text: Largest Square in a Binary Matrix with at most K 1 s for multiple Queries | Java implementation to find the largest square in the matrix such that it contains atmost K 1 's ; Function to find the largest square in the matrix such that it contains atmost K 1 's ; Precomputing the countDP prefix sum of the matrix ; Loop to solve Queries ; Calculating the maximum possible distance of the centre from edge ; Calculating the number of 1 s in the submatrix ; Driver Code
code: import java . util . * ; class GFG { static int MAX = 100 ; static void largestSquare ( int matrix [ ] [ ] , int R , int C , int q_i [ ] , int q_j [ ] , int K , int Q ) { int [ ] [ ] countDP = new int [ R ] [ C ] ; countDP [ 0 ] [ 0 ] = matrix [ 0 ] [ 0 ] ; for ( int i = 1 ; i < R ; i ++ ) countDP [ i ] [ 0 ] = countDP [ i - 1 ] [ 0 ] + matrix [ i ] [ 0 ] ; for ( int j = 1 ; j < C ; j ++ ) countDP [ 0 ] [ j ] = countDP [ 0 ] [ j - 1 ] + matrix [ 0 ] [ j ] ; for ( int i = 1 ; i < R ; i ++ ) for ( int j = 1 ; j < C ; j ++ ) countDP [ i ] [ j ] = matrix [ i ] [ j ] + countDP [ i - 1 ] [ j ] + countDP [ i ] [ j - 1 ] - countDP [ i - 1 ] [ j - 1 ] ; for ( int q = 0 ; q < Q ; q ++ ) { int i = q_i [ q ] ; int j = q_j [ q ] ; int min_dist = Math . min ( Math . min ( i , j ) , Math . min ( R - i - 1 , C - j - 1 ) ) ; int ans = - 1 ; for ( int k = 0 ; k <= min_dist ; k ++ ) { int x1 = i - k , x2 = i + k ; int y1 = j - k , y2 = j + k ; int count = countDP [ x2 ] [ y2 ] ; if ( x1 > 0 ) count -= countDP [ x1 - 1 ] [ y2 ] ; if ( y1 > 0 ) count -= countDP [ x2 ] [ y1 - 1 ] ; if ( x1 > 0 && y1 > 0 ) count += countDP [ x1 - 1 ] [ y1 - 1 ] ; if ( count > K ) break ; ans = 2 * k + 1 ; } System . out . print ( ans + "NEW_LINE"); } } public static void main ( String [ ] args ) { int matrix [ ] [ ] = { { 1 , 0 , 1 , 0 , 0 } , { 1 , 0 , 1 , 1 , 1 } , { 1 , 1 , 1 , 1 , 1 } , { 1 , 0 , 0 , 1 , 0 } } ; int K = 9 , Q = 1 ; int q_i [ ] = { 1 } ; int q_j [ ] = { 2 } ; largestSquare ( matrix , 4 , 5 , q_i , q_j , K , Q ) ; } }