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.)

暴力解就是兩層loop然後每個grid往上下左右搜尋，所以time complexity是O(n^3)。

// Brute Force
bool canExpand(vector<vector<int> >& matrix, int N, int x, int y, int k) {
    if (x - k < 0 || y - k < 0 || x + k >= N || y + k >= N) return false;
    return matrix[x + k][y] && matrix[x - k][y] && matrix[x][y + k] && matrix[x][y - k];
}
int orderOfLargestPlusSign(int N, vector<vector<int>>& mines) { // time: O(N^3); space: O(N^2)
    int res = 0;
    vector<vector<int> > matrix(N, vector<int>(N, 1));
    for (auto& mine : mines) matrix[mine[0]][mine[1]] = 0;
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            int k = 0;
            while (canExpand(matrix, N, i, j, k)) ++k;
            res = max(res, k);
        }
    }
    return res;
}

利用一個hashset把0的位置存起來，然後一個2D DP table存每個位置能形成的最大＋符號的order是多少。

// Dynamic programming
int orderOfLargestPlusSign(int N, vector<vector<int>>& mines) { // time: O(N^2); space: O(N^2)
    int res = 0, cnt = 0;
    vector<vector<int> > dp(N, vector<int>(N, 0));
    unordered_set<int> s;
    for (auto& mine : mines) s.insert(mine[0] * N + mine[1]);
    for (int j = 0; j < N; ++j) {
        cnt = 0;
        // Up
        for (int i = 0; i < N; ++i) {
            cnt = s.count(i * N + j) ? 0 : cnt + 1;
            dp[i][j] = cnt;
        }
        cnt = 0;
        // Down
        for (int i = N - 1; i >= 0; --i) {
            cnt = s.count(i * N + j) ? 0 : cnt + 1;
            dp[i][j] = min(dp[i][j], cnt);
        }
    }
    for (int i = 0; i < N; ++i) {
        cnt = 0;
        // Left
        for (int j = 0; j < N; ++j) {
            cnt = s.count(i * N + j) ? 0 : cnt + 1;
            dp[i][j] = min(dp[i][j], cnt);
        }
        cnt = 0;
        // Right
        for (int j = N - 1; j >= 0; --j) {
            cnt = s.count(i * N + j) ? 0 : cnt + 1;
            dp[i][j] = min(dp[i][j], cnt);
        }
    }
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            res = max(res, dp[i][j]);
        }
    }
    return res;
}

其實也不需要hashset輔助，直接初始化dp table為所有都是1，然後把mines的位置設為0，之後scan時，不為0的grid代表不是mine的地方。

// Dynamic Programming
int orderOfLargestPlusSign(int N, vector<vector<int>>& mines) { // time: O(N^2); space: O(N^2)
    int res = 0, cnt = 0;
    vector<vector<int> > dp(N, vector<int>(N, 1));
    for (auto& mine : mines) dp[mine[0]][mine[1]] = 0;
    for (int j = 0; j < N; ++j) {
        cnt = 0;
        // Up
        for (int i = 0; i < N; ++i) {
            if (dp[i][j]) cnt += 1;
            else cnt = 0;
            dp[i][j] = cnt;
        }
        cnt = 0;
        // Down
        for (int i = N - 1; i >= 0; --i) {
            if (dp[i][j]) cnt += 1;
            else cnt = 0;
            dp[i][j] = min(dp[i][j], cnt);
        }
    }
    for (int i = 0; i < N; ++i) {
        cnt = 0;
        // Left
        for (int j = 0; j < N; ++j) {
            if (dp[i][j]) cnt += 1;
            else cnt = 0;
            dp[i][j] = min(dp[i][j], cnt);
        }
        cnt = 0;
        // Right
        for (int j = N - 1; j >= 0; --j) {
            if (dp[i][j]) cnt += 1;
            else cnt = 0;
            dp[i][j] = min(dp[i][j], cnt);
        }
    }
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            res = max(res, dp[i][j]);
        }
    }
    return res;
}

上述的DP解法經過整理讓code更為簡潔。

// Dynamic programming
int orderOfLargestPlusSign(int N, vector<vector<int>>& mines) { // time: O(N^2); space: O(N^2)
    int res = 0;
    vector<vector<int> > dp(N, vector<int>(N, N));
    for (auto& mine : mines) dp[mine[0]][mine[1]] = 0;
    for (int i = 0; i < N; ++i) {
        int l = 0, r = 0, u = 0, d = 0;
        for (int j = 0, k = N - 1; j < N && k >= 0; ++j, --k) {
            dp[i][j] = min(dp[i][j], l = (dp[i][j]) ? l + 1 : 0);
            dp[i][k] = min(dp[i][k], r = (dp[i][k]) ? r + 1 : 0);
            dp[j][i] = min(dp[j][i], u = (dp[j][i]) ? u + 1 : 0);
            dp[k][i] = min(dp[k][i], d = (dp[k][i]) ? d + 1 : 0);
        }
    }
    for (int l = 0; l < N * N; ++l) {
        res = max(res, dp[l / N][l % N]);
    }
    return res;
}