You’re in a coding interview. The interviewer says: “Given an integer array, return an array where each element is the product of everything except itself. Oh, and you can’t use division.” You know the trick: prefix and suffix products. But now you need to explain the two-pass structure out loud, trace through an example without losing track of running variables, and handle the follow-up when they ask you to optimize to O(1) extra space, all while someone watches. That’s a very different challenge.

This walkthrough covers how that conversation unfolds: the clarifying questions that signal constraint awareness, the elegant left-right decomposition interviewers want to hear, and the initialization bugs that catch even prepared candidates.

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The Problem

Given an integer array nums, return an array answer such that answer[i] is equal to the product of all elements in nums except nums[i].

You must write an algorithm that runs in O(n) time and does not use the division operation.

(LeetCode #238)

Example

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

Output [24, 12, 8, 6]

For index 0, the product of all other elements is 2 * 3 * 4 = 24. For index 1, it’s 1 * 3 * 4 = 12. For index 2, it’s 1 * 2 * 4 = 8. And for index 3, it’s 1 * 2 * 3 = 6. Every element in the output represents the product of the entire array with that one position excluded.

Already comfortable with the solution? Practice it in a mock interview →

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What Interviewers Are Actually Testing

This problem is really about your ability to recognize a pattern, reason about prefix/suffix relationships, and optimize for space under constraints.

Skill	What Interviewers Observe
Constraint awareness	Do you immediately acknowledge the no-division rule and O(n) requirement?
Prefix/suffix pattern recognition	Can you decompose the problem into two independent passes?
In-place optimization	Do you know how to reduce O(n) extra space to O(1) when prompted?
Edge case thinking	Do you account for zeros, single elements, or negative numbers?
Communication	Can you explain why the two-pass approach works before writing code?

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Step 1: Clarifying Questions

Before writing anything, pause and ask these questions. They demonstrate rigor and occasionally reveal constraints that change your approach.

1. Can the array contain zeros? This matters enormously. Two zeros, one zero, or no zeros each produce different output patterns worth discussing.

2. Can the array contain negative numbers? Negative numbers don’t affect the algorithm, but asking shows you’re thinking about the full input domain rather than just happy-path integers.

3. What’s the minimum array length? The problem guarantees at least 2 elements, but confirming this lets you skip a bounds-check conversation and shows you read constraints carefully.

4. Is O(1) extra space required, or is O(n) auxiliary space acceptable? The standard version allows O(n) space (the output array doesn’t count). Some interviewers add a follow-up asking you to optimize to O(1). Knowing which version you’re solving upfront saves you from over-engineering too early.

5. Should the output be a new array, or can we modify the input in place? Clarifying this shows you think about mutation and side effects, an important quality signal for production-minded engineers.

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Step 2: A First (Non-Optimal) Idea

The most natural first instinct is the brute-force nested loop: for each index i, multiply together every element that isn’t at position i.

def product_except_self_brute(nums):
    n = len(nums)
    answer = []
    for i in range(n):
        product = 1
        for j in range(n):
            if j != i:
                product *= nums[j]
        answer.append(product)
    return answer

Approach	Time	Space
Brute force (nested loop)	O(n²)	O(n)
Division-based	O(n)	O(n)

This is correct, but it runs in O(n²) time. With an array of 100,000 elements, that’s 10 billion operations.

You might also think: “Why not compute the total product and divide by nums[i] for each position?” That’s O(n) and intuitive, but the problem explicitly forbids division. If you mention this approach, frame it as something you considered and ruled out, not as your answer.

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Step 3: The Key Insight

For any index i, the product of everything except nums[i] equals the product of everything to its left multiplied by the product of everything to its right. These are called prefix products and suffix products. Build prefix products in a left pass, then fold in suffix products in a right pass using a single running variable, and you get O(n) time with O(1) extra space.

In other words:

answer[i] = (product of nums[0..i-1]) * (product of nums[i+1..n-1])

Once you see this decomposition, the algorithm writes itself:

• Build a prefix product array: prefix[i] = product of all elements before index i.
• Build a suffix product array: suffix[i] = product of all elements after index i.
• The answer at each index is just prefix[i] * suffix[i].

This is O(n) time and O(n) space. But here’s the elegant optimization: you don’t need to store both arrays. You can build the prefix products directly into the output array, then make a second pass from right to left, multiplying in the suffix product on the fly using a single running variable. That brings extra space down to O(1) (excluding the output).

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Step 4: Optimal Strategy

1. Initialize an answer array of the same length as nums, filled with 1s.
2. Left pass: Traverse from left to right. Maintain a running left_product starting at 1. At each index i, set answer[i] = left_product, then multiply left_product by nums[i]. After this pass, answer[i] holds the product of all elements to the left of i.
3. Right pass: Traverse from right to left. Maintain a running right_product starting at 1. At each index i, multiply answer[i] by right_product, then multiply right_product by nums[i]. After this pass, answer[i] holds the product of all elements to the left AND right of i.
4. Return answer.

