product of Array Except Self problem important for coding interviews?
- How can the Product of Array Except Self problem be solved using a brute force approach?
- What is the optimised approach for solving the Product of Array Except Self problem without using division?
- Can you provide implementations of the Product of Array Except Self problem in different programming languages?

## What Is the Product of Array Except Self Problem?

The biggest question that comes to mind is: What is the Product of Array Except Self problem, and why is it essential to understand it before going to an interview?

The answer to the above question is quite straightforward. The Product of Array Except Self problem is part of the Blind 75. If you’re not familiar with it, you should research it. Briefly, the Blind 75 consists of 75 core problems in data structures and algorithms (DSA) that are foundational for most interview-related DSA questions. If you can solve these Blind 75 problems, you will have a significant amount of knowledge to tackle any DSA problem in an interview.

Understanding how to solve the Product of Array Except Self problem is essential for several reasons:

- Algorithmic Thinking: It enhances your ability to think through complex problems and come up with efficient solutions.
- Array Manipulation: It reinforces concepts related to array manipulation and operations.
- Optimization Techniques: It introduces you to optimization techniques such as avoiding redundant calculations and space-time trade-offs.

The **Product of Array Except Self** problem requires you to return an array output such that output[i] is equal to the product of all elements of the input array except nums[i]. If you’re feeling confused, don’t worry. Below, I have included examples to help you understand the Product of Array Except Self problem.

### Example

Input: [1,2,3,4]

Output: [24,12,8,6]

## Explanation of Product of Array Except Self Problem:

As we’ve shown in the example of this problem, let’s now understand how to solve it and determine the best approach for selecting it in an interview. In this section, I will explain the logic behind the Product of Array Except Self problem. Stay with me to perform well in your interview.

The Product of Array Except Self problem requires calculating the product of all elements in an array except the one at the current index for each index.

In other words, you need to multiply all the elements of the array, but the condition is that you must exclude the element at the current index from the multiplication. Essentially, you skip the current index element and multiply the remaining elements of the array. This is the simplest definition of the problem.

Input: [1,2,3,4]

For index 0: 2*3*4 = 24

For index 1: 1*3*4 = 12

For index 2: 1*2*4 = 8

For index 3: 1*2*3 = 6

Output: [24,12,8,6]

If you still find this difficult to understand, take a look at the image below, which illustrates how the Product of Array Except Self problem works.

## Approaches to Solve the Problem

I hope you now understand what to look for in the Product of Array Except Self problem after reading the explanation provided above.

I will share both possible ways to achieve the output in the upcoming sections and discuss which approach is best and why.

### Brute Force Approach to Solve Product of Array Except Self Problem

The simplest way to solve this problem is by using a brute force approach, where we calculate the product of all elements for each index except the current one. However, this approach has a time complexity of O(n^2) and is not efficient for large arrays.

Check the pseudocode below to understand why the brute-force approach is not the ideal way to solve the Product of Array Except Self problem in an interview.

```
// Pseudo Code:
for i from 0 to n-1:
product = 1
for j from 0 to n-1:
if i != j:
product *= nums[j]
output[i] = product
```

### Optimized Approach Solve Product of Array Except Self Problem Without Division

A more efficient way to solve this problem is to use two additional arrays (left and right) to store the product of all elements to the left and right of each index. This approach has a time complexity of O(n) and a space complexity of O(n).

Since we can manage both time and space complexity to O(n), therefore this is the best way to solve the problem.

#### Steps to Solve Product of Array Except Self Problem Without Division:

We have four easy steps to implement the optimized approach without using division. Check the steps below, and if you still find it difficult to understand, please review the explanation section or jump directly to the code part to grasp the solution in your own way.

- Create two arrays left and right.
- Traverse the array to fill left[i] with the product of all elements to the left of i.
- Traverse the array to fill right[i] with the product of all elements to the right of i.
- Construct the output array where output[i] = left[i] * right[i].

Let’s implement the above steps in various programming languages such as Python, Java, JavaScript, and C++. This way, you can adapt the solution to your preferred language.

