0 votes 0 votes c=0 for i $\leftarrow$ 1 to n for j $\leftarrow$ i+1 to n for k $\leftarrow$ 1 to j c = c + 1 The value of c is O(n$^{2}$) O(n$^{2}$.log n) O(n$^{3}$) O(n$^{3}$.log n) Algorithms ace-test-series time-complexity + – Rishabh Agrawal asked Jan 18, 2019 retagged Jun 24, 2022 by makhdoom ghaya Rishabh Agrawal 295 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments muthu kumar commented Jan 19, 2019 i edited by muthu kumar Jan 19, 2019 reply Follow Share I think C is correct. some one verify! Variables n=1 n=2 n=3 T(N) = O(n*n*n) i 1 2 3 n j 0 1 2 (n-1) k 0 1 3+2 (n-1)+(n-2)+(n-3)+.....+1 0 votes 0 votes Rishabh Agrawal commented Jan 19, 2019 reply Follow Share And for third loop 'k'? 0 votes 0 votes muthu kumar commented Jan 19, 2019 reply Follow Share Now check bro 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes i think o(n3) is the answer Ruchi Vora answered Jun 5, 2019 Ruchi Vora comment Share Follow See all 0 reply Please log in or register to add a comment.