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Python Implementation

def product_except_self(nums: list[int]) -> list[int]:
    n = len(nums)
    answer = [1] * n

    # Left pass: answer[i] = product of all elements to the left of i
    left_product = 1
    for i in range(n):
        answer[i] = left_product
        left_product *= nums[i]  # Update for the next position

    # Right pass: multiply in the product of all elements to the right of i
    right_product = 1
    for i in range(n - 1, -1, -1):
        answer[i] *= right_product
        right_product *= nums[i]  # Update for the next position (moving left)

    return answer

Let’s trace through nums = [1, 2, 3, 4]:

After left pass: answer = [1, 1, 2, 6]

• Index 0: nothing to the left → 1
• Index 1: 1 to the left → 1
• Index 2: 1*2 to the left → 2
• Index 3: 1*2*3 to the left → 6

After right pass: answer = [24, 12, 8, 6]

• Index 3: multiply by 1 (nothing to the right)
• Index 2: multiply by 4
• Index 1: multiply by 4*3 = 12
• Index 0: multiply by 4*3*2 = 24

Clean, correct, and no division anywhere.

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Time & Space Complexity

Pass	Time Complexity	Extra Space
Left pass	O(n)	O(1), single running variable
Right pass	O(n)	O(1), single running variable
Total	O(n)	O(1) (output array excluded)

Time: Two sequential passes over the array, each visiting every element exactly once. Total: O(n).

Space: The output array itself is O(n), but by convention (and as specified in the problem) it doesn’t count toward the space complexity. The only auxiliary space used is two integer variables (left_product and right_product).

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Common Interview Mistakes

1. Proposing the division approach without acknowledging the constraint. Some candidates jump straight to “compute total product, divide each time,” which is wrong here. Reading constraints carefully and ruling out forbidden approaches explicitly is a strong signal.

2. Forgetting to initialize left_product and right_product to 1. It sounds trivial, but initializing the running product to 0 is a subtle bug that produces all-zero output. Always think about the identity element for your operation: for multiplication, it’s 1.

3. Off-by-one in the right pass. The right pass must start from n - 1 and go down to 0. Starting from n or stopping at 1 will silently skip an index and produce wrong output.

4. Using two full prefix/suffix arrays instead of optimizing to O(1). The two-array version is correct and acceptable in many interviews, but if an interviewer asks “can you do it with O(1) extra space?”, you should be ready with the running-variable optimization. Practicing both versions pays off.

5. Not tracing through the example by hand before declaring it done. This problem has a subtle two-pass structure that’s easy to convince yourself is correct when it isn’t. Always do a quick dry run on [1, 2, 3, 4] before moving on.

6. Coding before explaining the two-pass insight. The key insight here is genuinely elegant. Interviewers want to hear you articulate it, “the answer at each index is just a left product times a right product,” before you touch the keyboard.

7. Forgetting the edge case where the array has exactly two elements. nums = [3, 4] should return [4, 3]. It works with the algorithm above, but it’s worth verifying mentally during the interview to show thoroughness.

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What a Strong Candidate Sounds Like

“My first thought is that for each index, the answer is just the product of everything to its left times the product of everything to its right. So if I can precompute prefix products and suffix products efficiently, I can combine them in linear time. I’ll use the output array itself for the prefix pass, then make a second right-to-left pass with a single running variable to fold in the suffix products. That gives me O(n) time and O(1) extra space, and no division anywhere. Let me just trace through the example quickly to verify before I code it up.”

This response shows: the insight was reasoned, not memorized; the candidate thought about space; they proactively verified their approach.

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Example Interview Dialogue

Interviewer: What if the array contains a zero?

Candidate: Good edge case. If there’s exactly one zero, then every position except that zero’s index gets a product of zero, because every product includes that zero. The one position that had the zero gets the product of all the other elements. My algorithm handles this correctly because the zero just propagates naturally through the multiplication passes.

Interviewer: What about two zeros?

Candidate: Then every position’s product is zero, because no matter which element you exclude, the remaining elements still include at least one zero. Again, the algorithm handles it without any special casing.

Interviewer: Can you do it with O(1) extra space?

Candidate: Yes, that’s exactly what the running variable optimization does. Instead of storing a full suffix array, I maintain a single right_product variable and update it as I traverse right to left. The output array itself is O(n), but we’re not counting that toward auxiliary space, so we’re at O(1) beyond the output.

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Related Problems to Practice

Product of Array Except Self is the entry point to the prefix/suffix aggregation family. These problems all use the same “left pass, right pass” decomposition:

• Trapping Rain Water (LeetCode #42). Uses the same left-pass / right-pass prefix pattern to compute water levels.
• Maximum Product Subarray (LeetCode #152). Extends product reasoning to subarrays with sign-change tracking.
• Find Pivot Index (LeetCode #724). Prefix sums version of the same “left aggregate vs. right aggregate” pattern.
• Product of the Last K Numbers (LeetCode #1352). Designs a data structure around running products, a direct extension.
• Subarray Sum Equals K (LeetCode #560). Prefix sums cousin that builds fluency in the broader prefix-aggregate family.

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Practice This in a Mock Interview

Reading through this explanation makes the two-pass prefix/suffix approach feel elegant and obvious. But this problem has a consistent failure mode in interviews: candidates who know the technique still initialize a variable to 0 instead of 1, traverse the right pass in the wrong direction, or stumble when the interviewer asks about the zero edge case.

The gap between understanding and performing is real. The only way to make the two-pass structure, the initialization, and the edge cases feel automatic is deliberate practice under realistic conditions: a timer, an interviewer asking questions, and immediate feedback on where you hesitated.

Start a mock interview for Product of Array Except Self on Intervu.

Further reading:

• How to Prepare for a Coding Interview in 2026, the complete roadmap and study plans
• Why LeetCode alone isn’t enough, and what to practice instead
• Practice any LeetCode problem as a live mock interview
• The Grind 75 Pathway, a structured study plan with AI practice links
• Browse all walkthroughs on GitHub, condensed solutions with code