LeetCode 764 - Largest Plus Sign

https://leetcode.com/problems/largest-plus-sign/solution/

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.)

X.

How can we improve our bruteforce? One way is to try to speed up the inner loop involving k, the order of the candidate plus sign. If we knew the longest possible arm length $L_u, L_l, L_d, L_r$ in each direction from a center, we could know the order $\min(L_u, L_l, L_d, L_r)$ of a plus sign at that center. We could find these lengths separately using dynamic programming.

Algorithm

For each (cardinal) direction, and for each coordinate (r, c) let's compute the count of that coordinate: the longest line of '1's starting from (r, c) and going in that direction. With dynamic programming, it is either 0 if grid[r][c] is zero, else it is 1 plus the count of the coordinate in the same direction. For example, if the direction is left and we have a row like 01110110, the corresponding count values are 01230120, and the integers are either 1 more than their successor, or 0. For each square, we want dp[r][c] to end up being the minimum of the 4 possible counts. At the end, we take the maximum value in dp.

• Time Complexity: $O(N^2)$, as the work we do under two nested for loops is $O(1)$.
• Space Complexity: $O(N^2)$, the size of dp.

    public int orderOfLargestPlusSign(int N, int[][] mines) {
        Set<Integer> banned = new HashSet();
        int[][] dp = new int[N][N];
       
        for (int[] mine: mines)
            banned.add(mine[0] * N + mine[1]);
        int ans = 0, count;
       
        for (int r = 0; r < N; ++r) {
            count = 0;
            for (int c = 0; c < N; ++c) {
                count = banned.contains(r*N + c) ? 0 : count + 1;
                dp[r][c] = count;
            }
           
            count = 0;
            for (int c = N-1; c >= 0; --c) {
                count = banned.contains(r*N + c) ? 0 : count + 1;
                dp[r][c] = Math.min(dp[r][c], count);
            }
        }
       
        for (int c = 0; c < N; ++c) {
            count = 0;
            for (int r = 0; r < N; ++r) {
                count = banned.contains(r*N + c) ? 0 : count + 1;
                dp[r][c] = Math.min(dp[r][c], count);
            }
           
            count = 0;
            for (int r = N-1; r >= 0; --r) {
                count = banned.contains(r*N + c) ? 0 : count + 1;
                dp[r][c] = Math.min(dp[r][c], count);
                ans = Math.max(ans, dp[r][c]);
            }
        }
       
        return ans;
    }

https://leetcode.com/problems/largest-plus-sign/discuss/113314/JavaC++Python-O(N2)-solution-using-only-one-grid-matrix

Algorithms: For each position (i, j) of the grid matrix, we try to extend in each of the four directions (left, right, up, down) as long as possible, then take the minimum length of 1's out of the four directions as the order of the largest axis-aligned plus sign centered at position (i, j).

---

Optimizations: Normally we would need a total of five matrices to make the above idea work -- one matrix for the grid itself and four more matrices for each of the four directions. However, these five matrices can be combined into one using two simple tricks:

1. For each position (i, j), we are only concerned with the minimum length of 1's out of the four directions. This implies we may combine the four matrices into one by only keeping tracking of the minimum length.
2. For each position (i, j), the order of the largest axis-aligned plus sign centered at it will be 0 if and only if grid[i][j] == 0. This implies we may further combine the grid matrix with the one obtained above.

---

Implementations:

1. Create an N-by-N matrix grid, with all elements initialized with value N.
2. Reset those elements to 0 whose positions are in the mines list.
3. For each position (i, j), find the maximum length of 1's in each of the four directions and set grid[i][j] to the minimum of these four lengths. Note that there is a simple recurrence relation relating the maximum length of 1's at current position with previous position for each of the four directions (labeled as l, r, u, d).
4. Loop through the grid matrix and choose the maximum element which will be the largest axis-aligned plus sign of 1's contained in the grid.

---

Solutions: Here is a list of solutions for Java/C++/Python based on the above ideas. All solutions run at O(N^2) time with O(N^2) extra space. Further optimizations are possible such as keeping track of the maximum plus sign currently available and terminating as early as possible if no larger plus sign can be found for current row/column.

Note: For those of you who got confused by the logic within the first nested for-loop, refer to andier's comment below for a more clear explanation.

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;
}




X. Brute Force

Intuition and Algorithm

For each possible center, find the largest plus sign that could be placed by repeatedly expanding it. We expect this algorithm to be $O(N^3)$, and so take roughly $500^3 = (1.25) * 10^8$ operations. This is a little bit too big for us to expect it to run in time.

• Time Complexity: $O(N^3)$, as we perform two outer loops ($O(N^2)$), plus the inner loop involving k is $O(N)$.
• Space Complexity: $O(\text{mines.length})$.

    public int orderOfLargestPlusSign(int N, int[][] mines) {

        Set<Integer> banned = new HashSet();

        for (int[] mine: mines)

            banned.add(mine[0] * N + mine[1]);

            

        int ans = 0;

        for (int r = 0; r < N; ++r) for (int c = 0; c < N; ++c) {

            int k = 0;

            while (k <= r && r < N-k && k <= c && c < N-k &&

                    !banned.contains((r-k)*N + c) &&

                    !banned.contains((r+k)*N + c) &&

                    !banned.contains(r*N + c-k) &&

                    !banned.contains(r*N + c+k))

                k++;

            

            ans = Math.max(ans, k);

        }

        return ans;

    }