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2174. Remove All Ones With Row and Column Flips II

Description

You are given a 0-indexed m x n binary matrix grid.

In one operation, you can choose any i and j that meet the following conditions:

  • 0 <= i < m
  • 0 <= j < n
  • grid[i][j] == 1

and change the values of all cells in row i and column j to zero.

Return the minimum number of operations needed to remove all 1's from grid.

 

Example 1:

Input: grid = [[1,1,1],[1,1,1],[0,1,0]]
Output: 2
Explanation:
In the first operation, change all cell values of row 1 and column 1 to zero.
In the second operation, change all cell values of row 0 and column 0 to zero.

Example 2:

Input: grid = [[0,1,0],[1,0,1],[0,1,0]]
Output: 2
Explanation:
In the first operation, change all cell values of row 1 and column 0 to zero.
In the second operation, change all cell values of row 2 and column 1 to zero.
Note that we cannot perform an operation using row 1 and column 1 because grid[1][1] != 1.

Example 3:

Input: grid = [[0,0],[0,0]]
Output: 0
Explanation:
There are no 1's to remove so return 0.

 

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 15
  • 1 <= m * n <= 15
  • grid[i][j] is either 0 or 1.

Solutions

  • class Solution {
        public int removeOnes(int[][] grid) {
            int m = grid.length, n = grid[0].length;
            int state = 0;
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (grid[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;
            while (!q.isEmpty()) {
                for (int k = q.size(); k > 0; --k) {
                    state = q.poll();
                    if (state == 0) {
                        return ans;
                    }
                    for (int i = 0; i < m; ++i) {
                        for (int j = 0; j < n; ++j) {
                            if (grid[i][j] == 0) {
                                continue;
                            }
                            int nxt = state;
                            for (int r = 0; r < m; ++r) {
                                nxt &= ~(1 << (r * n + j));
                            }
                            for (int c = 0; c < n; ++c) {
                                nxt &= ~(1 << (i * n + c));
                            }
                            if (!vis.contains(nxt)) {
                                vis.add(nxt);
                                q.offer(nxt);
                            }
                        }
                    }
                }
                ++ans;
            }
            return -1;
        }
    }
    
  • class Solution {
    public:
        int removeOnes(vector<vector<int>>& grid) {
            int m = grid.size(), n = grid[0].size();
            int state = 0;
            for (int i = 0; i < m; ++i)
                for (int j = 0; j < n; ++j)
                    if (grid[i][j])
                        state |= (1 << (i * n + j));
            queue<int> q{ {state} };
            unordered_set<int> vis{ {state} };
            int ans = 0;
            while (!q.empty()) {
                for (int k = q.size(); k > 0; --k) {
                    state = q.front();
                    q.pop();
                    if (state == 0) return ans;
                    for (int i = 0; i < m; ++i) {
                        for (int j = 0; j < n; ++j) {
                            if (grid[i][j] == 0) continue;
                            int nxt = state;
                            for (int r = 0; r < m; ++r) nxt &= ~(1 << (r * n + j));
                            for (int c = 0; c < n; ++c) nxt &= ~(1 << (i * n + c));
                            if (!vis.count(nxt)) {
                                vis.insert(nxt);
                                q.push(nxt);
                            }
                        }
                    }
                }
                ++ans;
            }
            return -1;
        }
    };
    
  • class Solution:
        def removeOnes(self, grid: List[List[int]]) -> int:
            m, n = len(grid), len(grid[0])
            state = sum(1 << (i * n + j) for i in range(m) for j in range(n) if grid[i][j])
            q = deque([state])
            vis = {state}
            ans = 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):
                            if grid[i][j] == 0:
                                continue
                            nxt = state
                            for r in range(m):
                                nxt &= ~(1 << (r * n + j))
                            for c in range(n):
                                nxt &= ~(1 << (i * n + c))
                            if nxt not in vis:
                                vis.add(nxt)
                                q.append(nxt)
                ans += 1
            return -1
    
    
  • func removeOnes(grid [][]int) int {
    	m, n := len(grid), len(grid[0])
    	state := 0
    	for i, row := range grid {
    		for j, v := range row {
    			if v == 1 {
    				state |= 1 << (i*n + j)
    			}
    		}
    	}
    	q := []int{state}
    	vis := map[int]bool{state: true}
    	ans := 0
    	for len(q) > 0 {
    		for k := len(q); k > 0; k-- {
    			state = q[0]
    			if state == 0 {
    				return ans
    			}
    			q = q[1:]
    			for i, row := range grid {
    				for j, v := range row {
    					if v == 0 {
    						continue
    					}
    					nxt := state
    					for r := 0; r < m; r++ {
    						nxt &= ^(1 << (r*n + j))
    					}
    					for c := 0; c < n; c++ {
    						nxt &= ^(1 << (i*n + c))
    					}
    					if !vis[nxt] {
    						vis[nxt] = true
    						q = append(q, nxt)
    					}
    				}
    			}
    		}
    		ans++
    	}
    	return -1
    }
    

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