given a particular split point , based on b0 , find the number of left 1s needed to satisfy the left condition more than half 0s . that is , for a split point at i for the section before we can add 1s until just before the number of zeros contained therein . it is the difference of 0s and 1s at i minus one . concode_field_sep PlaceHolder placeHolder concode_field_sep int num0sr concode_elem_sep int[] initB0 concode_elem_sep int num1sr concode_elem_sep void print concode_elem_sep int last1 concode_elem_sep int right0 concode_elem_sep int solution concode_elem_sep void main concode_elem_sep int first0 concode_elem_sep int num1s concode_elem_sep int num0s
int function ( int [ ] arg0 , int arg1 ) { return Math . max ( 0 , num0s ( arg0 , arg1 ) - num1s ( arg0 , arg1 ) - 1 ) ; }