0 votes 0 votes A slight modification to binary search algorithm where it is split into sets of size one thirds and two thirds instead of equal sizes. What will be the recurence relation in worst case A_i_$_h asked Oct 30, 2017 A_i_$_h 216 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Anu007 commented Oct 30, 2017 reply Follow Share T(n) = T(2n/3) +1 i.e. T(n) = T(n/1.5) +1 = O(log3/2n) 1 votes 1 votes Rupendra Choudhary commented Oct 30, 2017 reply Follow Share in worst case , every time 1/3 size array would be eliminated so T(n)=T(2n/3)+c and in best case every time array of size 2/3 is eliminated so T(n)=T(n/3)+c 2 votes 2 votes Please log in or register to add a comment.