238_product_of_array_except_self.md

Product of Array Except Self

Problem Number: 238 Difficulty: Medium Category: Array, Prefix Sum LeetCode Link: https://leetcode.com/problems/product-of-array-except-self/

Problem Description

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.

Example 1:

Input: nums = [1,2,3,4]
Output: [24,12,8,6]
Explanation: answer[0] = 3*4*2 = 24, answer[1] = 1*3*4 = 12, answer[2] = 1*2*4 = 8, answer[3] = 1*2*3 = 6

Example 2:

Input: nums = [-1,1,0,-3,3]
Output: [0,0,9,0,0]
Explanation: answer[0] = 1*0*(-3)*3 = 0, answer[1] = (-1)*0*(-3)*3 = 0, answer[2] = (-1)*1*(-3)*3 = 9, answer[3] = (-1)*1*0*3 = 0, answer[4] = (-1)*1*0*(-3) = 0

Constraints:

• 2 <= nums.length <= 10^5
• -30 <= nums[i] <= 30
• The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

Follow up: Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)

My Approach

I used a Two-Pass approach to calculate the product without using division. The key insight is to use two passes: first to calculate left products, then to multiply with right products.

Algorithm:

1. Initialize result array with 1s
2. First pass (left to right): Calculate left products
   • For each index i, store product of all elements to the left
3. Second pass (right to left): Multiply with right products
   • For each index i, multiply with product of all elements to the right
4. Return the result array

Solution

The solution uses two-pass approach to calculate products without division. See the implementation in the solution file.

Key Points:

• Uses two passes through the array
• First pass calculates left products
• Second pass multiplies with right products
• Avoids division operation
• O(n) time complexity

Time & Space Complexity

Time Complexity: O(n)

• Two passes through the array
• Each pass performs constant time operations
• Total: O(n)

Space Complexity: O(1) (excluding output array)

• Uses only a constant amount of extra space
• Output array doesn't count for space complexity

Key Insights

1. Two-Pass Approach: Using two passes avoids the need for division.

2. Left and Right Products: Each element's product is left_product × right_product.

3. In-place Calculation: Can calculate products in-place using the result array.

4. No Division: The approach works without using division operation.

5. Linear Time: Achieves O(n) time complexity with two passes.

6. Constant Space: Uses only O(1) extra space (excluding output).

Mistakes Made

1. Division Approach: Initially might try to use division, which is not allowed.

2. Brute Force: Using nested loops to calculate each product.

3. Complex Logic: Overcomplicating the two-pass approach.

4. Wrong Order: Not understanding the order of the two passes.

Related Problems

• Trapping Rain Water (Problem 42): Two-pass approach for water trapping
• Candy (Problem 135): Two-pass approach for candy distribution
• Gas Station (Problem 134): Circular array with two-pass logic
• Container With Most Water (Problem 11): Two-pointer approach

Alternative Approaches

1. Prefix and Suffix Arrays: Use separate arrays for left and right products - O(n) time, O(n) space
2. Single Pass with Division: Use division (if allowed) - O(n) time, O(1) space
3. Recursive: Use recursion to calculate products - O(n) time, O(n) space

Common Pitfalls

1. Division Usage: Using division when it's not allowed.
2. Brute Force: Using nested loops for O(n²) complexity.
3. Complex Logic: Overcomplicating the two-pass approach.
4. Wrong Order: Not understanding the order of passes.
5. Space Inefficiency: Using O(n) extra space when O(1) is possible.

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Note: This is a two-pass problem that demonstrates efficient product calculation without division.