原题链接在这里：https://leetcode.com/problems/largest-1-bordered-square/
题目：
Given a 2D grid of 0s and 1s, return the number of elements in the largest square subgrid that has all 1s on its border, or 0 if such a subgrid doesn't exist in the grid.
Example 1:
Input: grid = [[1,1,1],[1,0,1],[1,1,1]] Output: 9
Example 2:
Input: grid = [[1,1,0,0]] Output: 1
Constraints:
1 <= grid.length <= 1001 <= grid[0].length <= 100grid[i][j]is0or1
题解：
For each cell in the grid, calculate its farest reach on top and left direction.
Then starting from l = Math.min(grid.length, grid[0].length) to l = 1, iterate grid to check if square with boarder l exist. If it does return l*l.
Time Complexity: O(m*n*(min(m,n))). m = grid.length. n = grid[0].length.
Space: O(m*n).
AC Java:
1 class Solution {
2 public int largest1BorderedSquare(int[][] grid) {
3 if(grid == null || grid.length == 0 || grid[0].length == 0){
4 return 0;
5 }
6
7 int m = grid.length;
8 int n = grid[0].length;
9 int [][] top = new int[m][n];
10 int [][] left = new int[m][n];
11 for(int i = 0; i<m; i++){
12 for(int j = 0; j<n; j++){
13 if(grid[i][j] > 0){
14 top[i][j] = i == 0 ? 1 : top[i-1][j]+1;
15 left[i][j] = j == 0 ? 1 : left[i][j-1]+1;
16 }
17 }
18 }
19
20 for(int l = Math.min(m, n); l>0; l--){
21 for(int i = 0; i+l-1<m; i++){
22 for(int j = 0; j+l-1<n; j++){
23 if(top[i+l-1][j] >= l
24 && top[i+l-1][j+l-1] >= l
25 && left[i][j+l-1] >= l
26 && left[i+l-1][j+l-1] >= l){
27 return l*l;
28 }
29 }
30 }
31 }
32
33 return 0;
34 }
35 }
Or after get top and left.
Iterate the grid, for each cell, get small = min(top[i][j], left[i][j]).
All l = small to 1 could be protential square boarder. But we only need to check small larger than global longest boarder since we only care about the largest.
Check top of grid[i][j-small+1] and left of grid[i-small+1][j]. If they are both larger than small, then it is a grid.
Time Complexity: O(n^3).
Space: O(n^2).
AC Java:
1 class Solution {
2 public int largest1BorderedSquare(int[][] grid) {
3 if(grid == null || grid.length == 0 || grid[0].length == 0){
4 return 0;
5 }
6
7 int m = grid.length;
8 int n = grid[0].length;
9 int [][] top = new int[m][n];
10 int [][] left = new int[m][n];
11 for(int i = 0; i<m; i++){
12 for(int j = 0; j<n; j++){
13 if(grid[i][j] > 0){
14 top[i][j] = i == 0 ? 1 : top[i-1][j]+1;
15 left[i][j] = j == 0 ? 1 : left[i][j-1]+1;
16 }
17 }
18 }
19
20 int res = 0;
21
22 for(int i = m-1; i>=0; i--){
23 for(int j = n-1; j>=0; j--){
24 int small = Math.min(top[i][j], left[i][j]);
25 while(small > res){
26 if(top[i][j-small+1] >= small && left[i-small+1][j] >= small){
27 res = small;
28 break;
29 }
30
31 small--;
32 }
33 }
34 }
35
36 return res*res;
37 }
38 }