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Problem solution · C++

Constrained Subsequence Sum

Constrained Subsequence Sum: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
25 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Constrained Subsequence Sum, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 25 lines of C++ from the credited upstream file 1425.cpp.
  • The implementation visibly relies on sequence storage, work queue, cached states.
  • 2 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeConstrained Subsequence Sum · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int constrainedSubsetSum(vector<int>& nums, int k) {    // dp[i] := the maximum the sum of non-empty subsequences in nums[0..i]    vector<int> dp(nums.size());    // dq stores dp[i - k], dp[i - k + 1], ..., dp[i - 1] whose values are > 0    // in decreasing order.    deque<int> dq;     for (int i = 0; i < nums.size(); ++i) {      if (dq.empty())        dp[i] = nums[i];      else        dp[i] = max(dq.front(), 0) + nums[i];      while (!dq.empty() && dq.back() < dp[i])        dq.pop_back();      dq.push_back(dp[i]);      if (i >= k && dp[i - k] == dq.front())        dq.pop_front();    }     return ranges::max(dp);  }};

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