# Largest Plus Sign

## Description

In a 2D `grid` from (0, 0) to (N-1, N-1), every cell contains a `1`, except those cells in the given list `mines` which are `0`. What is the largest axis-aligned plus sign of `1`s contained in the grid? Return the order of the plus sign. If there is none, return 0.

An "axis-aligned plus sign of `1`s of order **k**" has some center `grid[x][y] = 1` along with 4 arms of length `k-1` going up, down, left, and right, and made of `1`s. This is demonstrated in the diagrams below. Note that there could be `0`s or `1`s beyond the arms of the plus sign, only the relevant area of the plus sign is checked for 1s.

**Examples of Axis-Aligned Plus Signs of Order k:**

```
Order 1:
000
010
000

Order 2:
00000
00100
01110
00100
00000

Order 3:
0000000
0001000
0001000
0111110
0001000
0001000
0000000
```

**Example 1:**

```
Input: N = 5, mines = [[4, 2]]
Output: 2
Explanation:
11111
11111
11111
11111
11011
In the above grid, the largest plus sign can only be order 2.  One of them is marked in bold.
```

**Example 2:**

```
Input: N = 2, mines = []
Output: 1
Explanation:
There is no plus sign of order 2, but there is of order 1.
```

**Example 3:**

```
Input: N = 1, mines = [[0, 0]]
Output: 0
Explanation:
There is no plus sign, so return 0.
```

**Note:**

1. `N` will be an integer in the range `[1, 500]`.
2. `mines` will have length at most `5000`.
3. `mines[i]` will be length 2 and consist of integers in the range `[0, N-1]`.
4. (Additionally, programs submitted in C, C++, or C# will be judged with a slightly smaller time limit.)

## Solutions

### DFS

```cpp
class Solution {
public:
    int orderOfLargestPlusSign(int N, vector<vector<int>>& mines) {
        vector<vector<int>> grid(N, vector<int>(N, 1));
        for(auto &p: mines){
            grid[p[0]][p[1]] = 0;
        }
        
        auto all_ones = [&](int i, int j, int k){
            if(i - k < 0 || i + k >= N || j - k < 0 || j + k >= N)
                return false;
            return grid[i - k][j] && grid[i + k][j] && grid[i][j - k] && grid[i][j + k];
        };
        int K = 0;
        for(int i = 0; i < N; ++i){
            for(int j = 0; j < N; ++j){
                if(grid[i][j] == 1){
                    int k = 1;
                    while(all_ones(i, j, k))
                        ++k;
                    K = max(k, K);
                }
            }
        }
        return K;
    }
};
```


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