Given an array of size n. Each element denotes the work assigned to some student. Taking some amount of work from a student and assigning it some other student will increase it by a factor of k.
Now we have to redistribute the work such that each student will do equal work. Determine minimum possible work value. And round it off to two decimal places.
A={2,8} K= 1.5 You can take 2.4 from 8 and give it to 2. A={ 2+2.4*1.5 , 8-2.4} Answer is 5.60.
How can we approach to this question. For n=2, I am able to do this simply by solving equations. But for n>2, how can we approach to this. I tried binary search. But I am getting Time limit Exceeded.
sort(a.begin(),a.end());
double low=a[0];
double high= a[n-1];
double res=INT_MAX;
double mid;
int i,j;
while(low<=high){
mid=(low+high)/(2.0);
i=0, j=n-1;
while(i<=j){
if((a[i]+k*a[j])==(k+1)*mid){
i++;
j--;
}
else if((a[i]+k*a[j])<(k+1)*mid){
high=mid;
break;
}
else{
low=mid;
break;
}
}
if(i>j){
res=mid;
high=mid;
}
}
return res;
Please give me suggestions how I can overcome with Time limit exceeded.
If I'm understanding correctly you have an array of doubles and you want to make all elements the same but, the condition is when you subtract you subtract normal but when you add you add multiplied by 1.5, it's basically calculating the average with a condition, so you want make an average and see if its close enough or not so, I made new variable difference that see if the average that we used is bigger or less than the real one, note that difference not give any real value just positive or negative
sort(a.begin(), a.end());
double low = a[0];
double high = a[n - 1];
double average, difference= 0;
average = (low + high) / 2;
do
{
if (difference > 0)
{
low = average;
average = average + high / 2;
}
else if (difference<0)
{
high = average;
average = average + low / 2;
}
for (int i = 0, difference = 0; i < n; i++)
if (a[i] < average)
difference = difference + (a[i]-average) * 1.5;
else
difference = difference + a[i] - average;
} while (difference > 0.01 || difference < -0.01);
The while will done when difference is under 0.01 that's mean average is closer than 0.01 (you can say its 0.01/n almost).
I hope I answered your question, it's my first time to answer question here.
I'd start with some math.
If you graph the values, and you found the target value X, then you would have valleys below X and mountains above X. The volume of the mountains above X, times k, must equal the volume of the valleys.
If you sort the elements, then calculating the net mountain-valley volumes is a linear process. Starting at a_0, all there is is mountain; this is a function of the sum of the a_is and k.
Going from a_i to a_{i+1} makes the a_0 to a_i valleys deeper by a_{i+1}-a_i, and makes the a_{i+1} to a_{last} mountains shallower by the same amount.
In that region the valley/mountain transformation is linear, if it is within that region you can do linear math to find the zero.
Walk from least to greatest, and find the point where the valleys match the mountains.
I have 30 vectors of sorted ints, the vectors range in size from 20 to 230. I am trying to measure how long a binary search will take to find a value in each, but the reported time speeds up as the vectors get larger. The time doesn't actually take longer, the reporting is wrong.
Here's my code:
int main() {
const unsigned int maxPower = 30; // 2^maxPower
long long n = 1 << maxPower; // n = 2^i
for (int i = 0; i <= maxPower; ++i) {
std::vector<long long> haystack = getVector(i); // returns a sorted vector of size i
long long needle = haystack.size()/2 + 1;
clock_t t1 = clock(); // start timer
binary_search(haystack, needle);
clock_t t2 = clock(); // end timer
clock_t dt = t2 - t1;
double clocks_per_rep = ((double)dt)/n;
double seconds = clocks_per_rep/CLOCKS_PER_SEC;
std::cout << seconds << std::endl;
}
return 0;
}
I've tried using high_resolution_clock as well, but couldn't get that to even display anything but 0's.
