## Interview Question

ALGORITHM:

1. take the colored squares as 0 and the non-colored squares as 1. then form a n*n matrix 'a'.

2. now we have to find the largest square matrix having all 1's.

3. fill up the 1st row of the result (n*n) matrix as the matrix 'a'.

4. then from the second row on wards

4.1. if you find a[i][j] == 0 then res[i][j] = 0.

4.2. else res[i][j] = min(res[i-1][j], res[i][j-1], res[i-1][j-1])+1.

5. after creating the 'res' matrix the size of largest uncolored square will be the largest element in the 'res' matrix.

i'm giving the program that will find the size of the sub-square-matrix having all 1's.

the program for that is

```
public static void main(String [] args){
int [][] arr = {{0,1,1,0,1,0},
{1,1,0,0,1,0},
{0,1,1,1,0,1},
{0,1,1,0,0,1},
{0,1,1,1,0,1}};
int col = arr[0].length;
int row = arr.length;
int [][] res = new int [row][col];
for(int i=0; i<col; i++){
res[0][i] = arr[0][i];
}
for(int i=0; i<row; i++){
res[i][0] = arr[i][0];
}
for(int i=1; i<row; i++){
for(int j=1; j<col; j++){
if(arr[i][j] == 1){
res[i][j] = Math.min(Math.min(res[i][j-1], res[i-1][j]),res[i-1][j-1]) + 1;
}
else{res[i][j] = 0;}
}
}
System.out.println("the distance matrix is : ");
for(int i=0; i<row; i++){
for(int j=0; j<col; j++){
System.out.print(res[i][j]+" ");
}
System.out.println("");
}
int max = Integer.MIN_VALUE;
for(int i=0; i<row; i++){
for(int j=0; j<col; j++){
max = Math.max(max, res[i][j]);
}
}
System.out.println("max square : "+max);
}
```

please don't allow this kind of people "anonymous" to post craps

- CAFEBABE January 07, 2012