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.lengthn == mat[0].length1 <= m <= 31 <= n <= 3mat[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;
}
}