Maximum Product Subarray LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Maximum Product Subarray in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemMaximum Product Subarray– LeetCode Problem

Maximum Product Subarray– LeetCode Problem

Problem:

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 Solutions
Maximum Product Subarray Solution in C++:
class Solution {
 public:
  int maxProduct(vector<int>& nums) {
    int ans = nums[0];
    int dpMin = nums[0];  // min so far
    int dpMax = nums[0];  // max so far

    for (int i = 1; i < nums.size(); ++i) {
      const int num = nums[i];
      const int prevMin = dpMin;  // dpMin[i - 1]
      const int prevMax = dpMax;  // dpMax[i - 1]
      if (num < 0) {
        dpMin = min(prevMax * num, num);
        dpMax = max(prevMin * num, num);
      } else {
        dpMin = min(prevMin * num, num);
        dpMax = max(prevMax * num, num);
      }
      ans = max(ans, dpMax);
    }

    return ans;
  }
};
Maximum Product Subarray Solution in Java:
class Solution {
 public:
  int maxProduct(vector<int>& nums) {
    int ans = nums[0];
    int dpMin = nums[0];  // min so far
    int dpMax = nums[0];  // max so far

    for (int i = 1; i < nums.size(); ++i) {
      const int num = nums[i];
      const int prevMin = dpMin;  // dpMin[i - 1]
      const int prevMax = dpMax;  // dpMax[i - 1]
      if (num < 0) {
        dpMin = min(prevMax * num, num);
        dpMax = max(prevMin * num, num);
      } else {
        dpMin = min(prevMin * num, num);
        dpMax = max(prevMax * num, num);
      }
      ans = max(ans, dpMax);
    }

    return ans;
  }
};
Maximum Product Subarray Solution in Python:
class Solution:
  def maxProduct(self, nums: List[int]) -> int:
    ans = nums[0]
    prevMin = nums[0]
    prevMax = nums[0]

    for i in range(1, len(nums)):
      mini = prevMin * nums[i]
      maxi = prevMax * nums[i]
      prevMin = min(nums[i], mini, maxi)
      prevMax = max(nums[i], mini, maxi)
      ans = max(ans, prevMax)

    return ans

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