How do you solve Constrained Subsequence Sum on LeetCode?

Constrained Subsequence Sum (LeetCode #1425) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Dynamic programming can optimize problems involving subsequences.

What is the brute force solution for Constrained Subsequence Sum?

The brute force approach for Constrained Subsequence Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences of the array and checking which ones meet the condition of having consecutive indices differing by at most k. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Constrained Subsequence Sum?

Explain your thought process clearly while coding. Use comments to clarify your logic, especially in complex sections. Practice problems with varying constraints to strengthen your understanding.

What are common mistakes when solving Constrained Subsequence Sum?

Not considering edge cases like all negative numbers. Failing to maintain the condition for valid subsequences.

#1425

Constrained Subsequence Sum

Hard

ArrayDynamic ProgrammingQueueSliding WindowHeap (Priority Queue)Monotonic QueueDynamic ProgrammingSliding WindowDeque

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to build up the maximum sums while ensuring the k-distance condition is satisfied. We maintain a sliding window of maximum sums to efficiently calculate the best possible sum at each index.

⚙️

Algorithm

4 steps

1Step 1: Initialize a dp array where dp[i] represents the maximum sum of a subsequence ending at index i.
2Step 2: Use a deque to maintain the maximum dp values within the last k indices.
3Step 3: For each index i, calculate dp[i] as nums[i] + max(dp[j]) for j in the valid range, and update the deque accordingly.
4Step 4: The result will be the maximum value in the dp array.

solution.py17 lines

1# Full working Python code
2from collections import deque
3
4def maxSubsequenceSum(nums, k):
5    n = len(nums)
6    dp = [0] * n
7    max_sum = float('-inf')
8    dq = deque()
9    for i in range(n):
10        dp[i] = nums[i] + (dq[0] if dq else 0)
11        max_sum = max(max_sum, dp[i])
12        while dq and dq[-1] < dp[i]:
13            dq.pop()
14        dq.append(dp[i])
15        if i >= k:
16            dq.popleft()
17    return max_sum

ℹ

Complexity note: The time complexity is linear because we process each element once, and the deque operations are amortized O(1). The space complexity is O(n) due to the dp array.

1Dynamic programming can optimize problems involving subsequences.
2Maintaining a sliding window of maximum values can significantly reduce time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.