## Intuition The problem requires transforming the initial state of a grid of strings into a final state using a specific operation. The key insight is that since the number of '0's and '1's are the same in both states, we can always rearrange the chips to achieve the desired configuration. The operation allows us to move a chip from one cell to an adjacent cell, either in the same row or column, which is a flexible enough operation to achieve the transformation. ## Approach 1. **Collect Chips**: First, we will collect all '0's and '1's from the initial grid into a central location, such as the top-left corner of the grid. This is done by iterating over each cell and moving chips to the top-left corner using the allowed operations. 2. **Distribute Chips**: Once all chips are collected, we will redistribute them to match the final grid configuration. This involves moving chips from the central location to their respective target positions in the final grid. 3. **Operations**: Each move involves taking the last character from a string in one cell and appending it to another cell. We ensure that each move is valid by checking adjacency and non-emptiness of the source cell. 4. **Efficiency**: The number of operations is bounded by \(4 \times s\), where \(s\) is the total number of chips. This is because each chip can be moved at most twice to collect and twice to distribute. ## Complexity - **Time Complexity**: \(O(n \times m \times \text{max\_len})\), where \(\text{max\_len}\) is the maximum length of a string in any cell. This is because we potentially move each chip twice across the grid. - **Space Complexity**: \(O(1)\), aside from input storage, since we are only maintaining a list of operations. ## Solution (C++) ```cpp #include using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n, m; cin >> n >> m; vector> initial(n, vector(m)); vector> final(n, vector(m)); for (int i = 0; i < n; ++i) for (int j = 0; j < m; ++j) cin >> initial[i][j]; for (int i = 0; i < n; ++i) for (int j = 0; j < m; ++j) cin >> final[i][j]; vector> operations; // Collect all chips to (0, 0) for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { while (!initial[i][j].empty()) { char chip = initial[i][j].back(); initial[i][j].pop_back(); if (i > 0) { operations.emplace_back(i + 1, j + 1, i, j + 1); initial[i - 1][j].push_back(chip); } else if (j > 0) { operations.emplace_back(i + 1, j + 1, i + 1, j); initial[i][j - 1].push_back(chip); } } } } // Distribute chips from (0, 0) to final positions for (int i = n - 1; i >= 0; --i) { for (int j = m - 1; j >= 0; --j) { for (char chip : final[i][j]) { if (i > 0) { operations.emplace_back(i, j + 1, i + 1, j + 1); } else if (j > 0) { operations.emplace_back(i + 1, j, i + 1, j + 1); } } } } cout << operations.size() << "\n"; for (auto [x1, y1, x2, y2] : operations) { cout << x1 << " " << y1 << " " << x2 << " " << y2 << "\n"; } return 0; } ``` This solution reads the input, performs the collection and distribution of chips as described, and outputs the sequence of operations needed to transform the initial grid into the final grid. Each operation is printed in the required format.

Chips Puzzle

Problem Details