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Breadth-First Search with Bitmask Encoding: In this approach, we use a BFS to explore all possible states of the matrix, representing each state as a bitmask. Each bit in the bitmask represents a cell in the matrix, allowing us to efficiently keep track of visited states. Starting from the initial matrix configuration, we flip each cell and its neighbors, transforming the state and pushing it into our BFS queue. By marking visited states, we avoid cycles and redundant operations. We continue this process until we either find a zero matrix or exhaust all possibilities, at which point we return -1.
Complexity: Time Complexity: O(N * 2^(M*N)), where N is the number of cells, as we are exploring all possible states. Space Complexity: O(2^(M*N)) for the visited set and queue.
1from collections import deque
2
3class Solution:
4 def minFlips(self, mat):
5 def flip(bitmask, r, c):
6 n, m = len(mat), len(mat[0])
7 for i, j in [(r, c), (r+1, c), (r-1, c), (r, c+1), (r, c-1)]:
8 if 0 <= i < n and 0 <= j < m:
9 bitmask ^= (1 << (i * m + j))
10 return bitmask
11
12 n, m = len(mat), len(mat[0])
13 start = 0
14 for i in range(n):
15 for j in range(m):
16 if mat[i][j]:
17 start |= (1 << (i * m + j))
18
19 queue = deque([(start, 0)])
20 visited = set()
21 visited.add(start)
22
23 while queue:
24 bitmask, steps = queue.popleft()
25 if bitmask == 0:
26 return steps
27 for i in range(n):
28 for j in range(m):
29 next_bitmask = flip(bitmask, i, j)
30 if next_bitmask not in visited:
31 visited.add(next_bitmask)
32 queue.append((next_bitmask, steps + 1))
33
34 return -1Python BFS Explanation: We start by encoding the matrix into a single integer called a bitmask. Each bit in the integer represents a cell in the matrix. We use BFS to traverse all possible flips in the matrix, starting from the initial configuration. If a state reaches a zero matrix (bitmask == 0), we return the number of steps taken. We track visited states to avoid reprocessing states.
The problem can be tackled using a Breadth-First Search (BFS) approach, where we treat each matrix state as a node in a graph. The challenge is to explore all possible ways to make the minimum number of transformations from the initial matrix to a zero matrix. To efficiently manage and transition between states, we encode the matrix as a bitmask. Each state represents a particular configuration of the matrix.
Every time we flip a cell, it impacts the cell itself and its neighbors. For each state, we systematically toggle each cell and add the new resulting state to the BFS queue, marking states as visited to prevent redundant checks.
Time Complexity: O(2^(m*n) * m * n). We consider all possible matrix states (2^(m*n)). Each state performs an O(m*n) flip operation.
Space Complexity: O(2^(m*n)). We need space for the queue and visited set, each holding up to 2^(m*n) states.
1#include <vector>
2#include <queue>
3#include <unordered_set>
using namespace std;
class Solution {
public:
int minFlips(vector<vector<int>>& mat) {
int m = mat.size(), n = mat[0].size();
auto matToBits = [&](const vector<vector<int>>& mat) {
int bits = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
bits |= (mat[i][j] << (i * n + j));
}
}
return bits;
};
function<int(int,int)> flip = [&](int bits, int i, int j) {
vector<pair<int, int>> directions = {{0, 0}, {0, 1}, {1, 0}, {0, -1}, {-1, 0}};
for (auto& [di, dj] : directions) {
int ni = i + di, nj = j + dj;
if (0 <= ni && ni < m && 0 <= nj && nj < n) {
bits ^= 1 << (ni * n + nj);
}
}
return bits;
};
int start = matToBits(mat);
if (start == 0) return 0;
queue<pair<int, int>> q;
unordered_set<int> visited;
q.emplace(start, 0);
visited.insert(start);
while (!q.empty()) {
auto [state, step] = q.front();
q.pop();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int newState = flip(state, i, j);
if (newState == 0) return step + 1;
if (visited.insert(newState).second) {
q.emplace(newState, step + 1);
}
}
}
}
return -1;
}
};An alternative approach to solving the problem is through Depth-First Search (DFS) with Iterative Deepening. In this method, we explore the state space in a systematic and recursive manner, delving deeper into each potential flip configuration. We begin with a depth limit and increment gradually, attempting not to exceed this limit. The matrix state is represented at each node recursively, experimenting with each cell-flip combination.
The DFS pathway explores all configurations actively till terminal state clearance (zero matrix or true end-node) is discovered or edge constraints reach limits.
Time Complexity: Exponential in the number of matrix cells in the worst case, due to exhaustive state exploration.
Space Complexity: O(2^(m*n)) due to recursion depth and visited set overhead.
1import java.util.HashSet;
2import java
This C++ solution implements a BFS where each node is a state of the matrix represented by an integer bitmask. We perform the BFS through a queue, flipping each cell and checking converted states iteratively. For each cell selection and flip operation, the bitmask toggles appropriate bits, representing the matrix. If we reach the goal state (all zeroes), we return the number of steps. If impossible, return -1.
This Java implementation employs an iterative deepening DFS. We maintain a depth-bound search, recursing deeply till reaching the current depth limit. 'dfs()' explores flipping combinations, checking at each step whether a zero matrix configuration is achievable within constraints. 'matToBits()' and 'flip()' manage state representation, while checking reached states recursively using a visited set to prevent repetition.