human
Largest Square in a Binary Matrix with at most K 1 s for multiple Queries | 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"
How can the above be solved in Python?

What is a code to largest Square in a Binary Matrix with at most K 1 s for multiple Queries and the other steps required?
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#Here is code to largest Square in a Binary Matrix with at most K 1 s for multiple Queries and the other steps required.
 def largestSquare ( matrix , R , C , q_i , q_j , K , Q ) :
	countDP = [ [ 0 for x in range ( C ) ] for x in range ( R ) ]
	countDP [ 0 ] [ 0 ] = matrix [ 0 ] [ 0 ]
	for i in range ( 1 , R ) :
		countDP [ i ] [ 0 ] = ( countDP [ i - 1 ] [ 0 ] + matrix [ i ] [ 0 ] )
	for j in range ( 1 , C ) :
		countDP [ 0 ] [ j ] = ( countDP [ 0 ] [ j - 1 ] + matrix [ 0 ] [ j ] )
	for i in range ( 1 , R ) :
		for j in range ( 1 , C ) :
			countDP [ i ] [ j ] = ( matrix [ i ] [ j ] + countDP [ i - 1 ] [ j ] + countDP [ i ] [ j - 1 ] - countDP [ i - 1 ] [ j - 1 ] )
	for q in range ( 0 , Q ) :
		i = q_i [ q ]
		j = q_j [ q ]
		min_dist = min ( i , j , R - i - 1 , C - j - 1 )
		ans = - 1
		for k in range ( 0 , min_dist + 1 ) :
			x1 = i - k
			x2 = i + k
			y1 = j - k
			y2 = j + k
			count = countDP [ x2 ] [ y2 ] ;
			if ( x1 > 0 ) :
				count -= countDP [ x1 - 1 ] [ y2 ]
			if ( y1 > 0 ) :
				count -= countDP [ x2 ] [ y1 - 1 ]
			if ( x1 > 0 and y1 > 0 ) :
				count += countDP [ x1 - 1 ] [ y1 - 1 ]
			if ( count > K ) :
				break
			ans = 2 * k + 1
		print ( ans )
matrix = [ [ 1 , 0 , 1 , 0 , 0 ] , [ 1 , 0 , 1 , 1 , 1 ] , [ 1 , 1 , 1 , 1 , 1 ] , [ 1 , 0 , 0 , 1 , 0 ] ]
K = 9
Q = 1
q_i = [ 1 ]
q_j = [ 2 ]
largestSquare ( matrix , 4 , 5 , q_i , q_j , K , Q )