LeetCode解题报告（519）-- 2174. Remove All Ones With Row and Column Flips II

Problem

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.

Analysis

这道题是Remove All Ones With Row and Column Flips的系列题，但是和之前这题的做法截然不同。前面提到了flipping的特点是只flip一次，flip两次都多余。但是这道题目是对所有的位置为1的格子，对其所在行和所在列变更为0（注意这里不是flip了，而是直接变成0），所以之前的性质在这里就不管用了。所以要变换思路，第一个提示是题目要求的是minimum operations，这种明显的提示一般就是BFS或者memorized searching，那我们就按照这个思路去做。

首先要解决的是状态问题，每一个状态都是一个矩阵，这样非常不利于我们做BFS或DFS。对于01矩阵来说，一个常见的做法就是进行编码，把矩阵打平成一维，然后利用位运算，使用一个32位整数表示即可。这里一个前提是题目给定了矩阵的大小不超过15。

搜索方法这里我采用的是DFS+记忆数组。对每个位置的更新都是通过位运算完成，记忆数组memo[i]表示的是i这个状态下变成0需要的最少的操作数量。

Solution

无。

Code

class Solution {
public:
    int removeOnes(vector<vector<int>>& grid) {
        memo = vector<int>(32768, INT_MAX);
        int state = 0;
        m = grid.size(), n = grid[0].size();
        
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j]) {
                    state |= (1 << (i * n + j));
                }
            }
        }
        return dfs(state);
    }
private:
    int m, n;
    vector<int> memo;
    int dfs(int state) {
        if (state == 0) {
            return 0;
        }
        if (memo[state] != INT_MAX) {
            return memo[state];
        }
        
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (state & (1 << (i * n + j))) {
                    int newState = state;
                    for (int k = 0; k < m; ++k) {
                        newState &= ~(1 << (k * n + j));
                    }
                    for (int k = 0; k < n; ++k) {
                        newState &= ~(1 << (i * n + k));
                    }
                    memo[state] = min(memo[state], 1 + dfs(newState));
                }
            }
        }
        return memo[state];
    }
};

Summary

这道题目实际上就是BFS或DFS的应用，难点在于对矩阵进行编码。这道题目的分享到这里，感谢你的支持！