human
Check if a string can be split into two strings with same number of K | Function to print the arrangement of characters.
Stores frequency of characters.
Count the character having frequency K.
Count the character having frequency greater than K and not equal to 2 K.
Case 1.
Case 2.
Case 3.
If all cases fail.
Driver Code"
How can the above be solved in Python?
 ... Give me a code to check if a string can be split into two strings with same number of K and the other steps required?
assistant
#Here is code to check if a string can be split into two strings with same number of K and the other steps required.
 def DivideString ( s , n , k ) :
	c = 0
	no = 1
	c1 = 0
	c2 = 0
	fr = [ 0 ] * 26
	ans = [ ]
	for i in range ( n ) :
		fr [ ord ( s [ i ] ) - ord ( ' a ' ) ] += 1
	for i in range ( 26 ) :
		if ( fr [ i ] == k ) :
			c += 1
		if ( fr [ i ] > k and fr [ i ] != 2 * k ) :
			c1 += 1
			ch = chr ( ord ( ' a ' ) + i )
		if ( fr [ i ] == 2 * k ) :
			c2 += 1
			ch1 = chr ( ord ( ' a ' ) + i )
	for i in range ( n ) :
		ans . append ( "1" )
	mp = { }
	if ( c % 2 == 0 or c1 > 0 or c2 > 0 ) :
		for i in range ( n ) :
			if ( fr [ ord ( s [ i ] ) - ord ( ' a ' ) ] == k ) :
				if ( s [ i ] in mp ) :
					ans [ i ] = '2'
				else :
					if ( no <= ( c // 2 ) ) :
						ans [ i ] = '2'
						no += 1
						mp [ s [ i ] ] = 1
		if ( c % 2 == 1 and c1 > 0 ) :
			no = 1
			for i in range ( n ) :
				if ( s [ i ] == ch and no <= k ) :
					ans [ i ] = '2'
					no += 1
		if ( c % 2 == 1 and c1 == 0 ) :
			no = 1
			flag = 0
			for i in range ( n ) :
				if ( s [ i ] == ch1 and no <= k ) :
					ans [ i ] = '2'
					no += 1
				if ( fr [ s [ i ] - ' a ' ] == k and flag == 0 and ans [ i ] == '1' ) :
					ans [ i ] = '2'
					flag = 1
		print ( " " . join ( ans ) )
	else :
		print ( " NO " )
if __name__ == ' _ _ main _ _ ' :
	S = " abbbccc "
	N = len ( S )
	K = 1
	DivideString ( S , N , K )