it denotes both pointer at same position and we don 't  have sufficient char to make  palindrome string
if left == right : NEW_LINE INDENT ans = False NEW_LINE break NEW_LINE DEDENT else : NEW_LINE INDENT for j in range ( right , n - left - 1 ) : NEW_LINE INDENT ( s [ j ] , s [ j + 1 ] ) = ( s [ j + 1 ] , s [ j ] ) NEW_LINE count += 1 NEW_LINE DEDENT DEDENT if ans : NEW_LINE return ( count ) NEW_LINE else : NEW_LINE return - 1 NEW_LINE