PROBLEM LINK:

Author:Utkarsh Saxena
Testers:Jakub Safin
Editorialist:Praveen Dhinwa

DIFFICULTY:

easy-medium

PREREQUISITES:

probability, expected value

PROBLEM:

Consider a string s = "aaabbaa". Let us call maximal sequence of continuous ones as a group. You have to find expected the number of groups in a string of length $N$ and alphabet size $K$. The final output required in the problem was twice of this value.

SOLUTION

Satyaki's solution.

Let $E(N, K)$ be the answer for given $N$ and $K$.

The recurrence is - $$E(1, K) = 1$$ We can keep the $N$-th character same as $N-1$-th character with probability $\frac{1}{K}$, in this case, the $E(N, K)$ remains same as of $E(N-1, K)$. In the other case when the $N$-th character is different than the $N-1$-th character, the number of groups will be 1 more.

$$E(N, K) = \frac{1}{K} E(N-1, K) \, + \, (1 - \frac{1}{K}) (1 + E(N-1, K))$$ $$= E(N-1, K) + (1 - \frac{1}{K})$$ $$= E(1, K) + (1 - \frac{1}{K}) * (N-1)$$ $$= N - \frac{N-1}{K}$$

Linearty of expectation based solution

The number of maximal groups is 1 + the number of character changes. The probability of each of change is $\frac{K-1}{K}$. By linearity of expectation, the number of groups will be $(N-1) \frac{K-1}{K}$.

So, the answer will be $1 + (N-1) \frac{K-1}{K}$.

Combintorics based solution

The number of strings for which the number of groups is $g$ is given by $K * (K - 1)^{g - 1} * \binom{N - 1}{g - 1}$. The combination $\binom{N - 1}{g - 1}$ denotes the selection of $g-1$ places where the character will change. The character of the first group can be selected in $K$ ways, then the second group in $K-1$ ways, third $K-1$ ways, and so on till the $g$-th group.

So, the expected number of groups will be $$\sum\limits_{g = 1}^{N} {g \cdot \frac{K * (K - 1)^{g - 1} * \binom{N - 1}{g - 1}}{K^N}}$$ $$\frac{1}{K^{N-1}} \sum\limits_{g = 1}^{N} {g * (K - 1)^{g - 1} * \binom{N - 1}{g - 1}}$$ By change of variable $g \rightarrow g + 1$ $$\frac{1}{K^{N-1}} \sum\limits_{g = 0}^{N - 1} {(g + 1) * (K - 1)^{g} * \binom{N - 1}{g}}$$

The term $$\sum\limits_{g = 0}^{N - 1} {(g + 1) * (K - 1)^{g} * \binom{N - 1}{g}} = \sum\limits_{g = 0}^{N - 1} {g * (K - 1)^{g} * \binom{N - 1}{g}} + \sum\limits_{g = 0}^{N - 1} {(K - 1)^{g} * \binom{N - 1}{g}}$$

Binomial theorem states $$(1 + x)^n = \sum\limits_{r = 0}^{n} \binom{n}{r} x^r$$ In this formula, if we replace $x$ by $K-1$, and $n$ by $N-1$, we get the second term of the above summation. So, the second term becomes $K^{N-1}$.

Let us see how to calculate the first term. By differentiating the binomial theorem by both the sides, we get.

$$n (1 + x)^{n-1} = \sum\limits_{r = 1}^{n} \binom{n}{r} r x^{r-1}$$

Multiplying both the sides by $x$, we get $$n x (1 + x)^{n-1} = \sum\limits_{r = 1}^{n} \binom{n}{r} r x^r$$ $$n x (1 + x)^{n-1} = \sum\limits_{r = 0}^{n} \binom{n}{r} r x^r$$

So, by replacing $x$ with $K-1$ and $n$ by $N-1$ again, we get the first term of the summation

$$(N-1) \cdot (K-1) \cdot K^{N-2}$$

So, the expected number of groups turn out to be $$1 + (N - 1) \frac{K-1}{K}$$.