# Java DFS solution beat %95

• Using DFS method.

1. Assigned a large value to all the positions with value 1 and don't have 0 neighbors
2. Start dfs search from positions whose value is 1
``````public class Solution {
public int[][] updateMatrix(int[][] matrix) {
if(matrix.length==0) return matrix;

for(int i = 0; i<matrix.length; i++)
for(int j = 0; j<matrix[0].length; j++)
if(matrix[i][j]==1&&!hasNeiberZero(i, j,matrix))
matrix[i][j] = matrix.length+matrix[0].length+1;

for(int i = 0; i<matrix.length; i++)
for(int j = 0; j<matrix[0].length; j++)
if(matrix[i][j]==1)
dfs(matrix, i, j, -1);

return matrix;
}
private void dfs(int[][] matrix, int x, int y, int val){
if(x<0||y<0||y>=matrix[0].length||x>=matrix.length||matrix[x][y]<=val)
return;

if(val>0) matrix[x][y] = val;

dfs(matrix, x+1, y, matrix[x][y]+1);
dfs(matrix, x-1, y, matrix[x][y]+1);
dfs(matrix, x, y+1, matrix[x][y]+1);
dfs(matrix, x, y-1, matrix[x][y]+1);

}
private boolean hasNeiberZero(int x, int y, int[][] matrix){
if(x>0&&matrix[x-1][y]==0) return true;
if(x<matrix.length-1&&matrix[x+1][y]==0) return true;
if(y>0&&matrix[x][y-1]==0) return true;
if(y<matrix[0].length-1&&matrix[x][y+1]==0) return true;

return false;
}
}
``````

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