Watch 3 video solutions for Maximum Product of Subsequences With an Alternating Sum Equal to K, a hard level problem involving Array, Hash Table, Dynamic Programming. This walkthrough by Tech Courses has 548 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You are given an integer array nums and two integers, k and limit. Your task is to find a non-empty subsequence of nums that:
k.limit.Return the product of the numbers in such a subsequence. If no subsequence satisfies the requirements, return -1.
The alternating sum of a 0-indexed array is defined as the sum of the elements at even indices minus the sum of the elements at odd indices.
Example 1:
Input: nums = [1,2,3], k = 2, limit = 10
Output: 6
Explanation:
The subsequences with an alternating sum of 2 are:
[1, 2, 3]
1 - 2 + 3 = 21 * 2 * 3 = 6[2]
The maximum product within the limit is 6.
Example 2:
Input: nums = [0,2,3], k = -5, limit = 12
Output: -1
Explanation:
A subsequence with an alternating sum of exactly -5 does not exist.
Example 3:
Input: nums = [2,2,3,3], k = 0, limit = 9
Output: 9
Explanation:
The subsequences with an alternating sum of 0 are:
[2, 2]
2 - 2 = 02 * 2 = 4[3, 3]
3 - 3 = 03 * 3 = 9[2, 2, 3, 3]
2 - 2 + 3 - 3 = 02 * 2 * 3 * 3 = 36The subsequence [2, 2, 3, 3] has the greatest product with an alternating sum equal to k, but 36 > 9. The next greatest product is 9, which is within the limit.
Constraints:
1 <= nums.length <= 1500 <= nums[i] <= 12-105 <= k <= 1051 <= limit <= 5000Problem Overview: You are given an array and a target K. Choose a subsequence whose alternating sum (add the first element, subtract the second, add the third, and so on) equals K. Among all valid subsequences, return the maximum possible product of its elements.
Approach 1: Brute Force Subsequence Enumeration (O(2^n) time, O(n) space)
The direct method generates every subsequence and computes its alternating sum and product. For each subset, iterate through elements in order, flipping the sign based on index parity (+ - + -). If the final alternating sum equals K, update the maximum product. This approach uses recursion or bitmask enumeration and requires storing the current product and alternating sum during traversal. With 2^n possible subsequences, the method becomes impractical once n grows beyond small limits.
Approach 2: Dynamic Programming with Hash Map States (O(n * S) time, O(S) space)
A more practical solution tracks reachable alternating sums while scanning the array once. Maintain DP states keyed by (alternating_sum, parity), where parity indicates whether the next element contributes positively or negatively. Each state stores the maximum product achievable for that configuration. For every number, iterate through existing states and update two possibilities: include the element (update sum based on parity and multiply the product) or skip it. A hash map efficiently stores only reachable sums, which avoids scanning large unused ranges. This technique combines ideas from dynamic programming and hash tables while iterating through the array.
The key insight is that the alternating sign depends only on how many elements were already chosen. Tracking parity makes the alternating calculation constant time, and storing the best product per state avoids recomputing dominated results.
Approach 3: Pruned DP with State Compression (O(n * S) time, O(S) space)
The DP map can grow if many sums are reachable. You can prune dominated states by keeping only the maximum product for each (sum, parity). Any state producing a smaller product for the same configuration can be discarded because future multiplications will never outperform the larger one. This keeps the state space compact and improves practical performance while maintaining the same asymptotic complexity.
Recommended for interviews: Interviewers expect the dynamic programming approach with a hash map. Starting with the brute force method shows you understand the subsequence search space. Transitioning to DP demonstrates the key optimization: storing intermediate alternating sums and reusing them instead of recomputing all subsequences.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Subsequence Enumeration | O(2^n) | O(n) | Small arrays or when demonstrating the baseline idea during interviews |
| Dynamic Programming with Hash Map | O(n * S) | O(S) | General case where many subsequences exist but alternating sums can be reused |
| Pruned DP with State Compression | O(n * S) | O(S) | Large inputs where limiting dominated states improves practical performance |