text: Binary search to find the number of possible values of the first element
code: while ( l <= r ) { var m = ( l + r ) / 2 ; if ( indices [ x ] [ m ] < i ) { first = m ; l = m + 1 ; } else { r = m - 1 ; } } l = 0 ; r = indices [ z ] . length - 1 ; var third = - 1 ;