human
Maximize median of a KxK sub | Function to determine if a given value can be median.
Stores the prefix sum array.
Traverse the matrix arr [ ] [ ].
Update Pre [ i + 1 ] [ j + 1 ].
If arr [ i ] [ j ] is less than or equal to mid.
Stores the count of elements should be less than mid.
Stores if the median mid can be possible or not.
Iterate over the range [ K , N ].
Iterate over the range [ K , N ].
Stores count of elements less than or equal to the value mid in submatrix with bottom right vertices at ( i , j ).
If X is less than or equal to required.
Return flag.
Function to find the maximum median of a subsquare of the given size.
Stores the range of the search space.
Iterate until low is less than high.
Stores the mid value of the range [ low , high ].
If the current median can be possible.
Update the value of low.
Update the value of high.
Return value stored in low as answer.
Driver Code"
How can the above be solved in Python?
 What is a code to maximize median of a KxK sub and the other steps required?
assistant
#Here is code to maximize median of a KxK sub and the other steps required.
 def isMaximumMedian ( arr , N , K , mid ) :
	Pre = [ [ 0 for x in range ( N + 5 ) ] for y in range ( N + 5 ) ]
	for i in range ( N ) :
		for j in range ( N ) :
			Pre [ i + 1 ] [ j + 1 ] = ( Pre [ i + 1 ] [ j ] + Pre [ i ] [ j + 1 ] - Pre [ i ] [ j ] )
			if ( arr [ i ] [ j ] <= mid ) :
				Pre [ i + 1 ] [ j + 1 ] += 1
	required = ( K * K + 1 ) // 2
	flag = 0
	for i in range ( K , N + 1 ) :
		for j in range ( K , N + 1 ) :
			X = ( Pre [ i ] [ j ] - Pre [ i - K ] [ j ] - Pre [ i ] [ j - K ] + Pre [ i - K ] [ j - K ] )
			if ( X < required ) :
				flag = 1
	return flag
def maximumMedian ( arr , N , K ) :
	low = 0
	high = 1000000009
	while ( low < high ) :
		mid = low + ( high - low ) // 2
		if ( isMaximumMedian ( arr , N , K , mid ) ) :
			low = mid + 1
		else :
			high = mid
	return low
if __name__ == " _ _ main _ _ " :
	arr = [ [ 1 , 5 , 12 ] , [ 6 , 7 , 11 ] , [ 8 , 9 , 10 ] ]
	N = len ( arr )
	K = 2
	print ( maximumMedian ( arr , N , K ) )