human
Make a lexicographically smallest palindrome with minimal changes | Function to create a palindrome.
Count the occurrences of every character in the string.
Create a string of characters with odd occurrences.
Change the created string upto middle element and update count to make sure that only one odd character exists ..
decrease the count of character updated.
Create three strings to make first half second half and middle one ..
characters with even occurrences.
fill the first half ..
character with odd occurrence.
fill the first half with half of occurrence except one.
For middle element.
create the second half by reversing the first half ..
Driver Code"
Solution in Python:
 ... Give me a code to make a lexicographically smallest palindrome with minimal changes and the other steps required?
assistant
#Here is code to make a lexicographically smallest palindrome with minimal changes and the other steps required.
 def palindrome ( s : str , n : int ) -> int :
	cnt = dict ( )
	R = [ ]
	for i in range ( n ) :
		a = s [ i ]
		if a in cnt :
			cnt [ a ] += 1
		else :
			cnt [ a ] = 1
	i = ' a '
	while i <= ' z ' :
		if i in cnt and cnt [ i ] % 2 != 0 :
			R += i
		i = chr ( ord ( i ) + 1 )
	l = len ( R )
	j = 0
	for i in range ( l - 1 , ( l // 2 ) - 1 , - 1 ) :
		if R [ i ] in cnt :
			cnt [ R [ i ] ] -= 1
		else :
			cnt [ R [ i ] ] = - 1
		R [ i ] = R [ j ]
		if R [ j ] in cnt :
			cnt [ R [ j ] ] += 1
		else :
			cnt [ R [ j ] ] = 1
		j += 1
	first , middle , second = " " , " " , " "
	i = ' a '
	while i <= ' z ' :
		if i in cnt :
			if cnt [ i ] % 2 == 0 :
				j = 0
				while j < cnt [ i ] // 2 :
					first += i
					j += 1
			else :
				j = 0
				while j < ( cnt [ i ] - 1 ) // 2 :
					first += i
					j += 1
				middle += i
		i = chr ( ord ( i ) + 1 )
	second = first
	second = ' ' . join ( reversed ( second ) )
	resultant = first + middle + second
	print ( resultant )
if __name__ == " _ _ main _ _ " :
	S = " geeksforgeeks "
	n = len ( S )
	palindrome ( S , n )