Quiz 02

Let $f(n) = 3n^2 \log n$. True or False: $f(n) = O(n^3)$.

True False

Solution

True.

Since $\log n$ grows more slowly than $n$, we have $3n^2 \log n \leq 3n^2 \cdot n = 3n^3$ for all $n \geq 1$, so $f(n) = O(n^3)$. (In fact, $f(n) = \Theta(n^2 \log n)$.)

Problem #033

Consider the function $f(n) = 5n^2 - 100n + 100.$ Which of the following asymptotic bounds are true? Choose all that apply.

Solution

$\Omega(n)$, $\Theta(n^2)$, $O(n^2)$, $\Omega(n^2)$, and $O(n^3)$. Since $f(n) = \Theta(n^2)$, it is $O$ of anything that grows at least as fast as $n^2$ and $\Omega$ of anything that grows at most as fast as $n^2$. It is not $O(n)$ (and so not $\Theta(n)$), and it is not $\Omega(n^3)$ (and so not $\Theta(n^3)$).

Problem #034

Let $f(n) = \displaystyle\frac{n^3 - 2n^2 + 100}{n + 10}$ True or False: $f(n) = O(n^4)$

True False

Solution

True.

The leading term of the numerator is $n^3$ and the leading term of the denominator is $n$, so $f(n) = \Theta(n^2)$. Anything that is $\Theta(n^2)$ is also $O(n^4)$, since $O$ only gives an upper bound.

Problem #035

Consider the function defined below:

\[ f(n) = \begin{cases} 3n, & \text{if $n$ is even;}\\ 5n ,& \text{if $n$ is odd.} \end{cases}\]

True or False: $f(n) = \Theta(n)$.

True False

Solution

True.

For every $n$, $3n \leq f(n) \leq 5n$, so $f(n)$ is bounded above and below by constant multiples of $n$. It doesn't matter that $f$ jumps between two different constants.

Problem #036

Consider the function $f(n) = \sin(12 n) \cdot\cos(n) + 2$. A plot of this function is shown below:

True or False: this function is $\Theta(1)$.

True False

Solution

True. This function is upper bounded by 3 and lower bounded by 1 for all $n$, and is therefore $\Theta(1)$.

Problem #037

Consider the function $f(n) = \sin(12n) \cdot\cos(n) + 2$. A plot of this function is shown below:

True or False: $f(n) = O(n^3)$.

True False

Solution

True. We saw that this function is $\Theta(1)$, but it is also correct to say that it is $O(n^3)$, though this is not a tight upper bound.

Problem #038

Suppose Algorithm A takes $\Theta(n^2)$ time, while Algorithm B takes $\Theta(2^n)$ time. True or False: there could exist an input on which Algorithm $B$ takes less time than Algorithm A.

True False

Solution

True.

Asymptotic notation only describes how the time grows for large $n$ and hides constants. For example, if A takes $1000n^2$ steps and B takes $2^n$ steps, then B is faster on every input of size $n = 10$(1024 steps vs. 100,000).

Problem #039

Suppose algorithm A takes $\Theta(n^2)$ time, while algorithm B takes $\Theta(n^3)$ time.

True or False: there must exist an input of size 100 on which Algorithm A takes less time than Algorithm B.

Solution

False. Algorithm A could have really large constants. For example, it could take 1,000,000 $n^2$ seconds to run, while algorithm B takes $.0001 n^3$ seconds to run. Algorithm B will still be better for $n = 100$.

Problem #040

Suppose bar and baz are two functions. Suppose bar's time complexity is $\Theta(n^3)$, while baz's time complexity is $\Theta(n^2)$.

Suppose boo is defined as below:



def boo(n):
    if n < 1000:
        bar(n)
    else:
        baz(n)

What is the asymptotic time complexity of boo?

Solution

Asymptotic time complexity concerns the time taken when $n$ is large. Therefore, it doesn't matter what happens when $n < 1000$. When $n \geq 1000$, the time taken is $\Theta(n^2)$, since that is the time taken by baz.

