What is the time complexity of Maximum Product of Subsequences With an Alternating Sum Equal to K?

The optimal solution for Maximum Product of Subsequences With an Alternating Sum Equal to K runs in O(n * m) time and uses O(m) space. The time complexity is O(n * m) where n is the number of elements in nums and m is the range of possible alternating sums. The space complexity is O(m) for the DP array.

What is the brute force solution for Maximum Product of Subsequences With an Alternating Sum Equal to K?

The brute force approach for Maximum Product of Subsequences With an Alternating Sum Equal to K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible non-empty subsequences of the array and checking each one for the required alternating sum and product conditions. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Maximum Product of Subsequences With an Alternating Sum Equal to K?

Maximum Product of Subsequences With an Alternating Sum Equal to K uses the following concepts: Array, Hash Table, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Product of Subsequences With an Alternating Sum Equal to K?

Explain your thought process clearly while coding. Discuss the trade-offs of different approaches. Be prepared to optimize your solution on the fly.

What are common mistakes when solving Maximum Product of Subsequences With an Alternating Sum Equal to K?

Failing to account for negative sums in the DP approach. Not initializing the DP array correctly.

#3509

Maximum Product of Subsequences With an Alternating Sum Equal to K

Hard

ArrayHash TableDynamic ProgrammingHash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(m)

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Intuition

Time O(n * m)Space O(m)

The optimal solution uses dynamic programming to efficiently track the maximum product for each possible alternating sum. This reduces the need to check all subsequences explicitly and allows us to build solutions incrementally.

⚙️

Algorithm

5 steps

1Step 1: Initialize a DP array where dp[sum] holds the maximum product for that alternating sum.
2Step 2: Iterate through each number in nums, updating the DP array based on the current number's contribution to possible sums.
3Step 3: For each number, update the DP array in reverse to avoid overwriting results from the same iteration.
4Step 4: After processing all numbers, check dp[k] for the maximum product that does not exceed the limit.
5Step 5: Return the result or -1 if no valid product was found.

solution.py17 lines

1# Full working Python code
2
3def maxProduct(nums, k, limit):
4    dp = {0: 1}
5    for num in nums:
6        new_dp = dp.copy()
7        for s in list(dp.keys()):
8            new_sum = s + num
9            new_product = dp[s] * num
10            if new_product <= limit:
11                new_dp[new_sum] = max(new_dp.get(new_sum, 0), new_product)
12            new_sum = s - num
13            new_product = dp[s] * num
14            if new_product <= limit:
15                new_dp[new_sum] = max(new_dp.get(new_sum, 0), new_product)
16        dp = new_dp
17    return dp.get(k, -1)

ℹ

Complexity note: The time complexity is O(n * m) where n is the number of elements in nums and m is the range of possible alternating sums. The space complexity is O(m) for the DP array.

1Dynamic programming allows us to build solutions incrementally.
2Tracking products alongside sums helps manage constraints effectively.

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