Talk:TADM2E 3.11

From Algorithm Wiki
Revision as of 19:33, 6 December 2017 by Idealamz (Talk | contribs)

Jump to: navigation, search

For the answer given in part 2, the space required seems to be the sum from 1 to n of n which is n(n+1)/2. This gives O(n**2) for space requirements. Doesn't it?

Why cant we just sort the input (array)? Will this not mean the smallest number between x and y is always x?

a counterexample for 3.11 b

for this input = [0,1,...,98,99] and this query i = 1, j = 99.

the suggested algorithm will always span in both sides - right and left, so according to the suggested solution it will always split the task to 2. that means that if h is the height of the tree the query time is going to be O(2^h)

now h = log(n) so it's 2^log(n) = n

for O(n) cost we can simply iterate over all the values in the given range and find the minimum...