Problem #041

Suppose bar and baz are two functions. Suppose bar's time complexity is $\Theta(n^3)$, while baz's time complexity is $\Theta(n^2)$.

Suppose foo is defined as below:



def foo(n):
    if n < 1_000:
        bar(n)
    else:
        baz(n)

What is the asymptotic time complexity of foo?

Solution

$\Theta(n^2)$

Problem #042

Suppose bar and baz are two functions. Suppose bar's asymptotic time complexity is $\Theta(n^4)$, while baz's is $\Theta(n)$.

Suppose foo is defined as below:



def foo(n):
    if n < 1_000_000:
        bar(n)
    else:
        baz(n)

What is the asymptotic time complexity of foo?

Solution

$\Theta(n)$ If you were to plot the function $T(n)$ that gives the time taken by foo as a function of $n$, you'd see something like the below:

This function starts off looking like $n^4$, but at $n = 1_000_000$, it "switches" to looking like $n$.

Since asymptotic time complexity is concerned with the behavior of the function as $n$ gets large, we can ignore the part where $n$ is "small" (in this case, less than $1{,}000{,}000$). So, asymptotically, this function is $\Theta(n)$.

Problem #043

Suppose bar_1, bar_2 and bar_3 are three functions. Suppose bar_1's time complexity is $\Theta(n)$, bar_2's time complexity is $\Theta(n^2)$, and bar_3's time complexity is $\Theta(n^3)$.

Suppose foo is defined as below:


def foo(n):
    if n < 2023:
        bar_1(n**3)
    elif n == 2023:
        bar_3(n**2)
    else:
        bar_2(n)

What is the asymptotic time complexity of foo?

Solution

Asymptotic time complexity concerns the time taken when $n$ is large. Therefore, it doesn't matter what happens when $n < 1000$. When $n \geq 1000$, the time taken is $\Theta(n^2)$, since that is the time taken by bar_2.

Problem #044

Suppose bar and baz are two functions. Suppose bar's time complexity is $\Theta(n^2)$, while baz's time complexity is $\Theta(n)$.

Suppose foo is defined as below:



def foo(n):
    # will be True if n is even, False otherwise
    is_even = (n % 2) == 0
    if is_even:
        bar(n)
    else:
        baz(n)

Let $T(n)$ be the time taken by foo on an input of sized $n$. True or False: $T(n) = \Theta(n^2)$.

True False

Solution

False.

This function is not $\Theta(n^2)$. For that matter, it is also not $\Theta(n)$. It is$O(n^2)$ and $\Omega(n)$, though.

This function cannot be $\Theta(n^2)$ because there are no positive constants $c, n_0$ such that $T(n) > c n^2$ for all $n > n_0$. You can see this by imagining the plot of the time taken by foo as a function of $n$. It "oscillates" between something that grows like $n$ and something that grows like $n^2$. If you tried to lower bound it with $cn^2$, $T(n)$ would eventually dip below $cn^2$, since $cn^2$ grows faster than $n$.

Problem #045

Consider the function $f(n) = \frac{n^8 + 2^{3 + \log n} - \sqrt n}{20 n^5 - n^2}$.

True or False: $f(n) = O(n^{4})$.

True False

Solution

True.

Note that $2^{3 + \log n} = 8 \cdot 2^{\log n} = 8n$, so the dominant term in the numerator is $n^8$, while the dominant term in the denominator is $20n^5$. Therefore $f(n) = \Theta(n^3)$, which is $O(n^4)$.

Problem #046

Consider the function $f(n) = \frac{n^8 + 2^{3 + \log n} - \sqrt n}{20 n^5 - n^2}$.

True or False: $f(n) = \Omega(n^2)$.

True False

Solution

True.

Since $2^{3 + \log n} = 8n$, the numerator is dominated by $n^8$ and the denominator by $20n^5$, so $f(n) = \Theta(n^3)$. Anything that is $\Theta(n^3)$ is also $\Omega(n^2)$, since $\Omega$ only gives a lower bound.

