Input: Array A consisting n distinct integers, and an integer k
Output: an integer l
l = 0
for i = 1 to n
for j = 1 to n
if A[i] == A[j] + k then l = l + 1
endfor
endfor
return l
How to reduce the complexity lower than O(n^2)
well, I just have ended up with a good solution better than O(n2) I think.
First Build a BST O(N log N)
and then Search O(Log N ).
If I can use any data structures, I choose hash-set structure for storing distinct values. Hash-set can typically create with O(n) time complexity and can access with O(1) time complexity.
B =hashset(A) // requires O(n) time
for i = 1 to n
if B.contains(A[i]+k) then l= l+1
endfor // also requires n×O(1) = O(n) time
return l
I think this code has O(n) time complexity.
I would start by sorting the list of values, then assuming k is positive, you know 2 things:
in your inner loop, you only need to search from A[i+1] and up, as you know that only values after A[i] can be greater than A[i]
you stop the searching process in the inner loop once A[j] > A[i]+k, as you know that all further values will also be greater than A[i]+k
For a small set of data, the overhead of the O(log(n)) search would probably result in the algorithm being more expensive than your n^2 approach. However, as the data grows, that overhead will decrease relative to the n^2, and the substantially smaller set of comparisons will give you an overall gain.
Obviously the same principle could be applied if k was negative, you just have to swizzle round some logic, or simply negate both k AND your values.
Also - just noticed, as your integers are distinct, you can also break out of the inner loop if you find a match, as you know there can only be 1 match per entry.
That's what I'd do anyway!
Edit: Here's some example code:
{
Input: Array A consisting n distinct integers, and a positive integer k
Output: an integer l
A = sortascending(A)
l = 0
for i = 1 to n
#calculate target value
target = A[i]+k
#iterate through remaining entries
for j = i+1 to n
if target == A[j] then
#got a match, inc l then bail out
l++
break
elseif target < A[j]
#now impossible for any further data points
#to match, so bail out
break
endif
endfor
endfor
return l
}
-Chris
Related
What is the running time complexity of fun()?
int fun(int n)
{
int count = 0;
for (int i = n; i > 0; i =i-2)
for (int j = 2; j < i; j=j*j)
for (int k=j; k>0; k=k/2)
count += 1;
return count;
}
is it O(n * lglgn * lglglgn)?
----
Edit:
j = loglog(i) times, big iterative value of j can be almost n(for example n=17,max(j)=16)
k= log(j), since the max value of j is at most n. then the max iterative times can be log(n)
so we can say that the big O of this question is O(n* lglgn * lgn)
Since the value of j and k depends on the previous iterative value (i, j), maybe there is better tight answer to this question.
We need to count this carefully, because the number of iterations of each inner loop depends in a non-trivial way on the outer loop's variable. Simply averaging for each loop and multiplying the results together will not give the right answer.
The outer loop runs O(n) times, because i counts from n down to 0 in constant steps.
The middle loop's values of j are 2, then 2*2 = 4, then 4*4 = 16, and on the m'th iteration, j = 2^2^m. The last iteration will be when 2^2^m >= i, in which case m >= log log i. So this runs O(log log i) times.
The innermost loop runs O(log j) times, because on the m'th iteration, k = j / 2^m. The last iteration will be when k <= 1, in which case m >= log j. So this runs O(log j) times.
However, it is not correct to multiply these together to get O(n * log log n * log log log n) - because i is not n on every iteration, and j is not log log n on every iteration. This gives an upper bound, but not a tight one. To calculate the true time complexity, you will need to write it as a double-summation, and simplify it algebraically.
As a simpler example to think about, consider the following code:
for(i = 1; i < n; i *= 2) {
for(j = 0; j < i; j += 1) {
// do something
}
}
The outer loop runs O(log n) times, and the inner loop runs O(i) times, but the overall complexity is actually O(n). To see this, count how many times // do something is reached; the first time the outer loop iterates it'll be 1, then it'll be 2, then 4, then 8, and so on up to n. This is a geometric progression with a sum <= 2n, giving a total number of steps which is O(n).
Note that if we naively multiply the two loops' complexities we get O(n log n) instead, which is an upper bound, but not a tight one.
Using Big O notation:
With the first outer loop we get O(N/2). You have a loop of N items and you are reducing it by 2 everytime, though you get a total of N/2 loops.
With the outter loop we get O(Log(I)).
With the most inner loop we have O(Log(J)), because you are dividing by 2 your iterator on every loop.
