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2174. Remove All Ones With Row and Column Flips II πŸ”’

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

Solution 1

Thinking

An operation picks a \(1\) and clears its whole row and column. The grid is at most \(8\times 8\), so configurations fit in an integer. Order among a set of moves does not matter; BFS finds the shortest sequence.

Pack the matrix into \(\textit{state}\). From a \(1\) cell, clear every bit on its row and column. Distance in this graph is the number of operations.

Start from the initial mask and stop at \(0\).

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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

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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;
    }
}

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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;
    }
};

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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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