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# Maximum Product Subarray LeetCode Solution

## Problem – Maximum Product Subarray

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

The test cases are generated so that the answer will fit in a 32-bit integer.

subarray is a contiguous subsequence of the array.

Example 1:

``````Input: nums = [2,3,-2,4]
Output: 6
Explanation: [2,3] has the largest product 6.``````

Example 2:

``````Input: nums = [-2,0,-1]
Output: 0
Explanation: The result cannot be 2, because [-2,-1] is not a subarray.``````

Constraints:

• `1 <= nums.length <= 2 * 104`
• `-10 <= nums[i] <= 10`
• The product of any prefix or suffix of `nums` is guaranteed to fit in a 32-bit integer.

### Maximum Product Subarray LeetCode Solution in Python

``````    def maxProduct(self, A):
B = A[::-1]
for i in range(1, len(A)):
A[i] *= A[i - 1] or 1
B[i] *= B[i - 1] or 1
return max(A + B)``````

### Maximum Product Subarray LeetCode Solution in C++

``````    int maxProduct(vector<int> A) {
int n = A.size(), res = A, l = 0, r = 0;
for (int i = 0; i < n; i++) {
l =  (l ? l : 1) * A[i];
r =  (r ? r : 1) * A[n - 1 - i];
res = max(res, max(l, r));
}
return res;
}``````

### Maximum Product Subarray LeetCode Solution in Java

``````    public int maxProduct(int[] A) {
int n = A.length, res = A, l = 0, r = 0;
for (int i = 0; i < n; i++) {
l =  (l == 0 ? 1 : l) * A[i];
r =  (r == 0 ? 1 : r) * A[n - 1 - i];
res = Math.max(res, Math.max(l, r));
}
return res;
}``````
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