## Implementations in Various Languages

### Python

```
def product_except_self(nums):
n = len(nums)
left = [1] * n
right = [1] * n
output = [1] * n
for i in range(1, n):
left[i] = left[i - 1] * nums[i - 1]
for i in range(n - 2, -1, -1):
right[i] = right[i + 1] * nums[i + 1]
for i in range(n):
output[i] = left[i] * right[i]
return output
nums = [1, 2, 3, 4]
print(product_except_self(nums))
```

### Java

```
public int[] productExceptSelf(int[] nums) {
int n = nums.length;
int[] left = new int[n];
int[] right = new int[n];
int[] output = new int[n];
left[0] = 1;
for (int i = 1; i < n; i++) {
left[i] = left[i - 1] * nums[i - 1];
}
right[n - 1] = 1;
for (int i = n - 2; i >= 0; i--) {
right[i] = right[i + 1] * nums[i + 1];
}
for (int i = 0; i < n; i++) {
output[i] = left[i] * right[i];
}
return output;
}
```

### JavaScript

```
function productExceptSelf(nums) {
let n = nums.length;
let left = new Array(n).fill(1);
let right = new Array(n).fill(1);
let output = new Array(n).fill(1);
for (let i = 1; i < n; i++) {
left[i] = left[i - 1] * nums[i - 1];
}
for (let i = n - 2; i >= 0; i--) {
right[i] = right[i + 1] * nums[i + 1];
}
for (let i = 0; i < n; i++) {
output[i] = left[i] * right[i];
}
return output;
}
let nums = [1, 2, 3, 4];
console.log(productExceptSelf(nums));
```

### C++

```
#include <vector>
#include <iostream>
using namespace std;
vector<int> productExceptSelf(vector<int>& nums) {
int n = nums.size();
vector<int> left(n, 1), right(n, 1), output(n, 1);
for (int i = 1; i < n; i++) {
left[i] = left[i - 1] * nums[i - 1];
}
for (int i = n - 2; i >= 0; i--) {
right[i] = right[i + 1] * nums[i + 1];
}
for (int i = 0; i < n; i++) {
output[i] = left[i] * right[i];
}
return output;
}
int main() {
vector<int> nums = {1, 2, 3, 4};
vector<int> result = productExceptSelf(nums);
for (int num : result) {
cout << num << " ";
}
return 0;
}
```

## Summarizing Product of Array Except Self Problem Solution

The Product of Array Except Self problem involves calculating an array where each element is the product of all other elements except the one at the current index. This problem is crucial for interviews and is part of the Blind 75 list.

## Key Points:

- Problem Definition: Compute an array where each element is the product of all other elements in the array.
- Example: For input [1,2,3,4], the output is [24,12,8,6].
- Approaches:
- Brute Force: Inefficient with O(n^2) complexity.
- Optimized Approach: Uses additional arrays to store products of elements to the left and right of each index, achieving O(n) time complexity.

This guide helps you master the problem and prepare effectively for coding interviews.

## FAQs:

### What is the “Product of Array Except Self” problem?

The Product of Array Except Self problem involves creating an output array such that each element in the output array is the product of all elements in the input array except the one at the current index. This problem is essential to understand as it is a part of the Blind 75 list of foundational algorithmic problems commonly encountered in interviews.

### Why is the Product of Array Except Self problem important for coding interviews?

Understanding this problem is crucial for coding interviews because it tests your ability to handle array manipulations and optimisation techniques. It also helps in enhancing algorithmic thinking and understanding space-time trade-offs.

### How can the Product of Array Except Self problem be solved using a brute force approach?

The brute force approach involves iterating through each element and calculating the product of all other elements for each index. However, this approach has a time complexity of O(n^2) and is inefficient for larger arrays.

### What is the optimised approach for solving the Product of Array Except Self problem without using division?

The optimised approach involves using two additional arrays (left and right) to store the products of all elements to the left and right of each index, respectively. This method reduces the time complexity to O(n) and uses O(n) space, making it more efficient compared to the brute force approach.

### Can you provide implementations of the Product of Array Except Self problem in different programming languages?

Yes, implementations are provided for Python, Java, JavaScript, and C++. Each implementation follows the same logic using arrays to store intermediate products and then computes the final result efficiently.