Maximum Product Subarray¶

Question¶

Given an integer array nums, find a contiguous non-empty subarray that has the largest product, and return that product.

Solution¶

Pattern¶

Prefix and Suffix Accumulation Compute the running product from left-to-right (prefix) and right-to-left (suffix). Reset the running product to \(1\) whenever a \(0\) is encountered. The global maximum will always be found during one of these two traversals.

How to Identify¶

The problem asks for the maximum/minimum of a contiguous subarray.
The operation is multiplication, which behaves non-monotonically (a negative times a negative becomes positive).
Sliding Window fails because shrinking the window doesn't predictably increase or decrease the product due to negative numbers and zeroes.

Description¶

Step-by-step explanation:

Validate the input (handle null/empty arrays).
Initialize prefix and suffix trackers to \(1\).
Initialize maxProduct to the first element of the array. (Using double for these trackers is highly recommended in languages like Java/C++ to prevent silent integer overflow during intermediate calculations).
Iterate through the array. In the same loop, update the prefix using nums[i] and the suffix using nums[n - 1 - i].
If either prefix or suffix becomes \(0\) (because we multiplied by a \(0\) in the array), we will evaluate that \(0\) against maxProduct, and then immediately reset the tracker back to \(1\) to evaluate the next independent segment of the array.
Continuously update maxProduct with the maximum of itself, the current prefix, and the current suffix.
Cast maxProduct back to an int and return it.

The Intuition¶

This solution relies on the mathematical properties of parity (even/odd) regarding negative numbers. Imagine an array with no zeroes: [a, b, c, d]. - Case 1: Even number of negatives. The product of the entire array is positive. Therefore, the maximum product is the entire array. Both the prefix and suffix passes will eventually calculate this total product. - Case 2: Odd number of negatives. The product of the entire array is negative. To maximize the product, we must "remove" exactly one negative number. We can only do this by chopping off a prefix (up to the first negative) or a suffix (down to the last negative). - If we remove the last negative number, the maximal subarray is a prefix of the array. - If we remove the first negative number, the maximal subarray is a suffix of the array. Therefore, the optimal subarray must touch either the left boundary or the right boundary. By calculating all prefixes and all suffixes, we mathematically guarantee finding it. Zeroes simply act as "walls" that reset this logic for independent sub-segments.

Complexity¶

Label	Worst	Average
Time Complexity	\(O(N)\)	\(O(N)\)
Space Complexity	\(O(1)\)	\(O(1)\)

Time Complexity¶

We iterate through the array of size \(N\) exactly once. Inside the loop, we perform \(O(1)\) constant-time multiplications and comparisons. Total time is strictly \(O(N)\).

Space Complexity¶

We use exactly three primitive variables (prefix, suffix, maxProduct). No auxiliary arrays are allocated. Total space is \(O(1)\).

Code¶

class Solution {
    public int maxProduct(int[] nums) {
        if (nums == null || nums.length == 0) return 0;

        int n = nums.length;
        double prefix = 1;
        double suffix = 1;
        double maxProduct = nums[0]; 

        for (int i = 0; i < n; i++) {
            if (prefix == 0) prefix = 1;
            if (suffix == 0) suffix = 1;

            prefix *= nums[i];
            suffix *= nums[n - 1 - i];

            maxProduct = Math.max(maxProduct, Math.max(prefix, suffix));
        }

        return (int) maxProduct;
    }
}

Caveats¶

Intermediate Integer Overflow: Multiplying many numbers can quickly exceed the 32-bit limit of int. Subarray products grow much faster than sums. In Java, long or double might be needed for intermediate calculations, though the result usually fits in int based on standard competitive constraints.
Sliding Window Anti-Pattern: Never use sliding window for subarray products with negative numbers. The loss of monotonicity means the left pointer doesn't mathematically know whether to shrink or stay to maximize the value.
Single Element 0: If the array is [-2, 0, -1], the answer is 0. The initialization to MIN_VALUE and the reset logic handle this.

Concepts to Think About¶

Kadane's Algorithm Adaptation (DP): Another \(O(1)\) space solution exists where you track both currentMax and currentMin at every step. When you encounter a negative number, you swap currentMax and currentMin. This is the canonical DP approach, but the Prefix/Suffix approach is generally considered easier to intuit mathematically.
Sub-segment Independence: Recognizing that a \(0\) acts as a reset wall is a foundational concept for many array algorithms. It allows you to break a complex global problem down into independent local problems.
Logarithmic Transformation: Could you turn this into a Max Subarray Sum problem using log(abs(x))? (Think about signs!).
Sign Tracking: The parity of negative numbers determines if the whole segment is positive or negative.
Zero as a Reset: Why does a zero break the "chain"?
Handling Large Numbers: Does double provide enough precision for integer products? (In most interview cases, yes, but worth mentioning).

Logical Follow-up¶

Question: Explain how you would solve this using Kadane's dynamic programming pattern instead of the prefix/suffix method. Is one better than the other? Solution: We maintain a currentMax and currentMin. At each step, if nums[i] is negative, multiplying it by our currentMax makes it very small, and multiplying it by our currentMin makes it very big. So, if nums[i] < 0, we swap currentMax and currentMin. Then, we update currentMax = max(nums[i], currentMax * nums[i]) and currentMin = min(nums[i], currentMin * nums[i]). We update our global max with currentMax. Both approaches take \(O(N)\) time and \(O(1)\) space. The Prefix/Suffix approach is slightly easier to mathematically prove during an interview, while Kadane's requires slightly less branching logic.

Question: What if the array contains fractional numbers between 0 and 1? Solution: The product logic stays the same, but the "maximum" might not involve more elements; multiplying by a fraction makes the product smaller. Kadane's variation handles this more robustly than prefix/suffix.

Question: What if you need to return the actual subarray, not just the product? Solution: Store the indices whenever maxProduct is updated during the prefix/suffix scan.