human
Place the prisoners into cells to maximize the minimum difference between any two | Function that returns true if the prisoners can be placed such that the minimum distance between any two prisoners is at least sep.
Considering the first prisoner is placed at 1 st cell.
If the first prisoner is placed at the first cell then the last_prisoner_placed will be the first prisoner placed and that will be in cell [ 0 ].
Checking if the prisoner can be placed at ith cell or not.
If all the prisoners got placed then return true.
Function to return the maximized distance.
Sort the array so that binary search can be applied on it.
Minimum possible distance for the search space.
Maximum possible distance for the search space.
To store the result.
Binary search.
If the prisoners can be placed such that the minimum distance between any two prisoners is at least mid.
Update the answer.
Driver code"
How can the above be solved in Python?

What is a code to place the prisoners into cells to maximize the minimum difference between any two and the other steps required?
assistant
#Here is code to place the prisoners into cells to maximize the minimum difference between any two and the other steps required.
 def canPlace ( a , n , p , sep ) :
	prisoners_placed = 1
	last_prisoner_placed = a [ 0 ]
	for i in range ( 1 , n ) :
		current_cell = a [ i ]
		if ( current_cell - last_prisoner_placed >= sep ) :
			prisoners_placed += 1
			last_prisoner_placed = current_cell
			if ( prisoners_placed == p ) :
				return True
	return False
def maxDistance ( cell , n , p ) :
	cell = sorted ( cell )
	start = 0
	end = cell [ n - 1 ] - cell [ 0 ]
	ans = 0
	while ( start <= end ) :
		mid = start + ( ( end - start ) // 2 )
		if ( canPlace ( cell , n , p , mid ) ) :
			ans = mid
			start = mid + 1
		else :
			end = mid - 1
	return ans
cell = [ 1 , 2 , 8 , 4 , 9 ]
n = len ( cell )
p = 3
print ( maxDistance ( cell , n , p ) )