Solve Product of Array Except Self Problem in 4 Easy Steps

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.

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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:

1. Algorithmic Thinking: It enhances your ability to think through complex problems and come up with efficient solutions.
2. Array Manipulation: It reinforces concepts related to array manipulation and operations.
3. 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.

1. Create two arrays left and right.
2. Traverse the array to fill left[i] with the product of all elements to the left of i.
3. Traverse the array to fill right[i] with the product of all elements to the right of i.
4. 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:

1. Problem Definition: Compute an array where each element is the product of all other elements in the array.
2. Example: For input [1,2,3,4], the output is [24,12,8,6].
3. 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: