CPMagical CP Library

Problem solution · C++

Minimum Operations to Write the Letter Y on a Grid

Minimum Operations to Write the Letter Y on a Grid: 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
28 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 Minimum Operations to Write the Letter Y on a Grid, 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

  • 28 lines of C++ from the credited upstream file 3071.cpp.
  • The implementation visibly relies on sequence storage.
  • 2 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 codeMinimum Operations to Write the Letter Y on a Grid · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int minimumOperationsToWriteY(vector<vector<int>>& grid) {    return min({getOperations(grid, 0, 1), getOperations(grid, 0, 2),                getOperations(grid, 1, 0), getOperations(grid, 1, 2),                getOperations(grid, 2, 0), getOperations(grid, 2, 1)});  }  private:  // Returns the number of operations to turn Y into a and non-Y into b.  int getOperations(const vector<vector<int>>& grid, int a, int b) {    const int n = grid.size();    const int mid = n / 2;    int operations = 0;    for (int i = 0; i < n; ++i)      for (int j = 0; j < n; ++j)        // For the 'Y' pattern, before the midpoint, check the diagonal and        // anti-diagonal. After the midpoint, check the middle column.        if ((i < mid && (i == j || i + j == n - 1)) || (i >= mid && j == mid)) {          if (grid[i][j] != a)            ++operations;        } else if (grid[i][j] != b) {          ++operations;        }    return operations;  };};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