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.
struct Solution;
impl Solution {
fn min_flips(mut mat: Vec<Vec<i32>>) -> i32 {
let n = mat.len();
let m = mat[0].len();
let mut res = std::usize::MAX;
Self::dfs(0, 0, &mut mat, &mut res, n, m);
if res == std::usize::MAX {
-1
} else {
res as i32
}
}
fn dfs(start: usize, k: usize, mat: &mut Vec<Vec<i32>>, min: &mut usize, n: usize, m: usize) {
if start == n * m {
if Self::ones(mat, n, m) == 0 {
*min = (*min).min(k);
}
} else {
let r = start / m;
let c = start % m;
Self::flip(r, c, mat, n, m);
Self::dfs(start + 1, k + 1, mat, min, n, m);
Self::flip(r, c, mat, n, m);
Self::dfs(start + 1, k, mat, min, n, m);
}
}
fn flip(i: usize, j: usize, mat: &mut Vec<Vec<i32>>, n: usize, m: usize) {
mat[i][j] = 1 - mat[i][j];
if i > 0 {
mat[i - 1][j] = 1 - mat[i - 1][j];
}
if j > 0 {
mat[i][j - 1] = 1 - mat[i][j - 1];
}
if i + 1 < n {
mat[i + 1][j] = 1 - mat[i + 1][j];
}
if j + 1 < m {
mat[i][j + 1] = 1 - mat[i][j + 1];
}
}
fn ones(mat: &[Vec<i32>], n: usize, m: usize) -> usize {
let mut res = 0;
for i in 0..n {
for j in 0..m {
if mat[i][j] == 1 {
res += 1;
}
}
}
res
}
}
#[test]
fn test() {
let mat = vec_vec_i32![[0, 0], [0, 1]];
let res = 3;
assert_eq!(Solution::min_flips(mat), res);
let mat = vec_vec_i32![[0]];
let res = 0;
assert_eq!(Solution::min_flips(mat), res);
let mat = vec_vec_i32![[1, 1, 1], [1, 0, 1], [0, 0, 0]];
let res = 6;
assert_eq!(Solution::min_flips(mat), res);
let mat = vec_vec_i32![[1, 0, 0], [1, 0, 0]];
let res = -1;
assert_eq!(Solution::min_flips(mat), res);
}