Product of Array Except Self Solution

The product of array except self problem requires computing the product of all elements in an array except the current element, which can be solved efficiently using prefix and suffix products with linear time complexity.

1. Problem Restatement

Clarifying questions to ask the interviewer:

1. Can we modify the input array in-place, or do we need to return a new array?
2. What is the expected time complexity constraint? O(n) or better?
3. Are we allowed to use division operation, or must we avoid it?
4. What is the maximum size of the input array?
5. Should we handle negative numbers and zeros in the input?

Restatement of the problem: Given an integer array nums, return an array answer such that answer[i] equals the product of all elements in nums except nums[i]. The problem requires computing products without using division and should be done in linear time.

Input/Output specification:

• Input: Integer array nums of length n (1 ≤ n ≤ 10⁵)
• Output: Integer array answer of same length where answer[i] = product of all elements except nums[i]
• Example: Input [1,2,3,4] → Output [24,12,8,6]

Constraints:

• No division operator allowed
• Must run in O(n) time complexity
• Space complexity should be O(1) excluding the output array
• Array can contain zeros and negative numbers

Assumptions:

• The input array contains at least one element
• We can modify the input array if needed (but prefer not to)
• The result will fit in a 32-bit integer

Data shape and hidden structure: The problem exhibits a prefix/suffix pattern where each element's result is the product of all elements to its left multiplied by all elements to its right. This suggests we can compute prefix products and suffix products separately, then combine them.

The key insight is that for any element at index i, the answer is prefix_product[i-1] * suffix_product[i+1], where prefix_product[i] is the product of all elements from index 0 to i, and suffix_product[i] is the product of all elements from index i to the end.

2. Pattern Recognition

Identifying the most likely pattern or patterns

The most likely pattern here is the prefix/suffix product pattern. This is a classic array manipulation problem where we need to compute products while avoiding division operations.

Why this problem fits that pattern

The problem asks us to compute the product of all elements except the current element, which naturally decomposes into:

• Product of all elements to the left of index i (prefix product)
• Product of all elements to the right of index i (suffix product)

The key wording that signals this pattern:

• "except self" - suggests we're excluding one element at a time
• "product of array except self" - indicates multiplication operations across subsets
• The constraint that we cannot use division suggests we need to build up products from left and right

Core insight needed to solve it

The core insight is that for any position i, the result equals: result[i] = (product of elements from 0 to i-1) × (product of elements from i+1 to n-1)

This can be solved efficiently by computing prefix products and suffix products separately, then combining them. The key is recognizing that we can compute these in two passes and avoid redundant calculations.

This pattern appears in many array problems where we need to process elements with constraints about what's included or excluded from calculations.

3. Approach Evolution

My first instinct is to solve this with a brute force approach where for each position, I calculate the product of all other elements. But that's O(n²) because for each element, I'd have to iterate through all other elements. Let me think about how to optimize this.

I notice that for any position i, the result is the product of all elements except nums[i]. If I think about it mathematically, this is equivalent to the total product divided by nums[i]. However, I can't simply divide by zero when nums[i] is 0, and I don't want to handle special cases for division.

I notice that if I maintain two arrays - one with prefix products and another with suffix products, I can compute the result in O(n) time. For each position i, the answer is prefix[i-1] * suffix[i+1]. This way I'm avoiding division and handling zeros properly.

Let me think about this more carefully. If I compute prefix[i] as the product of all elements from index 0 to i, and suffix[i] as the product of all elements from index i to n-1, then for any element at position j, the result is prefix[j-1] * suffix[j+1].

This approach gives me O(n) time and O(n) space complexity. But I can do even better - I can compute the result in-place by using the output array as one of my working arrays, reducing space complexity to O(1) (excluding the output array).

The key insight is that I can build the prefix products in one pass, and then in a second pass, I can build the suffix products while computing the final result. For each position i, I want prefix[i-1] * suffix[i+1]. I can compute prefix products and store them in the result array, then compute suffix products while updating the result array.

The invariant I need to maintain is that at any point during the second pass, the result array contains the prefix products up to position i-1, and I'm multiplying by the suffix product from position i+1 onwards.

// First pass: compute prefix products and store in result array
// Second pass: compute suffix products and multiply with existing prefix values

4. Data Structure Choice

For this problem, I'll use a primary array and two auxiliary arrays to store intermediate results.

Primary Data Structures

1. int[] result - The main output array that will hold our final answer
2. int[] leftProducts - Auxiliary array storing products of all elements to the left of each index
3. int[] rightProducts - Auxiliary array storing products of all elements to the right of each index

Why Each Data Structure Is Needed

The leftProducts array is essential because it allows us to compute the product of all elements to the left of each position in O(1) time during the final multiplication step. Similarly, rightProducts enables O(1) access to products of elements to the right.

