[Java] - Pure Brute Approach
\nclass Solution {\n    public int orderOfLargestPlusSign(int n, int[][] mines) {\n        \n    \tif(n*n==mines.length)\n    \t\treturn 0;\n        \n    \tif(
```\nclass Solution {\n    public int orderOfLargestPlusSign(int n, int[][] mines) {\n        \n    \tif(n*n==mines.length)\n    \t\treturn 0;\n        \n    \tif(n<3)\n            return 1;\n    \t\n    \tint arr[][]=new int[n][n];\n    \t\n    \tint i,j,k;\n    \t\n    \tfor(i=0;i<n;i++)\n    \t\tArrays.fill(arr[i], 1);\n    \t\n    \tfor(i=0;i<mines.length;i++)\n    \t\tarr[mines[i][0]][mines[i][1]]=0;\n    \t\n        int tm=n;\n        if(tm%2==0)\n            tm--;\n    \t\t\n    \tfor(i=tm;i>=3;i-=2)\n    \t{\n    \t\tfor(j=0;j<=n-i;j++)\n    \t\t{\n    \t\t\tfor(k=i/2;k<n-i/2;k++)\n    \t\t\t{\n    \t\t\t\t int l,m;\n    \t\t\t\t //checking the vertical 1\'s of plus sign\n    \t\t\t\t for(l=0;l<i;l++)\n    \t\t\t\t {\n    \t\t\t\t\t if(arr[l+j][k]!=1)\n    \t\t\t\t\t\t break;\n    \t\t\t\t }\n    \t\t\t\t if(l<i)\n                        continue;\n    \t\t\t\t //checking the horizontal 1\'s of plus sign\n    \t\t\t\t for(m=0;m<i;m++)\n    \t\t\t\t {\n    \t\t\t\t\t if(arr[j+i/2][k-i/2+m]!=1)\n    \t\t\t\t\t\t break;\n    \t\t\t\t }\n    \t\t\t\t if(m==i)\n                         return i/2+1;\n    \t\t\t}\n    \t\t}\n    \t}\n    \t\n\t    return 1;\n    }\n}\n```\nAny Questions?\nPlease do upvote:))
['Java']