EDIT: My original problem has been solved, but now I have a new one: there's hardly any variation in my timings, even when some take much longer than others. Here's my new code and output:
int main() {
const unsigned int maxPower = 30; // 2^maxPower
long long n = 1 << maxPower; // n = 2^i
for (int i = 0; i <= maxPower; ++i) {
std::vector<long long> haystack = getVector(i); // returns a sorted vector of size i
long long needle = haystack.size()/2 + 1;
clock_t t1 = clock(); // start timer
ternary_search(haystack, needle);
clock_t t2 = clock(); // end timer
clock_t dt = t2 - t1;
double seconds = (double)dt/CLOCKS_PER_SEC;
std::cout << seconds << std::endl;
}
return 0;
}
1e-06
1e-06
1e-06
0
1e-06
1e-06
0
0
1e-06
0
0
1e-06
0
0
0
0
1e-06
1e-06
1e-06
4e-06
3e-06
3e-06
1e-06
2e-06
3e-06
3e-06
3e-06
2e-06
3e-06
3e-06
3e-06
Assuming no bias in search, dt is expected to be O(log(N)). When you compute
double clocks_per_rep = ((double)dt)/n;
clock_per_rep is O(log(N)/N). It makes sense that the value goes down as N goes up.
It makes sense to me to omit the division by n.
clock_t dt = t2 - t1;
double seconds = (double(dt))/CLOCKS_PER_SEC;
A binary search takes worst case O(log N). In your case that means 1 to 30 comparisons. Best case the first comparison is already right and returns directly.
You are trying to measure the time it takes to make 1-30 comparisons, so basically no time at all. I'm a bit surprised you get anything but 0 and 1e-06. Which seems to be the accuracy of your clock. Your cpu is probably clocked more than 1000 times faster.
You should do a loop to do many searches and then average that per search to get better accuracy. You are also only testing finding one specific value. Have you carefully constructed that choice to be a worst case? If not why not search for every value in the vector once?
My suggestion would be to use 3 loop:
N = 1
Outer loop: loop until the time difference between start and finish is >1s.
N = 2 * N
start clock
Middle loop: repeat N times
Inner loop: search for every value in the vector once
end clock here
My best guess as to why the output takes exponentially longer each time is that getVector(i) takes that time. You have to initialize the whole vector and each pass it's twice as big. So this is kind of expected. But it can probably be optimized a lot. Likely you are growing the vector instead of initializing it with the right size.
I have gone through Google and Stack Overflow search, but nowhere I was able to find a clear and straightforward explanation for how to calculate time complexity.
What do I know already?
Say for code as simple as the one below:
char h = 'y'; // This will be executed 1 time
int abc = 0; // This will be executed 1 time
Say for a loop like the one below:
for (int i = 0; i < N; i++) {
Console.Write('Hello, World!!');
}
int i=0; This will be executed only once.
The time is actually calculated to i=0 and not the declaration.
i < N; This will be executed N+1 times
i++ This will be executed N times
So the number of operations required by this loop are {1+(N+1)+N} = 2N+2. (But this still may be wrong, as I am not confident about my understanding.)
OK, so these small basic calculations I think I know, but in most cases I have seen the time complexity as O(N), O(n^2), O(log n), O(n!), and many others.
How to find time complexity of an algorithm
You add up how many machine instructions it will execute as a function of the size of its input, and then simplify the expression to the largest (when N is very large) term and can include any simplifying constant factor.
For example, lets see how we simplify 2N + 2 machine instructions to describe this as just O(N).
Why do we remove the two 2s ?
We are interested in the performance of the algorithm as N becomes large.
Consider the two terms 2N and 2.
What is the relative influence of these two terms as N becomes large? Suppose N is a million.
Then the first term is 2 million and the second term is only 2.
For this reason, we drop all but the largest terms for large N.
So, now we have gone from 2N + 2 to 2N.
Traditionally, we are only interested in performance up to constant factors.
This means that we don't really care if there is some constant multiple of difference in performance when N is large. The unit of 2N is not well-defined in the first place anyway. So we can multiply or divide by a constant factor to get to the simplest expression.
So 2N becomes just N.
This is an excellent article: Time complexity of algorithm
The below answer is copied from above (in case the excellent link goes bust)
The most common metric for calculating time complexity is Big O notation. This removes all constant factors so that the running time can be estimated in relation to N as N approaches infinity. In general you can think of it like this:
statement;
Is constant. The running time of the statement will not change in relation to N.
for ( i = 0; i < N; i++ )
statement;
Is linear. The running time of the loop is directly proportional to N. When N doubles, so does the running time.
for ( i = 0; i < N; i++ ) {
for ( j = 0; j < N; j++ )
statement;
}
Is quadratic. The running time of the two loops is proportional to the square of N. When N doubles, the running time increases by N * N.
while ( low <= high ) {
mid = ( low + high ) / 2;
if ( target < list[mid] )
high = mid - 1;
else if ( target > list[mid] )
low = mid + 1;
else break;
}
Is logarithmic. The running time of the algorithm is proportional to the number of times N can be divided by 2. This is because the algorithm divides the working area in half with each iteration.
void quicksort (int list[], int left, int right)
{
int pivot = partition (list, left, right);
quicksort(list, left, pivot - 1);
quicksort(list, pivot + 1, right);
}
Is N * log (N). The running time consists of N loops (iterative or recursive) that are logarithmic, thus the algorithm is a combination of linear and logarithmic.
In general, doing something with every item in one dimension is linear, doing something with every item in two dimensions is quadratic, and dividing the working area in half is logarithmic. There are other Big O measures such as cubic, exponential, and square root, but they're not nearly as common. Big O notation is described as O ( <type> ) where <type> is the measure. The quicksort algorithm would be described as O (N * log(N )).
Note that none of this has taken into account best, average, and worst case measures. Each would have its own Big O notation. Also note that this is a VERY simplistic explanation. Big O is the most common, but it's also more complex that I've shown. There are also other notations such as big omega, little o, and big theta. You probably won't encounter them outside of an algorithm analysis course. ;)
Taken from here - Introduction to Time Complexity of an Algorithm
1. Introduction
In computer science, the time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the string representing the input.
2. Big O notation
The time complexity of an algorithm is commonly expressed using big O notation, which excludes coefficients and lower order terms. When expressed this way, the time complexity is said to be described asymptotically, i.e., as the input size goes to infinity.
For example, if the time required by an algorithm on all inputs of size n is at most 5n3 + 3n, the asymptotic time complexity is O(n3). More on that later.
A few more examples:
1 = O(n)
n = O(n2)
log(n) = O(n)
2 n + 1 = O(n)
3. O(1) constant time:
An algorithm is said to run in constant time if it requires the same amount of time regardless of the input size.
Examples:
array: accessing any element
fixed-size stack: push and pop methods
fixed-size queue: enqueue and dequeue methods
4. O(n) linear time
An algorithm is said to run in linear time if its time execution is directly proportional to the input size, i.e. time grows linearly as input size increases.
Consider the following examples. Below I am linearly searching for an element, and this has a time complexity of O(n).
int find = 66;
var numbers = new int[] { 33, 435, 36, 37, 43, 45, 66, 656, 2232 };
for (int i = 0; i < numbers.Length - 1; i++)
{
if(find == numbers[i])
{
return;
}
}
More Examples:
Array: Linear Search, Traversing, Find minimum etc
ArrayList: contains method
Queue: contains method
5. O(log n) logarithmic time:
An algorithm is said to run in logarithmic time if its time execution is proportional to the logarithm of the input size.
Example: Binary Search
Recall the "twenty questions" game - the task is to guess the value of a hidden number in an interval. Each time you make a guess, you are told whether your guess is too high or too low. Twenty questions game implies a strategy that uses your guess number to halve the interval size. This is an example of the general problem-solving method known as binary search.
6. O(n2) quadratic time
An algorithm is said to run in quadratic time if its time execution is proportional to the square of the input size.
Examples:
Bubble Sort
Selection Sort
Insertion Sort
7. Some useful links
Big-O Misconceptions
Determining The Complexity Of Algorithm
Big O Cheat Sheet
Several examples of loop.
O(n) time complexity of a loop is considered as O(n) if the loop variables is incremented / decremented by a constant amount. For example following functions have O(n) time complexity.