Time and Space Impact

• Time Complexity: O(n) for building each auxiliary array, plus O(n) for the final multiplication
• Space Complexity: O(n) for storing the two auxiliary arrays plus O(1) additional space (excluding output array)
• The auxiliary arrays are necessary to avoid recomputing products multiple times, which would increase time complexity to O(n²)

Helper Structure

No additional helper structures are needed beyond the three arrays. The solution naturally decomposes into computing left and right products separately, which are then multiplied together to get the final result.

This approach elegantly separates concerns: left products, right products, and their combination. The arrays serve as memoization tables that eliminate redundant computation while maintaining clean, readable logic.

int[] result = new int[n];
int[] leftProducts = new int[n]; 
int[] rightProducts = new int[n];

5. Algorithm Walkthrough

Let me trace through the algorithm with the example array [1, 2, 3, 4]:

Step 1: Calculate left products

• i=0: left[0] = 1 (no elements to the left)
• i=1: left[1] = left[0] * nums[0] = 1 * 1 = 1
• i=2: left[2] = left[1] * nums[1] = 1 * 2 = 2
• i=3: left[3] = left[2] * nums[2] = 2 * 3 = 6
• left array becomes [1, 1, 2, 6]

Step 2: Calculate right products and combine

• i=3: right = 1 (no elements to the right), result[3] = left[3] * right = 6 * 1 = 6
• i=2: right = right * nums[3] = 1 * 4 = 4, result[2] = left[2] * right = 2 * 4 = 8
• i=1: right = right * nums[2] = 4 * 3 = 12, result[1] = left[1] * right = 1 * 12 = 12
• i=0: right = right * nums[1] = 12 * 2 = 24, result[0] = left[0] * right = 1 * 24 = 24

Final result: [24, 12, 8, 6]

The algorithm correctly handles edge cases like:

• Single element array: returns [1]
• Empty array: returns empty array
• Arrays with zeros: properly computes products without division

Each position's result is the product of all elements except itself, computed efficiently by combining left and right products.

6. Java Solution

public class ProductOfArrayExceptSelf {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] result = new int[n];
        
        // Step 1: Calculate left products
        result[0] = 1;
        for (int i = 1; i < n; i++) {
            result[i] = result[i - 1] * nums[i - 1];
        }
        
        // Step 2: Calculate right products and multiply with left
        int rightProduct = 1;
        for (int i = n - 1; i >= 0; i--) {
            result[i] = result[i] * rightProduct;
            rightProduct *= nums[i];
        }
        
        return result;
    }
}

This solution uses the concept of prefix and suffix products to compute the result in two passes. First, we calculate the product of all elements to the left of each index, then multiply by the product of all elements to the right. This approach avoids using division and handles zeros correctly without additional space complexity.

The solution works by:

1. Building left products array in-place using the result array
2. Maintaining a running right product variable to avoid extra space
3. Combining left and right products for final result

This is an optimal O(n) time and O(1) space solution that meets all requirements.

7. Complexity Analysis

Time Complexity: O(n)

The algorithm makes exactly two passes through the array:

1. First pass: O(n) to compute left products
2. Second pass: O(n) to compute right products and combine with left products

Since we're doing constant work for each element in both passes, the total time complexity is O(n + n) = O(n).

Space Complexity: O(1)

The space complexity is O(1) additional space (not counting the output array). We only use a constant number of variables:

• result array (required for output, not counted in additional space)
• rightProduct variable (single integer)
• Loop index variables

The algorithm avoids using extra arrays for left or right products, instead computing values on-the-fly and storing results directly in the output array.

Why These Bounds Are Correct

The O(n) time complexity is optimal because we must examine each element at least once to compute the product of all other elements. Any algorithm that solves this problem must make at least n operations.

The O(1) space complexity is achieved by leveraging the output array itself as a storage mechanism, avoiding the need for additional O(n) space structures like prefix or suffix arrays.

This solution is optimal in both time and space complexity, achieving the best possible bounds for this problem. The approach avoids the O(n²) brute force solution where each element requires computing n products, and also avoids the O(n) extra space solution that would use prefix/suffix arrays.

The solution is near-optimal because it's impossible to do better than O(n) time complexity for this problem, and the space usage is already at the theoretical minimum for an in-place algorithm.

8. Edge Cases & Pitfalls

Empty and Minimal Inputs

• Empty array: The problem constraints don't explicitly forbid this, but if nums is empty, we should return an empty array. Our solution handles this correctly by returning new int[0].
• Single element: If the array has only one element, the result should be [1] since there's no other element to multiply by. Our solution correctly handles this case.

Duplicate Values and Repeated States

• Zero values: The presence of zeros creates special cases. If there's exactly one zero, all elements except that zero's position will be 0, and the zero position will have the product of all non-zero elements. If there are multiple zeros, all results should be 0.
• Negative numbers: The solution correctly handles negative numbers since multiplication is associative and commutative.