Problem #047

Let

\[ f(n) = \frac{n^3 - 2n + 100}{\sqrt n}\cdot\frac{n}{n^3 + n^2 + \sin n}\cdot n^2 \]

True or False: $f(n) = \Theta(n^2)$.

True False

Solution

False.

Keeping only the dominant terms, the first factor is $\Theta(n^3 / \sqrt n) = \Theta(n^{2.5})$ and the second is $\Theta(n / n^3) = \Theta(1/n^2)$. Multiplying by $n^2$ gives $f(n) = \Theta(n^{2.5})$, which is not $\Theta(n^2)$(it is not $O(n^2)$).

Problem #048

Let

\[ f(n) = \frac{n^3 + 10 n}{\sqrt{n} - 100}\times\frac{ \log(n^2 + 10)}{n(n+1)}\]

Write $f(n)$ in asymptotic notation in the simplest terms possible.

Solution

$f(n) = \Theta(\sqrt n \log n)$

Problem #049

Let

\[ f(n) = 5 n \log n + \frac{n^3 + 5}{n + 2 + |\sin \pi n|} + n \sqrt n \]

Write $f$ in asymptotic notation in as simplest terms possible.

Solution

$f(n) = \Theta(n^2)$

Problem #050

Let

Define $f(n) = f_1(n) \times f_2(n) \times f_3(n)$. What is $f(n)$, in asymptotic notation, in simplest terms possible?

Solution

$\Theta(n^3)$

Problem #051

Let $f(n) = n \cdot(\sin{n} + 1 )$. A plot of this function is shown below.

True or False: $f(n) = \Theta(n)$.

True False

Solution

False.

Remember that for $f(n)$ to be $\Theta(n)$, it has to be both upper- and lower-bounded by some positive constant times $n$. This function is upper bounded by $O(n)$, but it doesn't have a lower bound of $\Omega(n)$. Therefore, it can't be $\Theta(n)$.

In other words, there is no positive constant $c$ and no positive number $N$ such that $f(n) > c n$ for all $n > N$.

If you had to describe this function in asymptotic notation, you could say that $f(n) = O(n)$(and that would be a tight upper bound), but there is no lower bound, since the function keeps coming back down to zero.

Problem #052

Consider the function $f(n) = n \times(\sin(n) + 1)$. A plot of this function is shown below:

True or False: this function is $O(1 + \sin n)$.

True False

Solution

False.

$f(n)$ grows arbitrarily large, while $\sin n + 1$ is bounded above by 2.

More formally, there is no constant $c$ such that $c (1 + \sin n)$ upper bounds $f(n)$ for all large $n$.

Problem #053

Consider the function $f(n) = n \times( \sin(n) + 1)$. A plot of this function is shown below:

True or False: $f(n) = O(n^3)$.

True False

Solution

True.

Since $\sin(n) + 1$ is always between 0 and 2, we have $0 \leq f(n) \leq 2n$, so $f(n) = O(n)$ and therefore also $O(n^3)$. (The oscillation only prevents $f$ from being $\Omega(n)$; it doesn't affect the upper bound.)

Problem #054

Let $f$ be the piecewise function defined as:

\[ f(n) = \begin{cases} 1, & \text{if $n$ is a multiple of 1 million},\\ n^2, &\text{otherwise}. \end{cases}\]

If you were to plot $f$, the function would look like $n^2$, but would have point discontinuities where it ``jumps'' back to 1 whenever $n$ is a multiple of 1 million.

True or False: $f(n) = \Theta(n^2)$.

True False

Solution

False.

$f(n)$is upper bounded by $O(n^2)$, but it doesn't have a lower bound of $\Omega(n^2)$, which is necessary for it to be $\Theta(n^2)$.

To see why, remember that for $f(n)$ to be $\Omega(n^2)$, there must be a positive constant $c$ so that $f(n)$ stays above $c\cdot n^2$ for all $n$ greater than some $n_0$, meaning that it goes above $c n^2$and it stays above $c n^2$ forever, after some point.

