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Problem solution · C++

Maximum Score From Grid Operations

Maximum Score From Grid Operations: 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
46 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 Score From Grid Operations, 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

  • 46 lines of C++ from the credited upstream file 3225.cpp.
  • The implementation visibly relies on sequence storage.
  • 5 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 Score From Grid Operations · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  long long maximumScore(vector<vector<int>>& grid) {    const int n = grid.size();    // prefix[j][i] := the sum of the first i elements in the j-th column    vector<vector<long>> prefix(n, vector<long>(n + 1));    // prevPick[i] := the maximum achievable score up to the previous column,    // where the bottommost selected element in that column is in row (i - 1)    vector<long> prevPick(n + 1);    // prevSkip[i] := the maximum achievable score up to the previous column,    // where the bottommost selected element in the column before the previous    // one is in row (i - 1)    vector<long> prevSkip(n + 1);     for (int j = 0; j < n; ++j)      for (int i = 0; i < n; ++i)        prefix[j][i + 1] = prefix[j][i] + grid[i][j];     for (int j = 1; j < n; ++j) {      vector<long> currPick(n + 1);      vector<long> currSkip(n + 1);      // Consider all possible combinations of the number of current and      // previous selected elements.      for (int curr = 0; curr <= n; ++curr)        for (int prev = 0; prev <= n; ++prev)          if (curr > prev) {            // 1. The current bottom is deeper than the previous bottom.            // Get the score of grid[prev..curr)[j - 1] for pick and skip.            const long score = prefix[j - 1][curr] - prefix[j - 1][prev];            currPick[curr] = max(currPick[curr], prevSkip[prev] + score);            currSkip[curr] = max(currSkip[curr], prevSkip[prev] + score);          } else {            // 2. The previous bottom is deeper than the current bottom.            // Get the score of grid[curr..prev)[j] for pick only.            const long score = prefix[j][prev] - prefix[j][curr];            currPick[curr] = max(currPick[curr], prevPick[prev] + score);            currSkip[curr] = max(currSkip[curr], prevPick[prev]);          }      prevPick = std::move(currPick);      prevSkip = std::move(currSkip);    }     return ranges::max(prevPick);  }};

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