LeetCode 764. Largest Plus Sign

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 1s contained in the grid? Return the order of the plus sign. If there is none, return 0.

An “axis-aligned plus sign of 1s 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 1s. This is demonstrated in the diagrams below. Note that there could be 0s or 1s 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.)

这道题对我来说挺坑的，我一开始不想用二位数组来保存mines的位置，而是使用set\unordered_set来保存，导致超时……后来又想了很久找到了另一种解法。

Brute force

对每一个点，都依次向外搜索最大的+，最坏时间复杂度为O(n³)。

class Solution {
    unordered_set<int64_t > mines;
    int _N;
public:
    int orderOfLargestPlusSign(int N, vector<vector<int>>& _mines) {
        _N = N;
        vector<vector<int>> zeros(N, vector<int>(N, 0));
        for (auto &v: _mines) {
            zeros[v[0]][v[1]] = 1;
        }

        int ans = 0;

        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                if (ans != 0 && (!valid(i - ans + 1, j) || !valid(i + ans - 1, j) || !valid(i, j - ans + 1) || !valid(i, j + ans - 1)))
                    continue;
                for (int k = 1; ; k++) {
                    int l = k - 1;
                    if (!valid(i - l, j) || zeros[i - l][j]) {
                        break;
                    }
                    if (!valid(i + l, j) || zeros[i + l][j]) {
                        break;
                    }
                    if (!valid(i, j - l) || zeros[i][j - l]) {
                        break;
                    }
                    if (!valid(i, j + l) || zeros[i][j + l]) {
                        break;
                    }

                    ans = max(ans, k);
                    if (N % 2 == 0 && ans * 2 == N) {
                        return ans;
                    }
                    else if (N % 2 == 1 && ans * 2 - 1 == N) {
                        return ans;
                    }
                }
            }
        }

        return ans;
    }

    bool valid (int x, int y) {
        return (x >= 0 && x < _N && y >= 0 && y < _N);
    }
};

DP

用四次二维DP，分别找到每个点的上下左右四个方向的最长的连续1的个数，然后对每一个点从这四个长度中选择最小值，就是该点的最大+大小。时间复杂度O(n²)。

要注意这个方法很容易超内存，所以我不得已把三维数组改成了二位数组循环利用四次。

class Solution {
    unordered_set<int32_t > mines;
    int _N;
public:
    int orderOfLargestPlusSign(int N, vector<vector<int>>& _mines) {
        _N = N;
        for (auto &v : _mines) {
            mines.insert((int32_t)v[0] << 16 | (int32_t)v[1]);
        }

        int ans = 0;
        vector<vector<int>> dp(N, vector<int>(N, 0));
        vector<vector<int>> len(N, vector<int>(N, INT_MAX));

        for (int i = 0; i < N; i++) {
            if (!isMine(i, 0)) {
                dp[i][0] = 1;
                len[i][0] = min(len[i][0], dp[i][0]);
                ans = max(ans, len[i][0]);
            }
        }
        for (int i = 0; i < N; i++) {
            for (int j = 1; j < N; j++) {
                if (isMine(i, j))
                    continue;

                dp[i][j] = dp[i][j - 1] + 1;
                len[i][j] = min(len[i][j], dp[i][j]);
            }
        }

        dp.assign(N, vector<int>(N, 0));
        for (int i = 0; i < N; i++) {
            if (!isMine(0, i)) {
                dp[0][i] = 1;
                len[0][i] = min(len[0][i], dp[0][i]);
                ans = max(ans, len[0][i]);
            }
        }

        for (int i = 1; i < N; i++) {
            for (int j = 0; j < N; j++) {
                if (isMine(i, j))
                    continue;

                dp[i][j] = dp[i - 1][j] + 1;
                len[i][j] = min(len[i][j], dp[i][j]);
            }
        }

        dp.assign(N, vector<int>(N, 0));
        for (int i = N - 1; i >= 0; i--) {
            if (!isMine(N - 1, i)) {
                dp[N - 1][i] = 1;
                len[N - 1][i] = min(len[N - 1][i], dp[N - 1][i]);
                ans = max(ans, len[N - 1][i]);
            }
        }

        for (int i = N - 2; i >= 0; i--) {
            for (int j = N - 1; j >= 0; j--) {
                if (isMine(i, j))
                    continue;

                dp[i][j] = dp[i + 1][j] + 1;
                len[i][j] = min(len[i][j], dp[i][j]);
            }
        }

        dp.assign(N, vector<int>(N, 0));
        for (int i = N - 1; i >= 0; i--) {
            if (!isMine(i, N - 1)) {
                dp[i][N - 1] = 1;
                len[i][N - 1] = min(len[i][N - 1], dp[i][N - 1]);
                ans = max(ans, len[i][N - 1]);
            }
        }

        for (int i = N - 1; i >= 0; i--) {
            for (int j = N - 2; j >= 0; j--) {
                if (isMine(i, j))
                    continue;

                dp[i][j] = dp[i][j + 1] + 1;
                len[i][j] = min(len[i][j], dp[i][j]);
                ans = max(ans, len[i][j]);
            }
        }

        return ans;
    }

    bool isMine (int x, int y) {
        return (bool)mines.count((int32_t)x << 16 | (int32_t)y);
    }

    bool valid (int x, int y) {
        return (x >= 0 && x < _N && y >= 0 && y < _N);
    }
};