Pick whatever positive $c$ you'd like -- the function $c\cdot n^2$ grows larger and larger as $n$ increases, and eventually becomes larger than 1. But the function $f(n)$ keeps dipping down to 1, meaning that it doesn't stay above $c\cdot n^2$ for all $n$ when $n$ is large. Therefore, $f(n)$ is not $\Omega(n^2)$, and therefore also not $\Theta(n^2)$.

Problem #055

What is the time complexity of the following function?


def foo(n):
    for i in range(n):
        for j in range(i**2): # <-- notice the bound!
            print(i + j)

Solution

$\Theta(n^3)$. The inner loop runs $i^2$ times, so the total number of iterations is $0^2 + 1^2 + \ldots + (n-1)^2 = \frac{(n-1)n(2n-1)}{6} = \Theta(n^3)$.

Problem #056

What is the time complexity of the following function? State your answer as a function of $n$ using asymptotic notation in the simplest form possible. (e.g., $\Theta(n)$)


import math

def boo(n):
    for i in range(n):
        for j in range(n):
            print(i + j)

    for i in range(math.floor(math.sqrt(n))):
        for j in range(math.log2(i), i * math.floor(math.log2(i + 10))):
            print(i + j)

Solution

The second loop looks complicated to analyze, but we can effectively ignore it. This is because the most i can ever be is $\sqrt n$, and so an upper bound for the number of iterations made by the second loop is $O(\sqrt n \log n)$. Since the first loop takes $\Theta(n^2)$, it will dominate the time complexity, and we do not need to worry about the time taken by the second loop.

Problem #057

Let $f(n) = n^{ \frac{1}{\log_2(n)} } \times n.$ Which of the following asymptotic bounds on $f$ is true?

Hint: you might want to review your log properties. One useful one may be $\log_b c = 1 / \log_c b$.

Solution

$\Theta(n)$. Using $\frac{1}{\log_2 n} = \log_n 2$, we get $n^{1/\log_2 n} = n^{\log_n 2} = 2$. So $f(n) = 2n = \Theta(n)$.

Problem #058

Let $f(n) = 12 \log_2(3^{(n^2 - 2n)} + 2^{(\log n)} - 10n^2 - \log_3 n).$ Which of the following asymptotic bounds on $f$ is true?

Solution

$\Theta(n^2)$. Inside the logarithm, the term $3^{(n^2 - 2n)}$ dominates the others (which grow at most like $n^2$), so for large $n$ the argument is between $\frac{1}{2} \cdot 3^{(n^2 - 2n)}$ and $2 \cdot 3^{(n^2 - 2n)}$. Taking $\log_2$, $f(n)$ is within a constant of $12 (n^2 - 2n) \log_2 3$, which is $\Theta(n^2)$.

Problem #059

Suppose $f_1(n)$ is $O(n^2)$ and $\Omega(n)$. Also suppose that $f_2(n) = \Theta(n^2)$.

Consider the function $f(n) = f_1(n) + f_2(n)$. True or false: it must be the case that $f(n) = \Theta(n^2)$.

True False

Solution

True.

The upper bound: both $f_1$ and $f_2$ are $O(n^2)$, so their sum is $O(n^2)$. The lower bound: $f_1$ is nonnegative (for large $n$), so $f_1(n) + f_2(n) \geq f_2(n)$, which is $\Omega(n^2)$.

Problem #060

Suppose $f(n)$ is both $O(n^2)$ and $\Omega(n)$, and suppose $g(n) = \Theta(n^3)$. What are the tightest possible asymptotic bounds that can be placed on $f + g$?

If the upper and lower bounds are the same, you may use $\Theta$. If the upper and lower bounds are different, you should state them separately using $O$ and $\Omega$.

Solution

$\Theta(n^3)$

Problem #061

Suppose $f(n)$ is both $O(n^2)$ and $\Omega(n)$, and suppose $g(n) = \Theta(n^3)$. What are the tightest possible asymptotic bounds that can be placed on the product, $f \times g$?

If the upper and lower bounds are the same, you may use $\Theta$; otherwise you should use $O$ and $\Omega$.

