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Maximum Product Subarray

Published at05 Jan 2026
Created by
Array Manipulation Hard 13 views
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This problem asks for the biggest product you can make from one continuous part of the list. A product means you multiply the numbers together, not add them. The chosen numbers must stay next to each other, because you are looking for a subarray.

This puzzle gets interesting because negative numbers can completely change the result. A negative number can make the product smaller, but two negatives together can suddenly turn the product positive again. Zero can also break a streak, because multiplying by zero resets everything to zero.

For example, in [2,3,-2,4], the best answer is 6 from the subarray [2,3]. If you keep going and include -2, the product changes in a bad way. But in [-2,3,-4], the full subarray gives 24, because -2 × 3 × -4 = 24. In [-2,0,-1], the best answer is 0, since the zero beats the other possible products.

The answer is only the largest product value. You do not need to return the subarray itself, just the best number you can get.

Example Input & Output

Example 1
Input
nums = [2,3,-2,4]
Output
6
Explanation

Example with input: nums = [2,3,-2,4]

Example 2
Input
nums = [-2,3,-4]
Output
24
Explanation

Example with input: nums = [-2,3,-4]

Example 3
Input
nums = [-2,0,-1]
Output
0
Explanation

Example with input: nums = [-2,0,-1]

Algorithm Flow

Recommendation Algorithm Flow for Maximum Product Subarray

Solution Approach

Find the maximum product subarray in an array that may contain negative numbers. Track both max and min products ending at each position because a negative times a negative produces a positive. For each element, compute the new max and min as the maximum and minimum of the element itself, element * previous max, and element * previous min.

function maxProduct(nums) {
  var maxProd = nums[0], minProd = nums[0], result = nums[0];
  for (var i = 1; i < nums.length; i++) {
    var a = maxProd * nums[i], b = minProd * nums[i];
    maxProd = Math.max(nums[i], a, b);
    minProd = Math.min(nums[i], a, b);
    result = Math.max(result, maxProd);
  }
  return result;
}

The min product is tracked because a negative number can become the max product when multiplied by another negative. The result is the global maximum across all positions.

Time complexity is O(n), space complexity is O(1).

Best Answers

java
class Solution {
    public int max_product(int[] nums) {
        if (nums.length == 0) return 0;
        int res = nums[0], maxP = nums[0], minP = nums[0];
        for (int i = 1; i < nums.length; i++) {
            int x = nums[i];
            if (x < 0) {
                int tmp = maxP; maxP = minP; minP = tmp;
            }
            maxP = Math.max(x, maxP * x);
            minP = Math.min(x, minP * x);
            res = Math.max(res, maxP);
        }
        return res;
    }
}