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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;
    }
};
  • 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;
        }
    }
    
    ############
    
    class Solution {
        public int minFlips(int[][] mat) {
            int m = mat.length, n = mat[0].length;
            int state = 0;
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (mat[i][j] == 1) {
                        state |= 1 << (i * n + j);
                    }
                }
            }
            Deque<Integer> q = new ArrayDeque<>();
            q.offer(state);
            Set<Integer> vis = new HashSet<>();
            vis.add(state);
            int ans = 0;
            int[] dirs = {0, -1, 0, 1, 0, 0};
            while (!q.isEmpty()) {
                for (int t = q.size(); t > 0; --t) {
                    state = q.poll();
                    if (state == 0) {
                        return ans;
                    }
                    for (int i = 0; i < m; ++i) {
                        for (int j = 0; j < n; ++j) {
                            int nxt = state;
                            for (int k = 0; k < 5; ++k) {
                                int x = i + dirs[k], y = j + dirs[k + 1];
                                if (x < 0 || x >= m || y < 0 || y >= n) {
                                    continue;
                                }
                                if ((nxt & (1 << (x * n + y))) != 0) {
                                    nxt -= 1 << (x * n + y);
                                } else {
                                    nxt |= 1 << (x * n + y);
                                }
                            }
                            if (!vis.contains(nxt)) {
                                vis.add(nxt);
                                q.offer(nxt);
                            }
                        }
                    }
                }
                ++ans;
            }
            return -1;
        }
    }
    
  • // 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;
        }
    };
    
  • class Solution:
        def minFlips(self, mat: List[List[int]]) -> int:
            m, n = len(mat), len(mat[0])
            state = sum(1 << (i * n + j) for i in range(m) for j in range(n) if mat[i][j])
            q = deque([state])
            vis = {state}
            ans = 0
            dirs = [0, -1, 0, 1, 0, 0]
            while q:
                for _ in range(len(q)):
                    state = q.popleft()
                    if state == 0:
                        return ans
                    for i in range(m):
                        for j in range(n):
                            nxt = state
                            for k in range(5):
                                x, y = i + dirs[k], j + dirs[k + 1]
                                if not 0 <= x < m or not 0 <= y < n:
                                    continue
                                if nxt & (1 << (x * n + y)):
                                    nxt -= 1 << (x * n + y)
                                else:
                                    nxt |= 1 << (x * n + y)
                            if nxt not in vis:
                                vis.add(nxt)
                                q.append(nxt)
                ans += 1
            return -1
    
    
    
  • func minFlips(mat [][]int) int {
    	m, n := len(mat), len(mat[0])
    	state := 0
    	for i, row := range mat {
    		for j, v := range row {
    			if v == 1 {
    				state |= 1 << (i*n + j)
    			}
    		}
    	}
    	q := []int{state}
    	vis := map[int]bool{state: true}
    	ans := 0
    	dirs := []int{0, -1, 0, 1, 0, 0}
    	for len(q) > 0 {
    		for t := len(q); t > 0; t-- {
    			state = q[0]
    			if state == 0 {
    				return ans
    			}
    			q = q[1:]
    			for i := 0; i < m; i++ {
    				for j := 0; j < n; j++ {
    					nxt := state
    					for k := 0; k < 5; k++ {
    						x, y := i+dirs[k], j+dirs[k+1]
    						if x < 0 || x >= m || y < 0 || y >= n {
    							continue
    						}
    						if (nxt & (1 << (x*n + y))) != 0 {
    							nxt -= 1 << (x*n + y)
    						} else {
    							nxt |= 1 << (x*n + y)
    						}
    					}
    					if !vis[nxt] {
    						vis[nxt] = true
    						q = append(q, nxt)
    					}
    				}
    			}
    		}
    		ans++
    	}
    	return -1
    }
    

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