Formatted question description: https://leetcode.ca/all/1284.html

1284. Minimum Number of Flips to Convert Binary Matrix to Zero Matrix (Hard)

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

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

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

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 don't need to change it.

Example 3:

Input: mat = [[1,1,1],[1,0,1],[0,0,0]]
Output: 6

Example 4:

Input: mat = [[1,0,0],[1,0,0]]
Output: -1
Explanation: Given matrix can't be a zero matrix

 

Constraints:

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

Related Topics:
Breadth-first Search

Solution 1. Bit vector + BFS

// OJ: https://leetcode.com/problems/minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix/
// Time: O(MN * 2^(MN))
// Space: O(2^(MN))
class Solution {
public:
    int minFlips(vector<vector<int>>& A) {
        int start = 0, M = A.size(), N = A[0].size(), step = 0, dirs[5] = {1,0,-1,0,1};
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                start |= (A[i][j] << (i * 3 + j));
            }
        }
        queue<int> q;
        unordered_set<int> s;
        q.push(start);
        s.insert(start);
        while (q.size()) {
            int cnt = q.size();
            while (cnt--) {
                int state = q.front();
                q.pop();
                if (state == 0) return step;
                for (int i = 0; i < 9; ++i) {
                    int next = state, r = i / 3, c = i % 3;
                    next ^= (1 << (r * 3 + c));
                    for (int j = 0; j < 4; ++j) {
                        int x = r + dirs[j], y = c + dirs[j + 1];
                        if (x < 0 || x >= M || y < 0 || y >= N) continue;
                        next ^= (1 << (x * 3 + y));
                    }
                    if (s.count(next) == 0) {
                        q.push(next);
                        s.insert(next);
                    }
                }
            }
            ++step;
        }
        return -1;
    }
};

Java

  • class Solution {
        public int minFlips(int[][] mat) {
            int rows = mat.length, columns = mat[0].length;
            int[][] zeroMatrix = new int[rows][columns];
            String zeroMatrixStr = matrixToString(zeroMatrix);
            final int WHITE = 0;
            final int GRAY = 1;
            final int BLACK = 2;
            Map<String, Integer> colorMap = new HashMap<String, Integer>();
            Queue<int[][]> queue = new LinkedList<int[][]>();
            queue.offer(mat);
            Queue<Integer> flipsQueue = new LinkedList<Integer>();
            flipsQueue.offer(0);
            String matStr = matrixToString(mat);
            colorMap.put(matStr, GRAY);
            while (!queue.isEmpty()) {
                int[][] curMatrix = queue.poll();
                int flip = flipsQueue.poll();
                String curStr = matrixToString(curMatrix);
                if (zeroMatrixStr.equals(curStr))
                    return flip;
                for (int i = 0; i < rows; i++) {
                    for (int j = 0; j < columns; j++) {
                        int[][] flipMatrix = flip(curMatrix, i, j);
                        String flipStr = matrixToString(flipMatrix);
                        int color = colorMap.getOrDefault(flipStr, WHITE);
                        if (color == WHITE) {
                            queue.offer(flipMatrix);
                            flipsQueue.offer(flip + 1);
                            colorMap.put(flipStr, GRAY);
                        }
                    }
                }
                colorMap.put(curStr, BLACK);
            }
            return -1;
        }
    
        public String matrixToString(int[][] mat) {
    		String str = "[";
    		int rows = mat.length;
    		for (int i = 0; i < rows; i++) {
    			int[] row = mat[i];
    			String rowStr = Arrays.toString(row);
    			if (i > 0)
    				str += ", ";
    			str += rowStr;
    		}
    		str += "]";
    		return str;
    	}
    
        public int[][] flip(int[][] mat, int row, int column) {
            int rows = mat.length, columns = mat[0].length;
            int[][] flipMat = new int[rows][columns];
            for (int i = 0; i < rows; i++) {
                for (int j = 0; j < columns; j++)
                    flipMat[i][j] = mat[i][j];
            }
            flipMat[row][column] = 1 - flipMat[row][column];
            if (row > 0)
                flipMat[row - 1][column] = 1 - flipMat[row - 1][column];
            if (row < rows - 1)
                flipMat[row + 1][column] = 1 - flipMat[row + 1][column];
            if (column > 0)
                flipMat[row][column - 1] = 1 - flipMat[row][column - 1];
            if (column < columns - 1)
                flipMat[row][column + 1] = 1 - flipMat[row][column + 1];
            return flipMat;
        }
    }
    
  • // OJ: https://leetcode.com/problems/minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix/
    // Time: O(MN * 2^(MN))
    // Space: O(2^(MN))
    class Solution {
    public:
        int minFlips(vector<vector<int>>& A) {
            int M = A.size(), N = A[0].size(), init = 0, step = 0, dirs[4][2] = { {0,1},{0,-1},{1,0},{-1,0} };
            for (int i = 0; i < M; ++i) {
                for (int j = 0; j < N; ++j) init |= A[i][j] << (i * N + j);
            }
            queue<int> q{ {init} };
            unordered_set<int> seen{init};
            auto flip = [&](int &state, int x, int y) { state ^= 1 << (x * N + y); };
            while (q.size()) {
                int cnt = q.size();
                while (cnt--) {
                    int u = q.front();
                    q.pop();
                    if (u == 0) return step;
                    for (int i = 0; i < M; ++i) {
                        for (int j = 0; j < N; ++j) {
                            int v = u;
                            flip(v, i, j);
                            for (auto &[dx, dy] : dirs) {
                                int a = i + dx, b = j + dy;
                                if (a < 0 || b < 0 || a >= M || b >= N) continue;
                                flip(v, a, b);
                            }
                            if (seen.count(v)) continue;
                            seen.insert(v);
                            q.push(v);
                        }
                    }
                }
                ++step;
            }
            return -1;
        }
    };
    
  • print("Todo!")
    

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