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 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.)

Solution

題目好長，翻譯過來就是在一個01稀疏矩陣中找由連續(xù)1組成的最大的十字架（十字架每條邊長度相等）。

3D-DP + HashSet, time O(n ^ 2), space O(n ^ 2)

三維DP，int[][][] dp = new int[n + 2][n + 2][4]，z維度表示四個方向。
注意對于坐標(i, j)來說，它依賴于上下左右四個方向的dp值，所以還需要從后往前遍歷。

由于mines是一組坐標，可以將其轉(zhuǎn)換成一維存在HashSet中，方便查詢。

class Solution {
    // up, left, down, right
    public static final int[][] DIRECTIONS = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};
    
    public int orderOfLargestPlusSign(int n, int[][] mines) {
        int[][][] dp = new int[n + 2][n + 2][4];
        Set<Integer> mineSet = new HashSet<>();
        for (int[] mine : mines) {
            mineSet.add(getIndex(mine[0], mine[1], n));
        }
        // traverse from up to bottom, left to right
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (mineSet.contains(getIndex(i, j, n))) {
                    continue;
                }     
                
                for (int k = 0; k < 2; ++k) {
                    dp[i + 1][j + 1][k] 
                         = 1 + dp[i + 1 + DIRECTIONS[k][0]][j + 1 + DIRECTIONS[k][1]][k];  
                }
            }
        }
        // traverse from bottom to up, right to left
        int max = 0;
        for (int i = n - 1; i >= 0; --i) {
            for (int j = n - 1; j >= 0; --j) {
                if (mineSet.contains(getIndex(i, j, n))) {
                    continue;
                }     
                
                for (int k = 2; k < 4; ++k) {
                    dp[i + 1][j + 1][k] 
                         = 1 + dp[i + 1 + DIRECTIONS[k][0]][j + 1 + DIRECTIONS[k][1]][k];  
                }
                // update max when 4 directions are done
                max = Math.max(min(dp[i + 1][j + 1]), max);
            }
        }
        
        return max;
    }
    
    public int getIndex(int i, int j, int n) {
        return i * n + j;
    }
    
    public int min(int[] a) {
        int min = Integer.MAX_VALUE;
        for (int n : a) {
            min = Math.min(n, min);
        }
        return min;
    }
}

2D-DP + HashSet, time O(n ^ 2), space O(n ^ 2)

下面這種寫法更清晰一點，推薦。

class Solution {
    public int orderOfLargestPlusSign(int n, int[][] mines) {
        int[][] dp = new int[n][n];
        Set<Integer> mineSet = new HashSet<>();
        for (int[] mine : mines) {
            mineSet.add(getIndex(mine[0], mine[1], n));
        }

        for (int i = 0; i < n; ++i) {
            int count = 0;
            for (int j = 0; j < n; ++j) {
                count = mineSet.contains(getIndex(i, j, n)) ? 0 : count + 1;
                dp[i][j] = count;
            }
            
            count = 0;
            for (int j = n - 1; j >= 0; --j) {
                count = mineSet.contains(getIndex(i, j, n)) ? 0 : count + 1;
                dp[i][j] = Math.min(count, dp[i][j]);
            }
        }
        
        int max = 0;
        
        for (int j = 0; j < n; ++j) {
            int count = 0;
            for (int i = 0; i < n; ++i) {
                count = mineSet.contains(getIndex(i, j, n)) ? 0 : count + 1;
                dp[i][j] = Math.min(count, dp[i][j]);
            }
            
            count = 0;
            for (int i = n - 1; i >= 0; --i) {
                count = mineSet.contains(getIndex(i, j, n)) ? 0 : count + 1;
                dp[i][j] = Math.min(count, dp[i][j]);
                max = Math.max(dp[i][j], max);
            }
        }
        
        return max;
    }
    
    public int getIndex(int i, int j, int n) {
        return i * n + j;
    }
}