// Here c is a positive integer constant
for (int i = 1; i <= n; i += c) {
// some O(1) expressions
}
for (int i = n; i > 0; i -= c) {
// some O(1) expressions
}
O(nc) time complexity of nested loops is equal to the number of times the innermost statement is executed. For example, the following sample loops have O(n2) time complexity
for (int i = 1; i <=n; i += c) {
for (int j = 1; j <=n; j += c) {
// some O(1) expressions
}
}
for (int i = n; i > 0; i += c) {
for (int j = i+1; j <=n; j += c) {
// some O(1) expressions
}
For example, selection sort and insertion sort have O(n2) time complexity.
O(log n) time complexity of a loop is considered as O(log n) if the loop variables is divided / multiplied by a constant amount.
for (int i = 1; i <=n; i *= c) {
// some O(1) expressions
}
for (int i = n; i > 0; i /= c) {
// some O(1) expressions
}
For example, [binary search][3] has _O(log n)_ time complexity.
O(log log n) time complexity of a loop is considered as O(log log n) if the loop variables is reduced / increased exponentially by a constant amount.
// Here c is a constant greater than 1
for (int i = 2; i <=n; i = pow(i, c)) {
// some O(1) expressions
}
//Here fun is sqrt or cuberoot or any other constant root
for (int i = n; i > 0; i = fun(i)) {
// some O(1) expressions
}
One example of time complexity analysis
int fun(int n)
{
for (int i = 1; i <= n; i++)
{
for (int j = 1; j < n; j += i)
{
// Some O(1) task
}
}
}
Analysis:
For i = 1, the inner loop is executed n times.
For i = 2, the inner loop is executed approximately n/2 times.
For i = 3, the inner loop is executed approximately n/3 times.
For i = 4, the inner loop is executed approximately n/4 times.
…………………………………………………….
For i = n, the inner loop is executed approximately n/n times.
So the total time complexity of the above algorithm is (n + n/2 + n/3 + … + n/n), which becomes n * (1/1 + 1/2 + 1/3 + … + 1/n)
The important thing about series (1/1 + 1/2 + 1/3 + … + 1/n) is around to O(log n). So the time complexity of the above code is O(n·log n).
References:
1
2
3
Time complexity with examples
1 - Basic operations (arithmetic, comparisons, accessing array’s elements, assignment): The running time is always constant O(1)
Example:
read(x) // O(1)
a = 10; // O(1)
a = 1,000,000,000,000,000,000 // O(1)
2 - If then else statement: Only taking the maximum running time from two or more possible statements.
Example:
age = read(x) // (1+1) = 2
if age < 17 then begin // 1
status = "Not allowed!"; // 1
end else begin
status = "Welcome! Please come in"; // 1
visitors = visitors + 1; // 1+1 = 2
end;
So, the complexity of the above pseudo code is T(n) = 2 + 1 + max(1, 1+2) = 6. Thus, its big oh is still constant T(n) = O(1).
3 - Looping (for, while, repeat): Running time for this statement is the number of loops multiplied by the number of operations inside that looping.
Example:
total = 0; // 1
for i = 1 to n do begin // (1+1)*n = 2n
total = total + i; // (1+1)*n = 2n
end;
writeln(total); // 1
So, its complexity is T(n) = 1+4n+1 = 4n + 2. Thus, T(n) = O(n).
4 - Nested loop (looping inside looping): Since there is at least one looping inside the main looping, running time of this statement used O(n^2) or O(n^3).
Example:
for i = 1 to n do begin // (1+1)*n = 2n
for j = 1 to n do begin // (1+1)n*n = 2n^2
x = x + 1; // (1+1)n*n = 2n^2
print(x); // (n*n) = n^2
end;
end;
Common running time
There are some common running times when analyzing an algorithm:
O(1) – Constant time
Constant time means the running time is constant, it’s not affected by the input size.
O(n) – Linear time
When an algorithm accepts n input size, it would perform n operations as well.
O(log n) – Logarithmic time
Algorithm that has running time O(log n) is slight faster than O(n). Commonly, algorithm divides the problem into sub problems with the same size. Example: binary search algorithm, binary conversion algorithm.
O(n log n) – Linearithmic time
This running time is often found in "divide & conquer algorithms" which divide the problem into sub problems recursively and then merge them in n time. Example: Merge Sort algorithm.
