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Problem solution · C++

Maximum Strength of K Disjoint Subarrays

Maximum Strength of K Disjoint Subarrays: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
37 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

Direct simulation

For Maximum Strength of K Disjoint Subarrays, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 37 lines of C++ from the credited upstream file 3077.cpp.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeMaximum Strength of K Disjoint Subarrays · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  long long maximumStrength(vector<int>& nums, int k) {    vector<vector<vector<long>>> mem(        nums.size(), vector<vector<long>>(k + 1, vector<long>(2, -1)));    return maximumStrength(nums, 0, k, /*fresh=*/true, mem);  }  private:  static constexpr long kMin = LONG_MIN / 2;   // Returns the maximum strength of nums[i..n) with k operations left, where  // `fresh` means we're starting a new subarray.  long maximumStrength(const vector<int>& nums, int i, int k, bool fresh,                       vector<vector<vector<long>>>& mem) {    if (nums.size() - i < k)      return kMin;    if (k == 0)      return 0;    if (i == nums.size())      return k == 0 ? 0 : kMin;    if (mem[i][k][fresh] != -1)      return mem[i][k][fresh];    // If it's not fresh, we can't skip the current number and consider it as a    // fresh start, since the case where it's fresh is already covered by    // `includeAndFreshStart`.    const long skip = fresh ? maximumStrength(nums, i + 1, k, true, mem) : kMin;    const long gain = (k % 2 == 0 ? -1 : 1) * static_cast<long>(nums[i]) * k;    const long includeAndContinue =        maximumStrength(nums, i + 1, k, false, mem) + gain;    const long includeAndFreshStart =        maximumStrength(nums, i + 1, k - 1, true, mem) + gain;    return mem[i][k][fresh] =               max(skip, max(includeAndContinue, includeAndFreshStart));  }};

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