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Minimum Number of Flips to Convert Binary Matrix to Zero Matrix

Given a m x n binary matrix mat. In one step, you can choose one cell and flip it and all the four neighbors of it if they exist (Flip is changing 1 to 0 and 0 to 1). A pair of cells are called neighbors if they share one edge.

Return the minimum number of steps required to convert mat to a zero matrix or -1 if you cannot.

A binary matrix is a matrix with all cells equal to 0 or 1 only.

A zero matrix is a matrix with all cells equal to 0.

 

Example 1:

Input: mat = [[0,0],[0,1]]
Output: 3
Explanation: One possible solution is to flip (1, 0) then (0, 1) and finally (1, 1) as shown.

Example 2:

Input: mat = [[0]]
Output: 0
Explanation: Given matrix is a zero matrix. We do not need to change it.

Example 3:

Input: mat = [[1,0,0],[1,0,0]]
Output: -1
Explanation: Given matrix cannot be a zero matrix.

 

Constraints:

  • m == mat.length
  • n == mat[i].length
  • 1 <= m, n <= 3
  • mat[i][j] is either 0 or 1.

Solution Explanation

This problem asks for the minimum number of steps to convert a binary matrix to a zero matrix. In each step, we can flip a cell and its four neighbors. The solution uses Breadth-First Search (BFS) to explore all possible states of the matrix.

Approach:

  1. State Representation: Each matrix state is represented as an integer. We use bit manipulation. Each bit in the integer corresponds to a cell in the matrix. A 1 indicates the cell is 1, and a 0 indicates the cell is 0. This allows efficient checking and modification of the matrix state.

  2. BFS Traversal: We start with the initial matrix state and perform a BFS. The queue stores matrix states, and the visited set tracks visited states to avoid cycles.

  3. Flipping Operation: For each state in the queue, we iterate through each cell. When a cell is selected for flipping, we flip the cell and its four neighbors (if they exist). This produces a new matrix state.

  4. Termination Condition: The BFS continues until we reach the zero matrix state (represented by the integer 0). The number of steps taken is the minimum number of flips. If the zero matrix is unreachable, we return -1.

Time Complexity Analysis:

The time complexity is difficult to express precisely. In the worst case, BFS explores all possible states of the matrix. The number of possible states is 2(m*n), where m and n are the dimensions of the matrix. However, due to the constraints (1 ≤ m, n ≤ 3), the search space is relatively small. In practice, the algorithm's performance is quite good even for larger matrices because many states are pruned due to the visited set. Therefore, while the theoretical worst-case complexity is exponential, the practical runtime is manageable for the given constraints.

Space Complexity Analysis:

The space complexity is dominated by the vis (visited) set, which stores visited states. In the worst case, it can store up to 2(m*n) states. Again, due to the constraints (1 ≤ m, n ≤ 3), the space used remains reasonably small. The queue also has a maximum size bounded by the number of possible states, but it's usually much smaller in practice because of early termination. So, the space complexity is also effectively exponential in the worst case but manageable for the problem's input limitations.

Code Explanation (Python):

The Python code directly implements the described approach. Key aspects are:

  • minFlips(mat): This function initiates the BFS.
  • state: The initial matrix state is calculated using bit manipulation.
  • q and vis: The queue and visited set for BFS.
  • dirs: An array defining the neighbor offsets (up, down, left, right).
  • Inner Loop: The inner loop iterates through all cells and flips cells and their neighbors.
  • ans: The number of steps taken.

The code in other languages (Java, C++, Go) follows the same logic and data structures, translating the core algorithm into their respective syntax. They all employ the same BFS approach with bit manipulation for state representation.