human
Split a string in equal parts such that all parts are palindromes | Function to return the frequency array for the given string.
Function to return the required count.
Add frequencies of the even appearing characters.
Count of the characters that appeared odd number of times.
If there are no characters with odd frequency.
If there are no characters with even frequency.
Only a single character with odd frequency.
More than 1 character with odd frequency string isn 't a palindrome.
All odd appearing characters can also contribute to the even length palindrome if one character is removed from the frequency leaving it as even.
If k palindromes are possible where k is the number of characters with odd frequency.
Current character can no longer be an element in a string other than the mid character.
If current character has odd frequency > 1 take two characters which can be used in any of the parts.
Update the frequency.
If not possible , then every character of the string will act as a separate palindrome.
Driver code"
Solution in Python:
 ... Can you provide a code to split a string in equal parts such that all parts are palindromes and the other steps required?
assistant
#Here is code to split a string in equal parts such that all parts are palindromes and the other steps required.
 def getFrequencies ( string ) :
	freq = [ 0 ] * 26
	for i in range ( len ( string ) ) :
		freq [ ord ( string [ i ] ) - ord ( ' a ' ) ] += 1
	return freq
def countMinParts ( string ) :
	n = len ( string )
	freq = getFrequencies ( string )
	oddFreq = [ ]
	evenFreq = [ ]
	sumEven = 0
	for i in range ( 26 ) :
		if freq [ i ] == 0 :
			continue
		if freq [ i ] % 2 == 0 :
			evenFreq . append ( freq [ i ] )
		else :
			oddFreq . append ( freq [ i ] )
	for i in range ( len ( evenFreq ) ) :
		sumEven += evenFreq [ i ]
	if len ( oddFreq ) == 0 :
		return 1
	if sumEven == 0 :
		if len ( oddFreq ) == 1 :
			return 1
		return 0
	i = 0
	while ( i < len ( oddFreq ) ) :
		if ( ( sumEven / 2 ) % len ( oddFreq ) == 0 ) :
			return len ( oddFreq )
		if ( oddFreq [ i ] == 1 ) :
			i += 1
			continue
		sumEven += 2
		oddFreq [ i ] = oddFreq [ i ] - 2
	return n
if __name__ == " _ _ main _ _ " :
	s = " noonpeep "
	print ( countMinParts ( s ) )