If we multiply the three complexities because they are nested:
O(N/2)*O(Log(I)) + O(Log(j)) ~ O(N/2*Log(I)*Log(J)) ~ O(N/2*Log^2(N)) ~ O(N*Log^2(N)).
We get a linearithmic complexity: O(N*Log^2(N))
The complexity of this code is O(log(n^2)*log(n), and i don't understand how we arrive to this result.
According to me, the nested while's big O should be just log(n) since its a while loop and we divide j by 4 everytime we enter the loop, and same for the initial while loop with i divided by 2. I especially dont understand which while loop has the O(log^2(n)) complexity
c = 0
i = n * n
while i > 0:
j = n
while j > 0:
c += 1
j = j//4
i = i//2
print c
I seem to be coming up with O(log_4(n)*log_2(n)) as the complexity. First, appreciate that the outer and inner while loops are not correlated. That is to say, the outer loop in i is independent of the inner loop in j. Here are the complexities of the outer and inner loop, in terms of n:
outer loop: O(log_2(n)). This is because the loop starts with n^2, and then decrements the counter by factors of 2, which is therefore log_2 behavior. As #chepner has commented:
O(log_2(n^2)) == 2*O(log_2(n)) = O(log_2(n))
inner loop: O(log_4(n)). This loop begins at n, and decrements the counter by factors of 4, which is log_4 behavior.
Your current guess is almost right, except that you might have missed the bases of the logarithms.
What is the time Complexity for below code:
1)
function(values,xlist,ylist):
sum =0
n=0
for r from 0 to xlist:
for c from 0 to ylist:
sum+= values[r][c]
n+1
return sum/n
2)
function PrintCharacters():
characters= {"a","b","c","d"}
foreach character in characters
print(character)
According to me the 1st code has O(xlist*ylist) complexity and 2nd code has O(n).
Is this right?
Big O notation to describe the asymptotic behavior of functions. Basically, it tells you how fast a function grows or declines
For example, when analyzing some algorithm, one might find that the time (or the number of steps) it takes to complete a problem of size n is given by
T(n) = 4 n^2 - 2 n + 2
If we ignore constants (which makes sense because those depend on the particular hardware the program is run on) and slower growing terms, we could say "T(n)" grows at the order of n^2 " and write:T(n) = O(n^2)
For the formal definition, suppose f(x) and g(x) are two functions defined on some subset of the real numbers. We write
f(x) = O(g(x))
(or f(x) = O(g(x)) for x -> infinity to be more precise) if and only if there exist constants N and C such that
|f(x)| <= C|g(x)| for all x>N
Intuitively, this means that f does not grow faster than g
If a is some real number, we write
f(x) = O(g(x)) for x->a
if and only if there exist constants d > 0 and C such that
|f(x)| <= C|g(x)| for all x with |x-a| < d
So for your case it would be
O(n) as |f(x)| > C|g(x)|
Reference from http://web.mit.edu/16.070/www/lecture/big_o.pdf
for r from 0 to xlist: // --> n time
for c from 0 to ylist: // n time
sum+= values[r][c]
n+1
}
function PrintCharacters():
characters= {"a","b","c","d"}
foreach character in characters --> # This loop will run as many time as there are characters suppose n characters than it will run time so O(n)
print(character)
Big O Notation gives an assumption when value is very big outer loop
will run n times and inner loop is running n times
Assume n -> 100 than total n^2 10000 run times
I am trying to calculate the time complexity of this function
Code
int Almacen::poner_items(id_sala s, id_producto p, int cantidad){
it_prod r = productos.find(p);
if(r != productos.end()) {
int n = salas[s - 1].size();
int m = salas[s - 1][0].size();
for(int i = n - 1; i >= 0 && cantidad > 0; --i) {
for(int j = 0; j < m && cantidad > 0; ++j) {
if(salas[s - 1][i][j] == "NULL") {
salas[s - 1][i][j] = p;
r->second += 1;
--cantidad;
}
}
}
}
else {
displayError();
return -1;
}
return cantidad;
}
the variable productos is a std::map and its find method has a time complexity of Olog(n) and other variable salas is a std::vector.
I calculated the time and I found that it was log(n) + nm but am not sure if it is the correct expression or I should leave it as nm because it is the worst or if I whould use n² only.
Thanks
The overall function is O(nm). Big-O notation is all about "in the limit of large values" (and ignores constant factors). "Small" overheads (like an O(log n) lookup, or even an O(n log n) sort) are ignored.