O(n2) – Quadratic time
Look Bubble Sort algorithm!
O(n3) – Cubic time
It has the same principle with O(n2).
O(2n) – Exponential time
It is very slow as input get larger, if n = 1,000,000, T(n) would be 21,000,000. Brute Force algorithm has this running time.
O(n!) – Factorial time
The slowest!!! Example: Travelling salesman problem (TSP)
It is taken from this article. It is very well explained and you should give it a read.
When you're analyzing code, you have to analyse it line by line, counting every operation/recognizing time complexity. In the end, you have to sum it to get whole picture.
For example, you can have one simple loop with linear complexity, but later in that same program you can have a triple loop that has cubic complexity, so your program will have cubic complexity. Function order of growth comes into play right here.
Let's look at what are possibilities for time complexity of an algorithm, you can see order of growth I mentioned above:
Constant time has an order of growth 1, for example: a = b + c.
Logarithmic time has an order of growth log N. It usually occurs when you're dividing something in half (binary search, trees, and even loops), or multiplying something in same way.
Linear. The order of growth is N, for example
int p = 0;
for (int i = 1; i < N; i++)
p = p + 2;
Linearithmic. The order of growth is n·log N. It usually occurs in divide-and-conquer algorithms.
Cubic. The order of growth is N3. A classic example is a triple loop where you check all triplets:
int x = 0;
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
for (int k = 0; k < N; k++)
x = x + 2
Exponential. The order of growth is 2N. It usually occurs when you do exhaustive search, for example, check subsets of some set.
Loosely speaking, time complexity is a way of summarising how the number of operations or run-time of an algorithm grows as the input size increases.
Like most things in life, a cocktail party can help us understand.
O(N)
When you arrive at the party, you have to shake everyone's hand (do an operation on every item). As the number of attendees N increases, the time/work it will take you to shake everyone's hand increases as O(N).
Why O(N) and not cN?
There's variation in the amount of time it takes to shake hands with people. You could average this out and capture it in a constant c. But the fundamental operation here --- shaking hands with everyone --- would always be proportional to O(N), no matter what c was. When debating whether we should go to a cocktail party, we're often more interested in the fact that we'll have to meet everyone than in the minute details of what those meetings look like.
O(N^2)
The host of the cocktail party wants you to play a silly game where everyone meets everyone else. Therefore, you must meet N-1 other people and, because the next person has already met you, they must meet N-2 people, and so on. The sum of this series is x^2/2+x/2. As the number of attendees grows, the x^2 term gets big fast, so we just drop everything else.
O(N^3)
You have to meet everyone else and, during each meeting, you must talk about everyone else in the room.
O(1)
The host wants to announce something. They ding a wineglass and speak loudly. Everyone hears them. It turns out it doesn't matter how many attendees there are, this operation always takes the same amount of time.
O(log N)
The host has laid everyone out at the table in alphabetical order. Where is Dan? You reason that he must be somewhere between Adam and Mandy (certainly not between Mandy and Zach!). Given that, is he between George and Mandy? No. He must be between Adam and Fred, and between Cindy and Fred. And so on... we can efficiently locate Dan by looking at half the set and then half of that set. Ultimately, we look at O(log_2 N) individuals.
O(N log N)
You could find where to sit down at the table using the algorithm above. If a large number of people came to the table, one at a time, and all did this, that would take O(N log N) time. This turns out to be how long it takes to sort any collection of items when they must be compared.
Best/Worst Case
You arrive at the party and need to find Inigo - how long will it take? It depends on when you arrive. If everyone is milling around you've hit the worst-case: it will take O(N) time. However, if everyone is sitting down at the table, it will take only O(log N) time. Or maybe you can leverage the host's wineglass-shouting power and it will take only O(1) time.
Assuming the host is unavailable, we can say that the Inigo-finding algorithm has a lower-bound of O(log N) and an upper-bound of O(N), depending on the state of the party when you arrive.
Space & Communication
The same ideas can be applied to understanding how algorithms use space or communication.
Knuth has written a nice paper about the former entitled "The Complexity of Songs".
Theorem 2: There exist arbitrarily long songs of complexity O(1).
