Introduction

The Constrained Subsequence Sum problem involves finding the maximum subsequence sum with a distance constraint.

Given:

  • an integer array
  • integer k

the task is to:

  • choose subsequence elements
  • adjacent selected elements must be at most k distance apart
  • maximize total sum

This problem is one of the most important applications of:

Dynamic Programming + Monotonic Deque

This problem helps in understanding:

  • dynamic programming optimization
  • monotonic queues
  • maximum range queries
  • constrained subsequences

Example

Input:arr = [10,2,-10,5,20]
k = 2

Output:
37

Explanation:
Choose:
10 + 2 + 5 + 20

Maximum subsequence sum:
37

Constraints

1 <= arr.length <= 10^5-10^4 <= arr[i] <= 10^4
1 <= k <= arr.length

Approach 1 : Brute Force Dynamic Programming

Explanation

The simplest way is:

  1. For every index:
    • check previous k positions
  2. Find best subsequence sum
  3. Build DP array

This works but repeated scanning increases complexity.

Steps

  1. Create DP array.
  2. Initialize current element.
  3. Check previous k positions.
  4. Update best subsequence sum.
  5. Return maximum value.

Dry Run

Input:[10,2,-10,5,20]
k = 2

DP:
dp[0] = 10
dp[1] = 2 + 10 = 12
dp[2] = -10 + 12 = 2
dp[3] = 5 + 12 = 17
dp[4] = 20 + 17 = 37

Final Result:
37

Brute Force DP Code


Complexity Analysis

Time Complexity: O(n * k) Explanation: Previous k positions are checked repeatedly. 
Space Complexity: O(n)Explanation: DP array is used.

Approach 2 : Monotonic Deque Optimization

Explanation

The optimized solution uses:

 DP + Monotonic Decreasing Deque

Idea:

  • dp[i] stores best subsequence sum ending at index i
  • previous maximum DP value within distance k is required

Deque stores:

  • indices of useful DP values
    in decreasing order

Rules:

  1. Remove expired indices
  2. Use front as maximum DP
  3. Remove smaller DP values

This gives:

  • O(n) solution

Steps

  1. Create DP array.
  2. Create deque.
  3. Remove expired indices.
  4. Use deque front as best previous sum.
  5. Remove smaller DP values.
  6. Insert current index.
  7. Return maximum DP value.

Dry Run

Input:[10,2,-10,5,20]k = 2dp[0] = 10
Deque:
[10]
dp[1] = 2 + 10 = 12

Deque:
[12]
dp[2] = -10 + 12 = 2

Deque:
[12,2]
dp[3] = 5 + 12 = 17

Deque:
[17]
dp[4] = 20 + 17 = 37

Final Result:
37

Monotonic Deque Code


Complexity Analysis

Time Complexity: O(n)Explanation: Each index enters and leaves deque at most once.
Space Complexity: O(n) Explanation: DP array is used.

Edge Cases

  1. All negative numbers
  2. Single element array
  3. k = 1
  4. k = array length
  5. Large positive values

Why This Problem is Important

Constrained Subsequence Sum helps in understanding:

  1. Dynamic programming optimization
  2. Monotonic deque
  3. Maximum range queries
  4. Efficient DP transitions
  5. Sliding window maximum concepts

It is one of the most important advanced deque and DP interview problems.

Real-World Applications

Monotonic deque and DP concepts are used in:

  1. Financial optimization
  2. Resource scheduling
  3. Streaming analytics
  4. Recommendation systems
  5. Path optimization problems

Common Mistakes

  1. Forgetting expired indices
  2. Incorrect deque ordering
  3. Wrong DP transition
  4. Missing standalone element case

Interview Tips

Interviewers often expect:

  1. DP optimization explanation
  2. Monotonic deque reasoning
  3. O(n) transition optimization

Always explain:

  • why deque stores decreasing DP values
  • how front gives maximum transition
  • why expired indices are removed

Related Questions

  1. Sliding Window Maximum
  2. Shortest Subarray with Sum ≥ K
  3. First Negative Integer in Window
  4. Jump Game VI
  5. Daily Temperatures

Final Takeaway

The Constrained Subsequence Sum problem is a fundamental advanced deque and dynamic programming problem that teaches monotonic queue optimization and efficient DP transition techniques. Understanding this problem builds a strong foundation for advanced sliding window and DP interview problems.