Actually, the O(n log n) sort case is a bit more complex. If you expect m to be typically the same sort of size as n, then O(nm + nlogn) == O(nm), if you expect n ≫ m, then O(nm + nlogn) == O(nlogn).
Incidentally, this is not a question about C++.
In general when using big O notation, you only leave the most dominant term when taking all variables to infinity.
n by itself is much larger than log n at infinity, so even without m you can (and generally should) drop the log n term, so O(nm) looks fine to me.
In non-theoretical use cases, it is sometimes important to understand the actual complexity (for non-infinite inputs), since sometimes algorithms that are slow at infinity can produce better results for shorter inputs (there are some examples where O(1) algorithms have such a terrible constant that an exponential algorithm does better in real life). quick sort is considered a practical example of an O(n^2) algorithm that often does better than it's O(n log n) counterparts.
Read about "Big O Notation" for more info.
let
k = productos.size()
n = salas[s - 1].size()
m = salas[s - 1][0].size()
your algorithm is O(log(k) + nm). You need to use a distinct name for each independent variable
Now it might be the case that there is a relation between k, n, m and you can re-label with a reduced set of variables, but that is not discernible from your code, you need to know about the data.
It may also be the case that some of these terms won't grow large, in which case they are actually constants, i.e. O(1).
E.g. you may know k << n, k << m and n ~= m , which allows you describe it as O(n^2)
So these are the for loops that I have to find the time complexity, but I am not really clearly understood how to calculate.
for (int i = n; i > 1; i /= 3) {
for (int j = 0; j < n; j += 2) {
... ...
}
for (int k = 2; k < n; k = (k * k) {
...
}
For the first line, (int i = n; i > 1; i /= 3), keeps diving i by 3 and if i is less than 1 then the loop stops there, right?
But what is the time complexity of that? I think it is n, but I am not really sure.
The reason why I am thinking it is n is, if I assume that n is 30 then i will be like 30, 10, 3, 1 then the loop stops. It runs n times, doesn't it?
And for the last for loop, I think its time complexity is also n because what it does is
k starts as 2 and keeps multiplying itself to itself until k is greater than n.
So if n is 20, k will be like 2, 4, 16 then stop. It runs n times too.
I don't really think I am understanding this kind of questions because time complexity can be log(n) or n^2 or etc but all I see is n.
I don't really know when it comes to log or square. Or anything else.
Every for loop runs n times, I think. How can log or square be involved?
Can anyone help me understanding this? Please.
Since all three loops are independent of each other, we can analyse them separately and multiply the results at the end.
1. i loop
A classic logarithmic loop. There are countless examples on SO, this being a similar one. Using the result given on that page and replacing the division constant:
The exact number of times that this loop will execute is ceil(log3(n)).
2. j loop
As you correctly figured, this runs O(n / 2) times;
The exact number is floor(n / 2).
3. k loop
Another classic known result - the log-log loop. The code just happens to be an exact replicate of this SO post;
The exact number is ceil(log2(log2(n)))
Combining the above steps, the total time complexity is given by
Note that the j-loop overshadows the k-loop.
Numerical tests for confirmation
JavaScript code:
T = function(n) {
var m = 0;
for (var i = n; i > 1; i /= 3) {
for (var j = 0; j < n; j += 2)
m++;
for (var k = 2; k < n; k = k * k)
m++;
}
return m;
}
M = function(n) {
return ceil(log(n)/log(3)) * (floor(n/2) + ceil(log2(log2(n))));
}
M(n) is what the math predicts that T(n) will exactly be (the number of inner loop executions):
n T(n) M(n)
-----------------------
100000 550055 550055
105000 577555 577555
110000 605055 605055
115000 632555 632555
120000 660055 660055
125000 687555 687555
130000 715055 715055
135000 742555 742555
140000 770055 770055
145000 797555 797555
150000 825055 825055
M(n) matches T(n) perfectly as expected. A plot of T(n) against n log n (the predicted time complexity):
I'd say that is a convincing straight line.
tl;dr; I describe a couple of examples first, I analyze the complexity of the stated problem of OP at the bottom of this post
In short, the big O notation tells you something about how a program is going to perform if you scale the input.
Imagine a program (P0) that counts to 100. No matter how often you run the program, it's going to count to 100 as fast each time (give or take). Obviously right?
Now imagine a program (P1) that counts to a number that is variable, i.e. it takes a number as an input to which it counts. We call this variable n. Now each time P1 runs, the performance of P1 is dependent on the size of n. If we make n a 100, P1 will run very quickly. If we make n equal to a googleplex, it's going to take a little longer.
