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Problem solution · C++

Count of Sub-Multisets With Bounded Sum

Count of Sub-Multisets With Bounded 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
35 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 Count of Sub-Multisets With Bounded 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

  • 35 lines of C++ from the credited upstream file 2902.cpp.
  • The implementation visibly relies on sequence storage, hash lookup, cached states.
  • 5 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 codeCount of Sub-Multisets With Bounded Sum · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int countSubMultisets(vector<int>& nums, int l, int r) {    constexpr int kMod = 1'000'000'007;    // dp[i] := the number of submultisets of `nums` with sum i    vector<long> dp(r + 1);    dp[0] = 1;    unordered_map<int, int> count;     for (const int num : nums)      ++count[num];     const int zeros = count[0];    count.erase(0);     for (const auto& [num, freq] : count) {      // stride[i] := dp[i] + dp[i - num] + dp[i - 2 * num] + ...      vector<long> stride = dp;      for (int i = num; i <= r; ++i)        stride[i] += stride[i - num];      for (int i = r; i > 0; --i)        if (i >= num * (freq + 1))          // dp[i] + dp[i - num] + dp[i - freq * num]          dp[i] = (stride[i] - stride[i - num * (freq + 1)]) % kMod;        else          dp[i] = stride[i] % kMod;    }     long ans = 0;    for (int i = l; i <= r; ++i)      ans = (ans + dp[i]) % kMod;    return ((zeros + 1) * ans) % kMod;  }};

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