What is the best approach to solve Product of Array Except Self?

The optimal approach for Product of Array Except Self is the "Optimal — O(1) Extra Space (Output array)" technique. It runs in O(n) time and uses O(1) space. Use the output array as the left-product array. Then do a second pass right-to-left, keeping a running right product in a single variable. Multiplies the right product into each answer in-place. No ex

What is the time complexity of Product of Array Except Self?

The optimal solution for Product of Array Except Self has a time complexity of O(n) and a space complexity of O(1), achieved using the Prefix Sum pattern.

Is Product of Array Except Self a Medium LeetCode problem?

Yes, LeetCode #238 (Product of Array Except Self) is rated Medium. It uses the Prefix Sum pattern and is commonly asked in interviews at Amazon, Microsoft, Apple.

How to solve Product of Array Except Self in Python, Java, C++, and JavaScript?

LumiChats provides complete solutions for Product of Array Except Self in Python, Java, C++, and JavaScript, along with step-by-step visual animations and time/space complexity analysis. All solutions are free to view.

#238MediumPrefix Sum

Product of Array Except Self

Build left and right product arrays without using division.

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Problem Statement

Given an integer array nums, return an array answer such that answer[i] equals the product of all elements of nums except nums[i]. You must write an algorithm that runs in O(n) time and without using the division operation.

Examples

Example 1

Input:nums = [1,2,3,4]

Output:[24,12,8,6]

Explanation: answer[0]=2×3×4=24, answer[1]=1×3×4=12...

Example 2

Input:nums = [-1,1,0,-3,3]

Output:[0,0,9,0,0]

Constraints

2 ≤ nums.length ≤ 10⁵
-30 ≤ nums[i] ≤ 30
No division allowed.

Solutions

Left & Right Pass (Two Extra Arrays)

TimeO(n)SpaceO(n)

Build a left[i] array where left[i] = product of all elements before i. Build a right[i] where right[i] = product of all elements after i. Answer[i] = left[i] * right[i].

Visual Animation

def productExceptSelf(nums: list[int]) -> list[int]:
    n = len(nums)
    left = [1] * n
    for i in range(1, n):
        left[i] = left[i-1] * nums[i-1]
    right = [1] * n
    for i in range(n-2, -1, -1):
        right[i] = right[i+1] * nums[i+1]
    return [left[i] * right[i] for i in range(n)]

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