# 1 Solving Recurrences - Stanford University

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tant time. Lines 4 and 5 take T (bn/2c)and T (dn/2e) time, since each of the subproblems has that many elements. The FindMaxCrossingSubarray procedure takes Θ(n) time, and the rest of the code takes Θ(1) time. SoT (n) Θ(1) T (bn/2c) T (dn/2e) Θ(n) Θ(1) 2T (n/2) Θ(n) (ignoring the floorsand ceilings).By case 2 of the master theorem, this recurrence has the solution T (n) Θ(n log n).8

CS 161 Lecture 333.1Jessica Su (some parts copied from CLRS)If we have extra timeChange of variables (substitution method) T (n) 2T (b nc) log n.For now don’t worry about rounding off values to be integers.Let m log n, i.e. n 2m . Then T (2m ) 2T (2m/2 ) m.Now let S(m) T (2m ). Then S(m) 2S(m/2) m.This recurrence has the solution S(m) O(m log m).So T (n) T (2m ) S(m) O(m log m) O(log n log log n).3.2Extra recursion tree problemConsider the recurrence T (n) T (n/3) T (2n/3) O(n). Let c represent the constantfactor in the O(n) term.The longest simple path from the root to a leaf is n (2/3)n (2/3)2 n · · · 1. Since(2/3)k n 1 when k log3/2 n, the height of the tree is log3/2 n.We get that each level costs at most cn, but as we go down from the root, more and moreinternal nodes are absent, so the costs become smaller. Fortunately we only care about an9

CS 161 Lecture 3Jessica Su (some parts copied from CLRS)upper bound.Based on this we can guess O(n log n) as an upper bound, and verify this by the substitutionmethod.when d c/(log 3 (2/3)).Note that we should be proving that the claim holds for the base case as well. You shoulddo this in your homework.10

and when to sell to maximize your pro t. Note that you cannot sell before you buy (so you can’t just sell on the highest day and buy on the lowest day). Naive strategy: Try all pairs of (buy, sell) dates, where the buy date must be before the sell date. This takes ( n2) time. bestProfit -MAX_INT bestBuyDate None

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