0 votes

How to solve such kind of questions ?

Can anybody tell what's is the concept behind this ??

someone provide me link so that I read it and understand the actual concept

0

Yes its should be 6 i forgot about the last access..

**My thought:**

In load factor **$\frac{n}{m}$** => I have total $'n'$ numbers inserted into a hash table which has $'m'$ slots.

So if we have to declare it as an unsuccessful probe, in the worst case scenario all the $'5'$ elements are one after the other i.e. we have to probe all those $n slots$ first where you will find all are mismatched and in the next attempt you find the blank after seeing this we say that the element is not in the hashtable

So total will be **n+1**=> worst case for unsuccessful attempts.

0 votes

Number of probes in unsuccessful search = $\displaystyle \frac{1}{1-x}$

where x= load factor of hash table

since, x = $\displaystyle \frac{5}{6 }$

Therefore using above formula , x= $\displaystyle \frac{5}{6 }$ , as no. of probes in unsuccessful search= 6.

where x= load factor of hash table

since, x = $\displaystyle \frac{5}{6 }$

Therefore using above formula , x= $\displaystyle \frac{5}{6 }$ , as no. of probes in unsuccessful search= 6.