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1284. Minimum Number of Flips to Convert Binary Matrix to Zero Matrix

Description

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.

Solutions

Solution 1

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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

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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) {