LeetCode 238: Product of Array Except Self

Date: January 20, 2026
Category: Arrays
Time Taken: 30 minutes
Difficulty: Medium

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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.

Link: Product of Array Except Self

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My Approach

Initial thought:
The first solution I thought of was a brute-force approach. It involved using nested for loops and a variable inside the first loop to keep track of the products of all values except the current one. The results would then be placed in the index that doesn't match the current value. This solution would have a time complexity of O(N²) and a space complexity of O(1) since the output array is not counted. However, LeetCode requires at least an O(N) solution, so we can't use this approach.

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Solution 1 (Separate prefix and suffix arrays):

• First, we define the answer, prefix, and suffix arrays, making them the same size as the nums array. The prefix and suffix arrays will be initialized with 1, while the answer array will be initialized with 0.

• Next, we use a for loop to store the product of all values before the current one in the prefix array. We start from the first index and go up to the length of nums.

• Similarly, we store the products of all values after the current index in the suffix array.

• Finally, we use another for loop to store the product of prefix[i] and suffix[i] in answer[i].

Solution Code

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        # Prefix and Suffix approach
        size = len(nums)
        answer = [0] * size
        prefix = [1] * size
        suffix = [1] * size

        for i in range(1, size):
            prefix[i] = prefix[i - 1] * nums[i - 1]

        for i in range(size - 2, -1, -1):
            suffix[i] = suffix[i + 1] * nums[i + 1]

        for i in range(size):
            answer[i] = suffix[i] * prefix[i]

        return answer

Complexity:

Time: O(N) Space: O(N)

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Solution 2 (Separate prefix and suffix arrays) [Optimized]:

• Similar to the above approach, we will create an answer array but won't use separate prefix and suffix arrays. Instead, we'll store the values directly in the answer array.

• Use a for loop to store the product of numbers before the current index directly in the answer array. This is the prefix loop.

• Use another loop starting from the last index len(nums) - 1 and go to index -1. Along the way, store the product of the numbers after the current index and multiply it with the product already in the answer array from the prefix loop.

Solution Code (No separate arrays)

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        size = len(nums)
        answer = [0] * size

        prefix = 1
        for i in range(size):
            answer[i] = prefix
            prefix *= nums[i]

        suffix = 1
        for i in range(size - 1, -1, -1):
            answer[i] *= suffix
            suffix *= nums[i]

        return answer

Complexity:

Time: O(N) Space: O(1)

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Key Takeaway: Understood how prefix and suffix problems work.

Pattern: Prefix Sum

Series: 90 Days of Data Engineering Progress: 12/90 problems completed

Tags: #DEQuest #LeetCode #Python #DataEngineering #BuildInPublic