Product of Array Except Self Algorithm Solution

The product of array except self problem requires computing the product of all elements in an array except the current element, without using division and in linear time complexity. This classic coding interview question tests understanding of prefix and suffix array concepts.

1. Pattern Restatement

Clarifying questions to ask the interviewer:

1. What are the constraints on input array size? (e.g., up to 10⁵ elements?)
2. Are duplicate values allowed in the input array?
3. Should we return a new array or modify the input array in-place?
4. Can we use division operation? (This would simplify but may not be allowed)
5. What is the expected time complexity for follow-up? O(n) or better?

Restate the pattern or problem type in your own words: The "Product of Array Except Self" problem asks us to compute an array where each element at index i contains the product of all elements in the input array except the one at index i. The challenge is to do this efficiently without using division, which would make the problem trivial.

Clarify the signal words and constraint shape that point to it:

• Signal words: "product", "except self", "array", "without division"
• Constraint shape: Array of integers, need O(n) time complexity, no division allowed
• The problem asks for a transformation where each output element depends on the entire input except one position

Note any assumptions you are making:

• Input array contains integers (positive, negative, zero)
• We can modify the input array or create a new result array
• Division is not allowed (which makes this a more interesting problem)
• The solution should be efficient with linear time complexity

Identify whether the pattern is best suited for static, streaming, or online scenarios if relevant: This is a static problem where the entire input array is available at once, so it's best suited for batch/static processing rather than streaming or online scenarios where data arrives incrementally.

2. When to Use This Pattern

Signal words and constraint shapes:

• "Product of array except self" or "array with all elements except current"
• Input is an array of integers where each element needs to be replaced by the product of all other elements
• "Do not use division" - a key constraint that rules out simple mathematical approaches
• "O(n) time complexity required" - indicates need for single-pass or linear-time solution

Why this pattern fits better than alternatives: This problem requires computing products in a way that avoids direct nested iteration, making it ideal for the prefix/suffix product pattern. Unlike brute force approaches that compute each product independently (O(n²)), this pattern leverages the mathematical property that we can build up products incrementally using previously computed values.

Conditions where variations apply:

• If the problem asks for products with division allowed, we could use a simple approach dividing total product by each element
• If only prefix or suffix products are needed, we can optimize to compute just one direction
• For streaming data where elements arrive over time, we'd need a sliding window or online variant

Core insight: The key insight is that for any element at index i, the result equals the product of all elements to its left multiplied by the product of all elements to its right. This allows us to compute prefix and suffix products separately, then combine them in O(n) time.

This pattern is particularly suited for static input arrays where we can make two passes - once to compute left products and once to compute right products, making it optimal for the constraints given.

3. Approach Evolution

Interview communication script:

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 need to iterate through the entire array again. Let me think about how to optimize this.

I notice that if I maintain prefix and suffix products, I can reduce it to O(n) time by computing them in two passes. The key insight is that for any element at index i, the result equals the product of all elements before i (prefix) multiplied by the product of all elements after i (suffix). This way, I don't recalculate overlapping products.

Let me think more carefully about the invariants:

• Prefix at index i represents product of elements from 0 to i-1
• Suffix at index i represents product of elements from i+1 to n-1
• For any element, the answer is prefix[i] * suffix[i]

This approach works because we're essentially decomposing the problem into two independent computations that can be done in linear time.

Brute-force approach: The naive solution would iterate through each position and calculate the product of all other elements by multiplying every element except the current one. This results in O(n²) time complexity because for each of the n elements, we perform n-1 multiplications.

Improved approach: The key optimization is to precompute prefix and suffix products. For each position i:

• Prefix[i] = product of elements from index 0 to i-1
• Suffix[i] = product of elements from index i+1 to n-1
• Result[i] = Prefix[i] × Suffix[i]

This reduces the problem to two linear passes, achieving O(n) time complexity.

Final approach with step-by-step reasoning:

1. First pass: Build prefix products array where prefix[i] = product of elements 0 to i-1
2. Second pass: Build suffix products array where suffix[i] = product of elements i+1 to n-1
3. Third pass: For each index, multiply prefix[i] × suffix[i] to get final result

Important invariants that must hold:

• Prefix[0] = 1 (no elements before index 0)
• Suffix[n-1] = 1 (no elements after index n-1)
• For any valid index i, result[i] = prefix[i] × suffix[i]
• All operations maintain integer overflow considerations for the problem constraints

4. Data Structure Choice

For the Product of Array Except Self problem, I'll use a primary array and two auxiliary arrays to store prefix and suffix products.

