Suppose we are given n keys, m hash table slots, and two simple uniform hash functions h1 and h2. Further, suppose our hashingscheme uses h1 for the odd keys and h2 for the even keys. What is the expected number of keys in a slot?*1 pointn/2mn/mm/n2n/m

To find the expected number of keys in a slot, we need to consider the probability distribution of the keys.

Since we have n keys and m hash table slots, each slot can be occupied by either an odd key or an even key.

Let's first consider the odd keys. Since we are using h1 for the odd keys, the probability of an odd key being hashed to a particular slot is 1/m.

Similarly, for the even keys, which are hashed using h2, the probability of an even key being hashed to a particular slot is also 1/m.

Now, let's calculate the expected number of odd keys in a slot. Since there are n/2 odd keys, the expected number of odd keys in a slot is (n/2) * (1/m) = n/(2m).

Similarly, the expected number of even keys in a slot is also n/(2m).

Therefore, the expected number of keys in a slot is the sum of the expected number of odd keys and the expected number of even keys, which is (n/(2m)) + (n/(2m)) = n/m.

Hence, the correct answer is n/m.

To find the expected number of keys in a slot, we need to consider the probability distribution of the keys.

Since we have n keys and m hash table slots, each slot can be occupied by either an odd key or an even key.

Let's first consider the odd keys. Since we are using h1 for the odd keys, the probability of an odd key being hashed to a particular slot is 1/m.

Similarly, for the even keys, since we are using h2 for the even keys, the probability of an even key being hashed to a particular slot is also 1/m.

Now, let's calculate the expected number of odd keys in a slot. Since there are n/2 odd keys, the expected number of odd keys in a slot is (n/2) * (1/m) = n/(2m).

Similarly, the expected number of even keys in a slot is also n/(2m).

Therefore, the expected number of keys in a slot is the sum of the expected number of odd keys and the expected number of even keys, which is (n/(2m)) + (n/(2m)) = n/m.

Hence, the correct answer is n/m.

To find the expected number of keys in a slot, we need to consider the probability distribution of the keys.

Since we have n keys and m hash table slots, each slot can be occupied by either an odd key or an even key.

Let's first consider the odd keys. Since we are using h1 for the odd keys, the probability of an odd key being hashed to a particular slot is 1/m.

Similarly, for the even keys, since we are using h2 for the even keys, the probability of an even key being hashed to a particular slot is also 1/m.

Now, let's calculate the expected number of odd keys in a slot. Since there are n/2 odd keys, the expected number of odd keys in a slot is (n/2) * (1/m) = n/(2m).

Similarly, the expected number of even keys in a slot is also n/(2m).

Therefore, the expected number of keys in a slot is the sum of the expected number of odd keys and the expected number of even keys, which is (n/(2m)) + (n/(2m)) = n/m.

Hence, the correct answer is n/m.