PROOF: (due to Casey and the Sunshine Band). Consider the songs Sk defined by (15), but with
V_k = 'That's the way,' U 'I like it, ' U
U = 'uh huh,' 'uh huh'
for all k.
For the mathematically-minded people: The master theorem is another useful thing to know when studying complexity.
O(n) is big O notation used for writing time complexity of an algorithm. When you add up the number of executions in an algorithm, you'll get an expression in result like 2N+2. In this expression, N is the dominating term (the term having largest effect on expression if its value increases or decreases). Now O(N) is the time complexity while N is dominating term.
Example
For i = 1 to n;
j = 0;
while(j <= n);
j = j + 1;
Here the total number of executions for the inner loop are n+1 and the total number of executions for the outer loop are n(n+1)/2, so the total number of executions for the whole algorithm are n + 1 + n(n+1/2) = (n2 + 3n)/2.
Here n^2 is the dominating term so the time complexity for this algorithm is O(n2).
Other answers concentrate on the big-O-notation and practical examples. I want to answer the question by emphasizing the theoretical view. The explanation below is necessarily lacking in details; an excellent source to learn computational complexity theory is Introduction to the Theory of Computation by Michael Sipser.
Turing Machines
The most widespread model to investigate any question about computation is a Turing machine. A Turing machine has a one dimensional tape consisting of symbols which is used as a memory device. It has a tapehead which is used to write and read from the tape. It has a transition table determining the machine's behaviour, which is a fixed hardware component that is decided when the machine is created. A Turing machine works at discrete time steps doing the following:
It reads the symbol under the tapehead.
Depending on the symbol and its internal state, which can only take finitely many values, it reads three values s, σ, and X from its transition table, where s is an internal state, σ is a symbol, and X is either Right or Left.
It changes its internal state to s.
It changes the symbol it has read to σ.
It moves the tapehead one step according to the direction in X.
Turing machines are powerful models of computation. They can do everything that your digital computer can do. They were introduced before the advent of digital modern computers by the father of theoretical computer science and mathematician: Alan Turing.
Time Complexity
It is hard to define the time complexity of a single problem like "Does white have a winning strategy in chess?" because there is a machine which runs for a single step giving the correct answer: Either the machine which says directly 'No' or directly 'Yes'. To make it work we instead define the time complexity of a family of problems L each of which has a size, usually the length of the problem description. Then we take a Turing machine M which correctly solves every problem in that family. When M is given a problem of this family of size n, it solves it in finitely many steps. Let us call f(n) the longest possible time it takes M to solve problems of size n. Then we say that the time complexity of L is O(f(n)), which means that there is a Turing machine which will solve an instance of it of size n in at most C.f(n) time where C is a constant independent of n.
Isn't it dependent on the machines? Can digital computers do it faster?
Yes! Some problems can be solved faster by other models of computation, for example two tape Turing machines solve some problems faster than those with a single tape. This is why theoreticians prefer to use robust complexity classes such as NL, P, NP, PSPACE, EXPTIME, etc. For example, P is the class of decision problems whose time complexity is O(p(n)) where p is a polynomial. The class P do not change even if you add ten thousand tapes to your Turing machine, or use other types of theoretical models such as random access machines.
A Difference in Theory and Practice
It is usually assumed that the time complexity of integer addition is O(1). This assumption makes sense in practice because computers use a fixed number of bits to store numbers for many applications. There is no reason to assume such a thing in theory, so time complexity of addition is O(k) where k is the number of bits needed to express the integer.
Finding The Time Complexity of a Class of Problems
The straightforward way to show the time complexity of a problem is O(f(n)) is to construct a Turing machine which solves it in O(f(n)) time. Creating Turing machines for complex problems is not trivial; one needs some familiarity with them. A transition table for a Turing machine is rarely given, and it is described in high level. It becomes easier to see how long it will take a machine to halt as one gets themselves familiar with them.
Showing that a problem is not O(f(n)) time complexity is another story... Even though there are some results like the time hierarchy theorem, there are many open problems here. For example whether problems in NP are in P, i.e. solvable in polynomial time, is one of the seven millennium prize problems in mathematics, whose solver will be awarded 1 million dollars.