Solution

$O(n^5)$ and $\Omega(n^4)$

Problem #062

Suppose $f_1(n)$ is $O(n^3)$ and $\Omega(n)$. Also suppose that $f_2(n) = O(n^2)$ and $\Omega(\sqrt n)$.

Consider the function $f(n) = f_1(n) + f_2(n)$. True or false: it must be the case that $f(n) = \Omega(n)$.

True False

Solution

True.

Both functions are nonnegative (for large $n$), so $f_1(n) + f_2(n) \geq f_1(n)$, and $f_1(n) = \Omega(n)$. Adding a nonnegative function can only make the sum bigger, so it can't hurt a lower bound.

Problem #063

Suppose that $f_1(n)$ is $O(n^3)$ and $\Omega(n)$. Also suppose that $f_2(n) = O(n^2)$ and $\Omega(\sqrt n)$.

Consider the function $g(n) = f_2(n) / f_1(n)$. Give the tightest possible upper bound on $g(n)$:

Solution

$O(n)$

Problem #064

Suppose $f_1(n)$ is $O(n^3)$ and $\Omega(n^2)$. Also suppose that $f_2(n)$ is $O(n^4)$ and $\Omega(n)$.

Consider the function $f(n) = f_1(n) + f_2(n)$. True or false: it must be the case that $f(n) = \Omega(n^2)$.

True False

Solution

True.

Both functions are nonnegative (for large $n$), so $f_1(n) + f_2(n) \geq f_1(n)$, and $f_1(n) = \Omega(n^2)$. The weaker lower bound on $f_2$ doesn't matter, since adding $f_2$ can only make the sum larger.

Problem #065

Suppose that $f_1(n)$ is $O(n^3)$ and $\Omega(n^2)$. Also suppose that $f_2(n)$ is $O(n^4)$ and $\Omega(n)$.

Consider the function $g(n) = f_2(n) / f_1(n)$. Give the tightest possible upper bound on $g(n)$:

Solution

$O(n^2)$

Problem #066

True or False. If $f = \Omega(n^5)$ and $f = O(n^7)$ then $f$ must be either $\Theta(n^5)$, or $\Theta(n^6)$, or $\Theta(n^7)$.

Solution

False. Counterexample: $f = n^{6.5}$. This is neither $\Theta(n^5)$ or $\Theta(n^6)$ or $\Theta(n^7)$; it is $\Theta(n^{6.5})$.

Problem #067

Suppose $f_1(n) = O(g_1(n))$ and $f_2(n) = \Omega(g_2(n))$. True or False: it is necessarily the case that $f_1 + f_2 = O(g_1(n))$.

True False

Solution

False.

We know nothing about how large $f_2$ can be, so it can dominate the sum. For example, take $f_1(n) = g_1(n) = g_2(n) = 1$ and $f_2(n) = n^2$. Then $f_2 = \Omega(1)$, but $f_1(n) + f_2(n) = 1 + n^2$ is not $O(1)$.

Problem #068

Suppose $f(n) = \Omega(n^3)$ and $g(n) = \Omega(n)$.

True or False: it is necessarily the case that $f/g = \Omega(n^2)$.

True False

Solution

False. Take $f(n) = n^3$ and $g(n) = n^3$. We can see from definition that $f(n) = \Omega(n^3)$ and $g(n) = \Omega(n)$. However,

\[\frac{f}{g}(n) = 1 \neq\Omega(n^2). \]

The key takeaway for this question is, since $g$ is on the denominator, we can make it grow arbitrarily fast, so that $f/g$ do not meet the requirements that we want.

Problem #069

Suppose $f_1(n)$ is $O(n^2)$ and $\Omega(n)$. Also suppose that $f_2(n) = \Theta(n^2)$.

Consider the function $g(n) = f_2(n) / f_1(n)$. True or false: it must be the case that $g(n) = \Omega(n)$.

True False

Solution

False.

