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Count special palindromes in a String | Function to count special Palindromic susbstring.
store count of special Palindromic substring.
it will store the count of continues same char.
traverse string character from left to right.
store same character count.
count smiler character.
Case : 1 so total number of substring that we can generate are : K * ( K + 1 ) / 2 here K is sameCharCount.
store current same char count in sameChar [ ] array.
increment i.
Case 2 : Count all odd length Special Palindromic substring.
if current character is equal to previous one then we assign Previous same character count to current one.
case 2 : odd length.
subtract all single length substring.
Driver Code"
Solution in Python:

Can you provide a code to count special palindromes in a String and the other steps required?
assistant
#Here is code to count special palindromes in a String and the other steps required.
 def CountSpecialPalindrome ( str ) :
	n = len ( str ) ;
	result = 0 ;
	sameChar = [ 0 ] * n ;
	i = 0 ;
	while ( i < n ) :
		sameCharCount = 1 ;
		j = i + 1 ;
		while ( j < n ) :
			if ( str [ i ] != str [ j ] ) :
				break ;
			sameCharCount += 1 ;
			j += 1 ;
		result += int ( sameCharCount * ( sameCharCount + 1 ) / 2 ) ;
		sameChar [ i ] = sameCharCount ;
		i = j ;
	for j in range ( 1 , n ) :
		if ( str [ j ] == str [ j - 1 ] ) :
			sameChar [ j ] = sameChar [ j - 1 ] ;
		if ( j > 0 and j < ( n - 1 ) and ( str [ j - 1 ] == str [ j + 1 ] and str [ j ] != str [ j - 1 ] ) ) :
			result += ( sameChar [ j - 1 ] if ( sameChar [ j - 1 ] < sameChar [ j + 1 ] ) else sameChar [ j + 1 ] ) ;
	return result - n ;
str = " abccba " ;
print ( CountSpecialPalindrome ( str ) ) ;