text: If previous character is 1 then the i - 1 th character and ith character can be decoded in a single character therefore , adding dp [ i - 1 ] .
code: if ( s [ i - 1 ] == '1' ) : NEW_LINE INDENT dp [ i + 1 ] = ( dp [ i + 1 ] + dp [ i - 1 ] ) % M NEW_LINE DEDENT