$f_1$ could be as large as $n^2$. For example, if $f_1(n) = f_2(n) = n^2$, then $g(n) = 1$, which is not $\Omega(n)$. (We only know $g(n) = \Omega(1)$ and $g(n) = O(n)$.)

Problem #070

True or False. If $f_1 = \Theta(g_1(n))$ and $f_2 = O(g_2(n))$ then $\frac{f_1}{f_2} = \Theta(g_1 / g_2)$.

Solution

False. Try breaking it with a counterexample. What if $f_1 = n^3$, $f_2 = n^2$, $g_1 = n^3$ and $g_2 = n^3$. All of the conditions are satisfied, but $f_1/f_2 = n$, while $g_1/g_2 = 1$, so $f_1 / f_2$ is not $\Theta(g_1/g_2)$.

Problem #071

Suppose $f_1(n) = O(g_1(n))$ and $f_2 = O(g_2(n))$. Define $f(n) = \min\{ f_1(n), f_2(n) \}$ and $g(n) = \min\{ g_1(n), g_2(n) \}$. True or false, it is necessarily the case that $f(n) = O(g(n))$.

True False

Solution

True.

For large $n$, $f_1(n) \leq c_1 g_1(n)$ and $f_2(n) \leq c_2 g_2(n)$. The minimum of $f_1$ and $f_2$ is at most each of them, so with $c = \max\{c_1, c_2\}$ we get $\min\{f_1(n), f_2(n)\}\leq c \, g_1(n)$ and $\leq c \, g_2(n)$, i.e., $f(n) \leq c \min\{g_1(n), g_2(n)\}$.

Problem #072

Suppose $f_1(n) = \Omega(g_1(n))$ and $f_2 = \Omega(g_2(n))$. Define $f(n) = \min\{ f_1(n), f_2(n) \}$ and $g(n) = \min\{ g_1(n), g_2(n)\}$.

True or false: it is necessarily the case that $f(n) = \Omega(g(n))$.

True False

Solution

True.

For large $n$, $f_1(n) \geq c_1 g_1(n)$ and $f_2(n) \geq c_2 g_2(n)$. Letting $c = \min\{c_1, c_2\}$, whichever of $f_1(n)$ or $f_2(n)$ is smaller is still at least $c$ times the corresponding $g$, which is at least $c \min\{g_1(n), g_2(n)\}$. So $f(n) \geq c \, g(n)$.

Problem #073

Suppose $f_1(n) = \Omega(g_1(n))$ and $f_2 = \Omega(g_2(n))$. Define $f(n) = \max\{ f_1(n), f_2(n) \}$ and $g(n) = \max\{ g_1(n), g_2(n)\}$.

True or false: it is necessarily the case that $f(n) = \Omega(g(n))$. In other words, it must be the case that $\max\{ f_1(n), f_2(n) \} = \Omega( \max\{ g_1(n), g_2(n) \}) $

True False

Solution

True.

For large $n$, $f_1(n) \geq c_1 g_1(n)$ and $f_2(n) \geq c_2 g_2(n)$. Letting $c = \min\{c_1, c_2\}$, $\max\{f_1(n), f_2(n)\}$ is at least both $c\, g_1(n)$ and $c\, g_2(n)$, so it is at least $c \max\{g_1(n), g_2(n)\}$.

Problem #076

What is the best case time complexity of the following code?


def insertion_sort(arr):
    """Sort `arr` in ascending order."""
    n = len(arr)
    for i in range(1, n):
        x = arr[i]
        j = i - 1
        # find where to place x
        while j >= 0 and x < arr[j]:
            arr[j+1] = arr[j]
            j -= 1
        arr[j+1] = x

Solution

$\Theta(n)$. The best case occurs when the array is already sorted. Notice that in this case, the while loop will never actually execute.

Problem #077

What is the worst case time complexity of the following code?


def insertion_sort(arr):
    """Sort `arr` in ascending order."""
    n = len(arr)
    for i in range(1, n):
        x = arr[i]
        j = i - 1
        # find where to place x
        while j >= 0 and x < arr[j]:
            arr[j+1] = arr[j]
            j -= 1
        arr[j+1] = x

Solution

$\Theta(n^2)$. The worst case occurs when the array is sorted in reverse order. In this case, the while-loop executes on every iteration of the for-loop, and must move x from its current position all the way to the beginning of the array.

