Recurrence relation: T(n) = T(n/2) + n -


hi there trying solve following recurrence relation telescoping stuck on last step.

t(n) = t(n/2) + n              t(1)=0 t(n/2) = t(n/4) + n/2 t(n/4) = t(n/8) + n/4 ... t(2) = t(1) + 2  t(n)= t(1) + n + n/2 + n/4

i think answer nlogn don't know how interpret above nlogn. can see doing logn steps n come from?

you have done absolutely correctly, not able find sum. got: n + n/2 + n/4 + ..., equal n * (1 + 1/2 + 1/4 + ...).

you got sum of geometric series, equal 2. therefore sum 2n. complexity o(n).

p.s. not called telescoping. telescoping in math when subsequent terms cancel each other.


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