Asymptotic Analysis (Solved Problem 1).

At this certain version, i is equals to n by 2 plus k minus one, which is equivalent to n. Now, allow me reword this over below in an appropriate manner.This is n by 2 plus k minus one equivalent to n. Let ' s address this. I can take the least top bound as n without any kind of issue, Loop carries out n by 2 plus one, which can be believed of as n by 2 times just. It is clear from this reality that this loophole is running n by 2 times.I can take the

top bound as n and can plainly claim that loophole is running with the time intricacy large O of n. So, this is the top bound I can take, which is n rather of n by 2.

At this certain version, i is equals to n by 2 plus k minus one, which is equivalent to n. Now, allow me reword this over right here in an appropriate manner.This is n by 2 plus k minus one equivalent to n. Let ' s fix this. By resolving this, this comes to be k minus one equivalent to 2 n minus n split by 2. I can take the least top bound as n without any kind of trouble, Loop implements n by 2 plus one, which can be believed of as n by 2 times just. It is clear that in model number k, j is equivalent to k which is equivalent to n by two.Why I am composing this as n by 2? It is clear from this truth that this loophole is running n by 2 times.I can take the

top bound as n and can plainly state that loophole is running with the time intricacy large O of n. So, this is the top bound I can take, which is n rather of n by 2.