764. Largest Plus Sign (Medium)

In a 2Dgridfrom (0, 0) to (N-1, N-1), every cell contains a1, except those cells in the given listmineswhich are0. What is the largest axis-aligned plus sign of1s contained in the grid? Return the order of the plus sign. If there is none, return 0.

An "axis-aligned plus sign of1sof order k" has some centergrid[x][y] = 1along with 4 arms of lengthk-1going up, down, left, and right, and made of1s. This is demonstrated in the diagrams below. Note that there could be0s or1s 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.)

Solution 1: DP O(n^2); O(n^2)

public int orderOfLargestPlusSign(int N, int[][] mines) {
    int[][] grid = new int[N][N];

    for (int i = 0; i < N; i++) {
        Arrays.fill(grid[i], N);
    }

    for (int[] m : mines) {
        grid[m[0]][m[1]] = 0;
    }

    for (int i = 0; i < N; i++) {
        for (int j = 0, k = N - 1, l = 0, r = 0, u = 0, d = 0; j < N; j++, k--) {
            grid[i][j] = Math.min(grid[i][j], l = (grid[i][j] == 0 ? 0 : l + 1));
            grid[i][k] = Math.min(grid[i][k], r = (grid[i][k] == 0 ? 0 : r + 1));
            grid[j][i] = Math.min(grid[j][i], u = (grid[j][i] == 0 ? 0 : u + 1));
            grid[k][i] = Math.min(grid[k][i], d = (grid[k][i] == 0 ? 0 : d + 1));
        }
    }

    int res = 0;

    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            res = Math.max(res, grid[i][j]);
        }
    }

    return res;
}