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LeetCode 3509: Maximum Product of Subsequences With an Alternating Sum Equal to K Solution

Master LeetCode problem 3509 (Maximum Product of Subsequences With an Alternating Sum Equal to K), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3509. Maximum Product of Subsequences With an Alternating Sum Equal to K

Hard
ArrayHash TableDynamic Programming

Problem Explanation

Explanation:

To solve this problem, we can use a dynamic programming approach. We will iterate through all possible subsequences of the input array nums and keep track of the maximum product of each subsequence that satisfies the alternating sum condition and has a product within the limit.

  1. We will use two dynamic programming arrays, maxProd and minProd, to keep track of the maximum and minimum product ending at index i.
  2. For each element at index i, we will calculate the maximum and minimum product ending at index i considering two cases:
    • Include the current element.
    • Start a new subsequence at the current element.
  3. We will also maintain a deque to keep track of the elements in the current subsequence to efficiently calculate the alternating sum.
  4. Finally, we will return the maximum product found within the limit.

Time Complexity:

  • The time complexity of this solution is O(n^2) where n is the length of the input array.

Space Complexity:

  • The space complexity of this solution is O(n) where n is the length of the input array.

:

Solution Code

class Solution {
    public int maxProduct(int[] nums, int k, int limit) {
        int n = nums.length;
        long[] maxProd = new long[n];
        long[] minProd = new long[n];
        maxProd[0] = nums[0];
        minProd[0] = nums[0];
        long result = nums[0];
        
        for (int i = 1; i < n; i++) {
            maxProd[i] = Math.max(Math.max(maxProd[i - 1] * nums[i], minProd[i - 1] * nums[i]), nums[i]);
            minProd[i] = Math.min(Math.min(maxProd[i - 1] * nums[i], minProd[i - 1] * nums[i]), nums[i]);
            
            if (maxProd[i] <= limit && (i % 2 == 0 && maxProd[i] % 2 == k || i % 2 != 0 && maxProd[i] % 2 != k)) {
                result = Math.max(result, maxProd[i]);
            }
        }
        
        return result <= 0 ? -1 : (int) result;
    }
}

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