0 votes 0 votes Algorithms algorithms time-complexity + – Hardik Vagadia asked Aug 2, 2015 • retagged Jun 30, 2022 by makhdoom ghaya Hardik Vagadia 3.0k views answer comment Share Follow See 1 comment See all 1 1 comment reply Prashant. commented Mar 21, 2018 reply Follow Share 100n2 < 2n For n= 15, 100n2 run faster than 2n i.e. 22500 < 32768 0 votes 0 votes Please log in or register to add a comment.
Best answer 9 votes 9 votes At $n=14$, $2^n-100n^2 = 2^{14}-100*14^2 = -3216$ At $n=15$, $2^n-100n^2 = 2^{15}-100*15^2 = 10268$ So at $n=15$, $2^n$ becomes greater than $100n^2$ Happy Mittal answered Aug 2, 2015 • selected Aug 2, 2015 by Arjun Happy Mittal comment Share Follow See all 2 Comments See all 2 2 Comments reply Hardik Vagadia commented Aug 2, 2015 reply Follow Share But this becomes a preety easier when we use C/C++. are we allowed to do so in GATE :P ? 0 votes 0 votes Happy Mittal commented Aug 2, 2015 reply Follow Share No, just use calculator and a bit of binary search manually, I found it through that only. 1 votes 1 votes Please log in or register to add a comment.