Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest square which diagonal is all 1 and others is 0.

Only consider the main diagonal situation.

For example, given the following matrix:

1 0 1 0 0
1 0 0 1 0
1 1 0 0 1
1 0 0 1 0

Return 9

Similiarly with Maximal Square, we can check each entry of 1 from scratch and get the max length of a square whose diagonal is all 1 and others are 0.

Again, this straightforward solution suffers the same NOT using previous check results to expediate our algorithm‘s runtime.

So we jump directly into the dynamic programming solution.

State: dp[i][j]: the max length of a square with only diagonal 1s whose bottom right corner is matrix[i][j].

Function: dp[i][j] = 0, if matrix[i][j] == 0;

　　　　 dp[i][j] = 1 + min {leftZeros[i][j], upZeros[i][j], dp[i - 1][j - 1]}, if matrix[i][j] == 1;

Similarly with Maxmial Square, when matrix[i][j] is 1, we also need to check its left, top, top left side.

leftZeros[i][j] is the max number of consecutive 0s to the left of matrix[i][j];

upZeros[i][j] is the max number of consecutive 0s to the top of matrix[i][j];

These two along with dp[i - 1][j - 1] decides the max length of a sqaure with only diagonal 1s whose bottom right corner is matrix[i][j].

 1 public class Solution {
 2     public int maxSquare2(int[][] matrix) {
 3         if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
 4             return 0;
 5         }
 6         int n = matrix.length; int m = matrix[0].length;
 7         int[][] leftZeros = new int[n][m];
 8         int[][] upZeros = new int[n][m];
 9         for(int i = 0; i < n; i++){
10             leftZeros[i][0] = 0;
11         }
12         for(int j = 0; j < m; j++){
13             upZeros[0][j] = 0;
14         }
15         for(int i = 0; i < n; i++){
16             for(int j = 1; j < m; j++){
17                 if(matrix[i][j - 1] == 0){
18                     leftZeros[i][j] = leftZeros[i][j - 1] + 1;
19                 }    
20                 else{
21                     leftZeros[i][j] = 0;
22                 }
23             }
24         }
25         for(int i = 1; i < n; i ++){
26             for(int j = 0; j < m; j++){
27                 if(matrix[i - 1][j] == 0){
28                     upZeros[i][j] = upZeros[i - 1][j] + 1;
29                 }
30                 else{
31                     upZeros[i][j] = 0;
32                 }                
33             }
34         }
35         int[][] dp = new int[n][m];
36         for(int i = 0; i < n; i++){
37             dp[i][0] = matrix[i][0];    
38         }
39         for(int j = 0; j < m; j++){
40             dp[0][j] = matrix[0][j];
41         }
42         for(int i = 1; i < n; i++){
43             for(int j = 1; j < m; j++){
44                 if(matrix[i][j] == 0){
45                     dp[i][j] = 0;
46                 }
47                 else{
48                     dp[i][j] = Math.min(Math.min(leftZeros[i][j], upZeros[i][j]), dp[i - 1][j - 1]) + 1;
49                 }
50             }
51         }
52         int max = 0;
53         for(int i = 0; i < n; i++){
54             for(int j = 0; j < m; j++){
55                 max = Math.max(max, dp[i][j]);    
56             }
57         }
58         return max * max;
59     }
60 }

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