Problem #078

What is the best case time complexity of the following function?


def foo(arr):
    """arr is an array of size n"""
    for x in arr:
        for y in arr:
            if sum([x,y]) == 5:
                return sum(arr)
    return False

Solution

$\Theta(n)$. In the best case, we return sum(arr) on the very first iteration.

Problem #079

What is the worst case time complexity of the following function?


def foo(arr):
    """arr is an array of size n"""
    for x in arr:
        for y in arr:
            if sum([x,y]) == 5:
                return sum(arr)
    return False

Solution

$\Theta(n^2)$. In the worst case, we never find x + y == 5, and go through all $\Theta(n^2)$ iterations just to return False at the end.

Problem #080

What is the best case time complexity of the following function in terms of $n$?


def foo(arr):
    """`arr` is a list containing n numbers."""
    previous = None
    n = len(arr)
    for x in arr:
        if x == previous:
            for i in range(n**2):
                print('Equal!')
            return
        previous = x

Solution

$\Theta(n)$

Problem #081

What is the best case time complexity of the following function in terms of $n$?


def foo(arr):
    """`arr` is a list containing n numbers."""
    previous = None
    n = len(arr)
    for x in arr:
        if x == previous:
            for i in range(n**2):
                print('Equal!')
            return
        previous = x

Solution

$\Theta(n)$

Problem #082

What is the worst case time complexity of the following function in terms of $n$?


def foo(arr):
    """`arr` is a list containing n numbers."""
    previous = None
    n = len(arr)
    for x in arr:
        if x == previous:
            for i in range(n**2):
                print('Equal!')
            return
        previous = x

Solution

$\Theta(n^2)$

Problem #084

The below code takes in an array of numbers and returns the index of the first maximum. What is this code's best case time complexity?


def index_of_maximum(arr):
    """`arr` is an array with n elements."""
    for i, x in enumerate(arr):
        if x == max(arr):
            return i

Solution

$\Theta(n)$

Problem #085

The below code takes in an array of numbers and returns the index of the first maximum.


def foo(arr):
    """`arr` is an array with n elements."""
    for i, x in enumerate(arr):
        if x == max(arr):
            return i

What is the worst case time complexity of the function?

Solution

$\Theta(n^2)$

Problem #086

The code below takes in an array of $n$ numbers and checks whether there is a pair of numbers in the array which, when added together, equal the maximum element of the array.

What is the best case time complexity of this code as a function of $n$? State your answer using asymptotic notation.


def exists_pair_summing_to_max(arr):
    n = len(arr)
    maximum = max(arr)
    for i in range(n):
        for j in range(i + 1, n):
            if arr[i] + arr[j] == maximum:
                return True
    return False

Solution

$\Theta(n)$

Problem #091

The code below takes in two lists, each containing $n$ integers, and determines if the any number in the second list appears at least $n/2$ times in the first list.


def boo(first_list, second_list):
    """first_list and second_list are both of size n"""
    n = len(first_list)
    for x in second_list:
        count = 0
        for y in first_list:
            if x == y:
                count += 1
            if count >= n // 2:
                return True
    return False

Part 1)

What is the best case time complexity of this code as a function of $n$? State your answer using asymptotic notation.

Solution

$\Theta(n)$

Part 2)

What is the worst case time complexity of this code as a function of $n$? State your answer using asymptotic notation.

Solution

$\Theta(n^2)$

Solution

The best case is when the first element of the second list appears at least $n/2$ times in the first list. In this case, the code will return True after iterating $n/2$ times through the inner loop, taking $\Theta(n)$ time total.

In the worst case, there is no element in the second list that appears $n/2$ times in the first. We make all $n$ iterations of the outer loop, and during each of the outer iterations, make $n$ iterations of the inner loop, for a total of $\Theta(n^2)$ time.

Problem #092