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Problem solution · C++

Disconnect Path in a Binary Matrix by at Most One Flip

Disconnect Path in a Binary Matrix by at Most One Flip: 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
27 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 Disconnect Path in a Binary Matrix by at Most One Flip, 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

  • 27 lines of C++ from the credited upstream file 2556.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 codeDisconnect Path in a Binary Matrix by at Most One Flip · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  bool isPossibleToCutPath(vector<vector<int>>& grid) {    if (!hasPath(grid, 0, 0))      return true;    // Reassign (0, 0) as 1.    grid[0][0] = 1;    return !hasPath(grid, 0, 0);  }  private:  // Returns true is there's a path from (0, 0) to (m - 1, n - 1).  // Also marks the visited path as 0 except (m - 1, n - 1).  bool hasPath(vector<vector<int>>& grid, int i, int j) {    if (i == grid.size() || j == grid[0].size())      return false;    if (i == grid.size() - 1 && j == grid[0].size() - 1)      return true;    if (grid[i][j] == 0)      return false;     grid[i][j] = 0;    // Go down first. Since we use OR logic, we'll only mark one path.    return hasPath(grid, i + 1, j) || hasPath(grid, i, j + 1);  }};

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