Given array of numbers $A[1..n]$. We search for element $x$ with linear search. Also we know that, element $x$ occurs in $A$ and probability of $x$ be in second half of the array is $3$ times of probability $x$ be in first half. What is the time complexity of linear search in average case?

Because of $x$ is in $A$, let $p$ be probability $x$ be in first half, so $$p+3p=1$$ so $p=\frac{1}{4}.$

Also I define Random variable $X$ as follow:

$$ X_i = \begin{cases} \frac{1}{4}& ,1\leq i\leq \frac{n}{2} \\ \frac{3}{4}& ,\frac{n}{2}<i\leq n \end{cases} $$ Let $$X=X_1+X_2+\dots+X_n.$$ Now the expected of $X$ is as follow

$$ E[X]=\frac{1}{2n}\sum_{i=1}^{\frac{n}{2}}X_i+\frac{3}{2n}\sum_{i=\frac{n}{2}+1}^{n} X_i $$ After solving above equality, $E[X]=\frac{10n+8}{16}$. But I think the answer is $\frac{5n}{8}$. What is my wrong?

  • $\begingroup$ Random variable defined with brace has only 2 different value, while in sum below it has $n$ different values. $\endgroup$
    – zkutch
    Commented May 21, 2022 at 14:08
  • $\begingroup$ @zkutch How I can fix it? $\endgroup$
    – jhjhb
    Commented May 21, 2022 at 14:10
  • $\begingroup$ Write down separately values and probabilities for $X$. What you wrote are probabilities only. $\endgroup$
    – zkutch
    Commented May 21, 2022 at 14:11
  • $\begingroup$ @zkutch I fix it. Are you agree? $\endgroup$
    – jhjhb
    Commented May 21, 2022 at 14:22
  • $\begingroup$ You need to write only values for random variable, not values times probabilities. This last is used in formula for expected value. $\endgroup$
    – zkutch
    Commented May 21, 2022 at 14:58

1 Answer 1


We must assume $n$ even. As the average of an arithmetic progression is the average of the extreme members, the number of comparisons for the first half is $\dfrac{n+2}4$, and that for the second half is $\dfrac{3n+2}4$.

Then taking the weighted sum,


Anyway, we can spare the very last comparison because we know that the element is in the array and we must deduct $\dfrac3{4n}$.

  • $\begingroup$ Can you explain more about your idea? Thank you. $\endgroup$
    – jhjhb
    Commented May 21, 2022 at 15:16
  • $\begingroup$ @jhjhb: which "idea", this is just a computation ? What is unclear ? $\endgroup$
    – user16034
    Commented May 21, 2022 at 15:17
  • $\begingroup$ You are agree that my idea is correct? $\endgroup$
    – jhjhb
    Commented May 21, 2022 at 15:19
  • $\begingroup$ But the i have doubt about that the answer is $\frac{3n}{4}$ or $\frac{5n}{8}$. $\endgroup$
    – jhjhb
    Commented May 21, 2022 at 15:20
  • $\begingroup$ @jhjhb: IMO, neither. By the way, there is no $\frac{3n}4$. $\endgroup$
    – user16034
    Commented May 21, 2022 at 16:12

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