Primary Data Structures

1. int[] result - The output array that will hold the final products

   • Size: n (same as input array)
   • Purpose: Stores the final result where each element is the product of all elements except self
   • Space: O(n) - required for output

2. int[] prefixProducts - Auxiliary array storing cumulative products from left to right

   • Size: n
   • Purpose: prefixProducts[i] = product of all elements from index 0 to i
   • Space: O(n) - for storing intermediate calculations

3. int[] suffixProducts - Auxiliary array storing cumulative products from right to left

   • Size: n
   • Purpose: suffixProducts[i] = product of all elements from index i to n-1
   • Space: O(n) - for storing intermediate calculations

Why These Choices Work

The key insight is that for any element at index i, the result equals: prefixProducts[i-1] * suffixProducts[i+1]

This approach uses the divide and conquer principle:

• Left products (prefix) are calculated from left to right
• Right products (suffix) are calculated from right to left
• Final answer is the product of corresponding prefix and suffix values

Space vs Time Trade-off

This approach uses O(n) extra space but achieves O(n) time complexity, which is optimal for this problem since we must produce n outputs. The auxiliary arrays eliminate the need for nested loops and avoid recalculating products multiple times.

Helper Structures

No additional helper structures are needed beyond the three arrays. The solution naturally builds prefix and suffix products in a single pass each, making the implementation clean and readable.

The choice of arrays over other data structures like stacks or queues is optimal because:

• Arrays provide O(1) random access needed for the final multiplication step
• We need to build up products sequentially from both directions
• The problem structure naturally maps to indexed array operations

5. Algorithm Walkthrough

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

Step 1: Calculate left products

• left[0] = 1 (no elements to the left)
• left[1] = 1 (product of elements to the left: 1)
• left[2] = 1 * 2 = 2 (product of elements to the left: 1,2)
• left[3] = 1 * 2 * 3 = 6 (product of elements to the left: 1,2,3)
• Result: left = [1, 1, 2, 6]

Step 2: Calculate right products and combine

• Start with right = 1 (no elements to the right of last element)
• result[3] = left[3] * right = 6 * 1 = 6
• right = right * nums[3] = 1 * 4 = 4
• result[2] = left[2] * right = 2 * 4 = 8
• right = right * nums[2] = 4 * 3 = 12
• result[1] = left[1] * right = 1 * 12 = 12
• right = right * nums[1] = 12 * 2 = 24
• result[0] = left[0] * right = 1 * 24 = 24
• Final result: [24, 12, 8, 6]

Key insights during walkthrough:

• At each position, we want the product of all elements except the current one
• We can achieve this by computing products from left and right separately
• The left product at position i is the product of all elements before i
• The right product at position i is the product of all elements after i
• Combining these gives us the desired result

Edge case handling:

• For single element array [5]: left = [1], right = 1, result = [1]
• For zero elements: handled by the algorithm logic
• When elements are zero, we correctly compute products excluding that element

The algorithm terminates because we process each element exactly once in both passes, and the final result array is fully populated.

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 combine
        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 O(1) extra space (excluding output array). First, we calculate all left products and store them in the result array. Then we traverse from right to left, maintaining a running product of elements to the right and combining it with the left products.

7. Complexity Analysis

Time Complexity: O(n)

The algorithm makes exactly two passes through the array:

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

Since we're doing two linear passes, the overall time complexity is O(n).

Space Complexity: O(1)

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

• result array (which is required for the output and not counted in space complexity)
• rightProduct variable (constant space)
• Loop index variables (constant space)

The solution avoids using extra arrays like leftProducts or rightProducts, which would have required O(n) space each.

Why These Bounds Are Correct

The bounds are tight and optimal because:

1. We must read each element at least once to compute the result (Ω(n) lower bound)
2. We must write each result element once (Ω(n) lower bound)
3. The algorithm achieves exactly this minimum work

The solution is optimal because:

• It's impossible to solve this problem in less than O(n) time since we need to examine all elements
• It uses the minimum possible additional space (O(1) extra space)
• The two-pass approach is necessary to compute the products correctly without knowing all elements in advance

This makes the solution near-optimal in both time and space complexity.

8. Edge Cases & Pitfalls

Empty and Minimal Inputs

• Empty array: Should return empty result (no multiplication possible)
• Single element: Should return array with one 1 (product of empty set is 1 by convention)
• Two elements: Should return array with swapped values (e.g., [2,3] → [3,2])

Duplicate Values and Repeated States

• Zero values: The presence of zeros creates special cases where most products become zero, but the position of zero matters
• Multiple zeros: If there are multiple zeros, all results should be 0 (since any product involving zero is zero)
• Negative numbers: Should work correctly with negative products, but care must be taken with signs

Boundary Conditions

• Large numbers: Integer overflow is possible when multiplying large numbers, though this is less of an issue with the O(n) approach
• Array size limits: For very large arrays, memory constraints could be an issue (though not typically a problem in coding interviews)

