Maximum Product Subarray – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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.
A 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
numsis guaranteed to fit in a 32-bit integer.
C++ Maximum Product Subarray LeetCode Solution
int maxProduct(vector<int> A) {
int n = A.size(), res = A[0], 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;
}
Java Maximum Product Subarray LeetCode Solution
public int maxProduct(int[] A) {
int n = A.length, res = A[0], 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;
}
Python 3 Maximum Product Subarray LeetCode Solution
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)
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