human
Minimum Number of Manipulations required to make two Strings Anagram Without Deletion of Character | Counts the no of manipulations required.
store the count of character.
iterate though the first String and update count.
iterate through the second string update char_count . if character is not found in char_count then increase count.
Driver code"
Solution in Python:
assistant
#Here is code to minimum Number of Manipulations required to make two Strings Anagram Without Deletion of Character and the other steps required.
 def countManipulations ( s1 , s2 ) :
	count = 0
	char_count = [ 0 ] * 26
	for i in range ( 26 ) :
		char_count [ i ] = 0
	for i in range ( len ( s1 ) ) :
		char_count [ ord ( s1 [ i ] ) - ord ( ' a ' ) ] += 1
	for i in range ( len ( s2 ) ) :
		char_count [ ord ( s2 [ i ] ) - ord ( ' a ' ) ] -= 1
	for i in range ( 26 ) :
		if char_count [ i ] != 0 :
			count += abs ( char_count [ i ] )
	return count
if __name__ == " _ _ main _ _ " :
	s1 = " ddcf "
	s2 = " cedk "
	print ( countManipulations ( s1 , s2 ) )