Common Implementation Mistakes

• Incorrect handling of zeros: Most common mistake is assuming all elements are non-zero and not properly accounting for zero positions
• Off-by-one errors: When computing prefix/suffix products, boundary conditions can cause index mismatches
• Not considering the mathematical identity: For arrays with one element, result should be [1], not [0]
• Mutating input array: If the problem requires preserving the input, this is a critical mistake
• Wrong result indexing: When building prefix and suffix arrays, mixing up indices can lead to incorrect results

Special Case Handling

The most critical edge case is handling zeros properly:

• If exactly one zero exists at index i, only the result at that index should be non-zero (equal to product of all other elements)
• If more than one zero exists, all results should be zero
• Care must be taken to avoid division by zero when computing products in alternative approaches

// Example of problematic handling:
// [1, 0, 3] - should return [0, 3, 0]
// But if we compute prefix products incorrectly, we might get [0, 0, 3]

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, but it's simple to understand and implement.

Division Approach: Compute the total product of all elements, then for each position divide the total by the element at that position. This works when there are no zeros in the array, but fails when there are zeros because division by zero is undefined. Additionally, this approach requires handling special cases for zero elements and can face precision issues with floating-point numbers.

Two-Pass Approach with Extra Space: Compute prefix products (products of elements from the start up to each position) and suffix products (products from each position to the end) in separate arrays. Then for each position, multiply the prefix product up to that position with the suffix product starting from the next position. This uses O(n) extra space but is intuitive and easy to debug.

Space-Optimized Two-Pass: Instead of storing prefix and suffix arrays, compute the result array in two passes using only one pass to build prefix products and another to build the final result by multiplying with suffix products. This reduces space usage but still requires multiple passes through the data.

The chosen solution using left and right products in a single pass is optimal because it achieves O(1) extra space complexity (excluding output array) and O(n) time complexity, making it the most efficient approach for this constraint.

The key difference between these alternatives and our chosen solution is that they either use more space or require multiple passes, while our approach maintains optimal time and space complexity in a single pass through the array.

10. Test Cases

public class ProductOfArrayExceptSelf {
    public int[] productExceptSelf(int[] nums) {
        int n = nums.length;
        int[] result = new int[n];
        
        // Left products
        result[0] = 1;
        for (int i = 1; i < n; i++) {
            result[i] = result[i-1] * nums[i-1];
        }
        
        // Right products
        int rightProduct = 1;
        for (int i = n - 1; i >= 0; i--) {
            result[i] *= rightProduct;
            rightProduct *= nums[i];
        }
        
        return result;
    }
}

Sample Test Cases

Example 1: Basic case Input: [1, 2, 3, 4] Expected Output: [24, 12, 8, 6]

• At index 0: product of 2×3×4 = 24
• At index 1: product of 1×3×4 = 12
• At index 2: product of 1×2×4 = 8
• At index 3: product of 1×2×3 = 6

Example 2: With zeros Input: [2, 3, 0, 4] Expected Output: [0, 0, 24, 0]

• Index 2 has zero so all products involving it are zero
• Index 0: 3×0×4 = 0 (but we exclude index 2's zero)
• Index 1: 2×0×4 = 0 (but we exclude index 2's zero)
• Index 3: 2×3×0 = 0 (but we exclude index 2's zero)

Example 3: With negative numbers Input: [-1, 2, -3, 4] Expected Output: [24, -12, 8, -6]

• Index 0: 2×(-3)×4 = 24
• Index 1: (-1)×(-3)×4 = 12 (but sign changes due to odd number of negatives)
• Index 2: (-1)×2×4 = -8 (but sign changes due to odd number of negatives)
• Index 3: (-1)×2×(-3) = 6 (but sign changes due to odd number of negatives)

Example 4: Boundary case - single element Input: [5] Expected Output: [1]

• Only one element, so product of all others is 1

Example 5: Boundary case - two elements Input: [3, 7] Expected Output: [7, 3]

• Index 0: 7 (product of all others)
• Index 1: 3 (product of all others)

11. Final Summary

The core idea is to compute the product of all elements except the current one in linear time by leveraging prefix and suffix products. This solution works efficiently by avoiding division operations and handling edge cases through careful array traversal, making it optimal for the given constraints.

The key insight is that for any element at index i, the result equals the product of all elements to its left (prefix) multiplied by the product of all elements to its right (suffix). By computing these values in two passes and storing them in the result array, we achieve O(n) time complexity with O(1) extra space (excluding output array). This approach is superior to brute force methods that would require O(n²) time and handles all edge cases including zeros gracefully through the multiplicative property of arrays.