Boundary Conditions

• Integer overflow: When multiplying many large numbers, integer overflow can occur. For example, if nums = [100000, 100000, 100000], intermediate products can exceed Integer.MAX_VALUE. However, since the problem asks for an integer array and doesn't specify overflow handling, we assume inputs are such that intermediate results fit in int. For production code, one might consider using long or handling overflow explicitly.

Common Implementation Mistakes

• Not handling zero properly: A common mistake is to treat zeros like any other number, which leads to incorrect results. For instance, if we compute product = product * nums[i] without checking for zero, we'll get 0 in all positions when there's a zero.
• Off-by-one errors: When computing prefix and suffix products, off-by-one errors can lead to incorrect indices being used. For example, computing prefix[i] = prefix[i-1] * nums[i-1] without proper bounds checking can cause array index out of bounds.
• Incorrect result assignment: When setting the final result, one might accidentally assign prefix or suffix values to wrong indices. For example, assigning result[i] = prefix[i] instead of prefix[i] * suffix[i].
• Not accounting for zero in suffix calculation: When computing the suffix array, if we don't properly handle zeros, we might compute incorrect products.

// Example of a problematic approach:
// int[] result = new int[nums.length];
// for (int i = 0; i < nums.length; i++) {
//     result[i] = prefix[i] * suffix[i]; // This is correct, but if we do:
//     result[i] = prefix[i]; // This would be wrong!
// }

9. Alternative Approaches

There are several alternative approaches to solving the Product of Array Except Self problem, each with different trade-offs:

Brute Force Approach: For each position, compute the product of all elements except the current one by iterating through the entire array. This approach has O(n²) time complexity and is inefficient for large inputs. It's useful to mention as a baseline but should be dismissed quickly due to poor performance.

Two-Pass with Division: Compute the total product of all elements and then divide by each element to get the result. This approach works for arrays without zeros but fails when there are zeros in the input array, as it would lead to division by zero errors. Even if we handle zeros specially, this approach requires two passes and doesn't work with negative numbers properly.

Prefix/Suffix Arrays: Create two separate arrays to store prefix products and suffix products. The result at index i is the product of prefix[i-1] and suffix[i+1]. This approach uses O(n) extra space for the two arrays but is intuitive and easy to understand. It's a valid alternative that trades space for simplicity.

Space-Optimized Single Pass: While our chosen approach uses O(1) extra space (excluding output), we could also consider maintaining running products from both directions simultaneously in a single pass. However, this approach is more complex to implement and debug without clear advantages over our two-pass solution.

Mathematical Approach with Logarithms: For arrays containing only positive numbers, we could use logarithmic properties to compute products. This approach is mathematically interesting but introduces floating-point precision errors and doesn't handle negative numbers or zeros properly.

The chosen solution is preferred because it's clean, efficient, and handles all edge cases including zeros and negative numbers without any mathematical approximations or special handling for specific number types.

10. Test Cases

public class ProductOfArrayExceptSelf {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] result = new int[n];
        
        // Handle edge cases
        if (nums == null || n == 0) {
            return result;
        }
        
        // Build left products
        result[0] = 1;
        for (int i = 1; i < n; i++) {
            result[i] = result[i-1] * nums[i-1];
        }
        
        // Build right products and combine
        int rightProduct = 1;
        for (int i = n - 1; i >= 0; i--) {
            result[i] *= rightProduct;
            rightProduct *= nums[i];
        }
        
        return result;
    }
}

Sample Test Cases:

Test Case 1: Basic case

• Input: [1, 2, 3, 4]
• Expected Output: [24, 12, 8, 6]
• Explanation: Each position contains product of all elements except self

Test Case 2: With zeros

• Input: [2, 3, 0, 4]
• Expected Output: [0, 0, 24, 0]
• Explanation: When one zero exists, only the position with zero gets non-zero product

Test Case 3: With negative numbers

• Input: [-1, 2, 3, -4]
• Expected Output: [24, -12, -8, 6]
• Explanation: Handles negative products correctly

Test Case 4: Single element

• Input: [5]
• Expected Output: [1]
• Explanation: Only one element, so result is 1

Test Case 5: Empty array

• Input: []
• Expected Output: []
• Explanation: Returns empty array as expected

Test Case 6: Array with duplicates

• Input: [2, 4, 2, 3]
• Expected Output: [24, 12, 24, 16]
• Explanation: Handles duplicate values correctly

These test cases cover all major scenarios including edge cases, zero handling, and negative numbers.

11. Final Summary

The core insight is that we can compute the product of all elements except self by calculating prefix products and suffix products separately, then combining them in a single pass. This approach works because each position's result equals the product of all elements to its left (prefix) multiplied by all elements to its right (suffix). The solution is optimal with O(n) time and O(1) extra space complexity, making it ideal for large inputs where memory efficiency matters. The key is recognizing that we don't need to store all prefix/suffix products separately, but can build the result incrementally while maintaining the necessary invariants.