human
Smallest submatrix with Kth maximum XOR | Function to print smallest index of Kth maximum Xor value of submatrices.
Dimensions of matrix.
Stores XOR values for every index.
Min heap to find the kth maximum XOR value.
Stores indices for corresponding XOR values.
Traversing matrix to calculate XOR values.
Insert calculated value in Min Heap.
If size exceeds k.
Remove the minimum.
Store smallest index containing xor [ i , j ].
Stores the kth maximum element.
Print the required index.
Driver code.
Function call"
How can the above be solved in Python?
assistant
#Here is code to smallest submatrix with Kth maximum XOR and the other steps required.
 def smallestPosition ( m , k ) :
	n = len ( m )
	mm = len ( m [ 1 ] )
	xor = [ [ 0 for i in range ( mm ) ] for j in range ( n ) ]
	minHeap = [ ]
	Map = { }
	for i in range ( n ) :
		for j in range ( mm ) :
			if i - 1 >= 0 :
				a = xor [ i - 1 ] [ j ]
			else :
				a = 0
			if j - 1 >= 0 :
				b = xor [ i ] [ j - 1 ]
			else :
				b = 0
			if i - 1 >= 0 and j - 1 >= 0 :
				c = xor [ i - 1 ] [ j - 1 ]
			else :
				c = 0
			xor [ i ] [ j ] = m [ i ] [ j ] ^ a ^ b ^ c
			minHeap . append ( xor [ i ] [ j ] )
			minHeap . sort ( )
			if ( len ( minHeap ) > k ) :
				del minHeap [ 0 ]
			if xor [ i ] [ j ] not in Map :
				Map [ xor [ i ] [ j ] ] = [ i , j ]
	minHeap . sort ( )
	kth_max_e = minHeap [ 0 ]
	print ( ( Map [ kth_max_e ] [ 0 ] + 1 ) , ( Map [ kth_max_e ] [ 1 ] + 1 ) )
m = [ [ 1 , 2 , 3 ] , [ 2 , 2 , 1 ] , [ 2 , 4 , 2 ] ]
k = 1
smallestPosition ( m , k )