Basically, the performance of P1 is dependent on how big n is, and this is what we mean when we say that P1 has time-complexity O(n).
Now imagine a program (P2) where we count to the square root of n, rather than to itself. Clearly the performance of P2 is going to be worse than P1, because the number to which they count differs immensely (especially for larger n's (= scaling)). You'll know by intuition that P2's time-complexity is equal to O(n^2) if P1's complexity is equal to O(n).
Now consider a program (P3) that looks like this:
var length= input.length;
for(var i = 0; i < length; i++) {
for (var j = 0; j < length; j++) {
Console.WriteLine($"Product is {input[i] * input[j]}");
}
}
There's no n to be found here, but as you might realise, this program still depends on an input called input here. Simply because the program depends on some kind of input, we declare this input as n if we talk about time-complexity. If a program takes multiple inputs, we simply call those different names so that a time-complexity could be expressed as O(n * n2 + m * n3) where this hypothetical program would take 4 inputs.
For P3, we can discover it's time-complexity by first analyzing the number of different inputs, and then by analyzing in what way it's performance depends on the input.
P3 has 3 variables that it's using, called length, i and j. The first line of code does a simple assignment, which' performance is not dependent on any input, meaning the time-complexity of that line of code is equal to O(1) meaning constant time.
The second line of code is a for loop, implying we're going to do something that might depend on the length of something. And indeed we can tell that this first for loop (and everything in it) will be executed length times. If we increase the size of length, this line of code will do linearly more, thus this line of code's time complexity is O(length) (called linear time).
The next line of code will take O(length) time again, following the same logic as before, however since we are executing this every time execute the for loop around it, the time complexity will be multiplied by it: which results in O(length) * O(length) = O(length^2).
The insides of the second for loop do not depend on the size of the input (even though the input is necessary) because indexing on the input (for arrays!!) will not become slower if we increase the size of the input. This means that the insides will be constant time = O(1). Since this runs in side of the other for loop, we again have to multiply it to obtain the total time complexity of the nested lines of code: `outside for-loops * current block of code = O(length^2) * O(1) = O(length^2).
The total time-complexity of the program is just the sum of everything we've calculated: O(1) + O(length^2) = O(length^2) = O(n^2). The first line of code was O(1) and the for loops were analyzed to be O(length^2). You will notice 2 things:
We rename length to n: We do this because we express
time-complexity based on generic parameters and not on the ones that
happen to live within the program.
We removed O(1) from the equation. We do this because we're only
interested in the biggest terms (= fastest growing). Since O(n^2)
is way 'bigger' than O(1), the time-complexity is defined equal to
it (this only works like that for terms (e.g. split by +), not for
factors (e.g. split by *).
OP's problem
Now we can consider your program (P4) which is a little trickier because the variables within the program are defined a little cloudier than the ones in my examples.
for (int i = n; i > 1; i /= 3) {
for (int j = 0; j < n; j += 2) {
... ...
}
for (int k = 2; k < n; k = (k * k) {
...
}
}
If we analyze we can say this:
The first line of code is executed O(cbrt(3)) times where cbrt is the cubic root of it's input. Since i is divided by 3 every loop, the cubic root of n is the number of times the loop needs to be executed before i is smaller or equal to 1.
The second for loop is linear in time because j is executed
O(n / 2) times because it is increased by 2 rather than 1 which
would be 'normal'. Since we know that O(n/2) = O(n), we can say
that this for loop is executed O(cbrt(3)) * O(n) = O(n * cbrt(n)) times (first for * the nested for).
The third for is also nested in the first for, but since it is not nested in the second for, we're not going to multiply it by the second one (obviously because it is only executed each time the first for is executed). Here, k is bound by n, however since it is increased by a factor of itself each time, we cannot say it is linear, i.e. it's increase is defined by a variable rather than by a constant. Since we increase k by a factor of itself (we square it), it will reach n in 2log(n) steps. Deducing this is easy if you understand how log works, if you don't get this you need to understand that first. In any case, since we analyze that this for loop will be run O(2log(n)) time, the total complexity of the third for is O(cbrt(3)) * O(2log(n)) = O(cbrt(n) *2log(n))
The total time-complexity of the program is now calculated by the sum of the different sub-timecomplexities: O(n * cbrt(n)) + O(cbrt(n) *2log(n))
As we saw before, we only care about the fastest growing term if we talk about big O notation, so we say that the time-complexity of your program is equal to O(n * cbrt(n)).