I'm trying to find the time complexity of a simple implementation of mandelbrot set. with following code
int main(){
int rows, columns, iterations;
rows = 22;
columns = 72;
iterations = 28;
char matrix[max_rows][max_columns];
for(int r = 0; r < rows; ++r){
for(int c = 0; c < columns; ++c){
complex<float> z;
int itr = 0;
while(abs(z) < 2 && ++itr < iterations)
z = pow(z, 2) + decltype(z)((float)c * 2 / columns - 1.5,
(float)r * 2 / rows - 1);
matrix[r][c]=(itr== iterations ? '*' : '.');
}
}
Now looking at above code i made some estimation for time complexity in terms of big O notation and want to know if it is correct or not
So we are creating a 2d array traversing it through nested loops and and at each element we are performing an operation and setting a value of that element, if we take n as input size we can say that greater the input the greater will be the complexity, so the time complexity for rowsxcolumns would be O(rxc) and then again we are traversing it for printout, so what would be the time complexity? is it O(rxc)+O(rxc) ? does the function itself have some effect on time complexity when we are doing multiplication and subtraction on rows and columns? If yes then how?
Almost, given r rows, c columns and i iterations then the running time is O(r*c*i). This should be trivial to see if abs(z)<2 is not there. But with this extra condition its not clear how many times will the inner while loop run in total. Yes, it will be less than r*c*i times, so O(r*c*i) is still the upper bound. But perhaps we might do better. Given that for any r,c you compute Mandelbrot set over the same domain with varying resolution then the while loop will run k*r*c*i times for some constant k which is somewhere between area-of-Mandelbrot-set-over-area-of-the-domain and 1 --> Running time of the code is Θ(r*c*i) and O(r*c*i) cannot be improved.
Had you computed the set over [-c,c]x[-r,r] domain with fixed resolution then for any |z|>2 the abs(z)<2 breaks after first iteration. Then O(r*c*i) would not be tight bound and this condition (as all loop conditions) should be considered if you want accurate estimation.
Please don't use malloc, std::vector is safer.
In big-O notation, O(rxc)+O(rxc) collapses to O(rxc).
Since the maximal iteration count is also an input variable, it has an influence on the complexity as well. In particular, the inner loop runs at most n iterations, therefore, your complexity is O(rxcxn).
All other operations are constant, in particular multiplication and addition of complex<float>. These operations by themselves are always O(1), which does not contribut to the overall complexity.
So, I really don't get Big O notation. I have been tasked with determining "O value" for this code segment.
for (int count =1; count < n; count++) // Runs n times, so linear, or O(N)
{
int count2 = 1; // Declares an integer, so constant, O(1)
while (count2 < count) // Here's where I get confused. I recognize that it is a nested loop, but does that make it O(N^2)?
{
count2 = count2 * 2; // I would expect this to be constant as well, O(N)
}
}
O(f(n))=g(n)
This implies that for some value k, f(n)>g(n) where n>k. This gives the upper bound for the function g(n).
When you are asked to find Big O for some code,
1) Try to count the number of computations being performed in terms of n and thus getting g(n).
2) Now try estimating the upper bound function of g(n). That will be your answer.
Lets apply this procedure to your code.
Lets count the number of computations made. The statements declaring and multiply by 2 take O(1) time. But these are executed repeatedly. We need to find how many times they are executed.
The outer loop executes for n times. Hence the first statement executes for n times. Now the number of times inner loop gets executed depends on value of n. For a given value of n it executes for logn times.
Now lets count the total number of computations performed,
log(1) + log(2) + log(3) +.... log(n) + n
Note that the last n is for the first statement. Simplifying the above series we get:
= log(1*2*3*...n) + n
= log(n!) + n
We have
g(n)=log(n!) + n
Lets guess the upper bound for log(n!).
Since,
1.2.3.4...n < n.n.n...(n times)
Hence,
log(n!) < log(n^n) for n>1
which implies
log(n!) = O(nlogn).
If you want a formal proof for this, check this out. Since nlogn increases faster than n , we therefore have:
O(nlogn + n) = O(nlogn)
Hence your final answer is O(nlogn).