int count=0;
do
{
count++;
n=n/2;
}while (n>1);
I'm having trouble seeing the pattern here.
Even when plugging in numbers for n and then plotting out each basic operation.
Thanks in advance!
Edit: I want the worst case here.
First step is to divide n with 2. So you get n/2. Now, you divide it again with 2, if n/2 > 1, and you get n/4. If n/4 > 1 you do it again, and you get n/8, or better to write it as n/(2^3)... Now if n/(2^3) > 1 you do it again, and you get n/(2^4)... So, if you do it k times, you get n/(2^k). How to calculate k to so you get n/(2^k) ≤ 1? Easy:
n/(2^k) ≤ 1
n ≤ 2^k
ln(n) ≤ k
So, your algorithm needs O(ln(n)) iterations to exit the loop.
In your code, k is count.
Related
int a = 0, i = N;
while (i > 0)
{
a += i;
i /= 2;
}
How will I calculate the time complexity of the code? Can anyone Explain?
Time complexity is basically the number of times a loop will run. Big O is the worst case complexity that a particular loop can have. For example, if linear search were being used to find K, which is, say the (n-1)th element of an array(0 indexed, starts with 0), the program would have to loop through the entire array to find the element. This would mean that the loop has to run n times in the worst case, giving linear search a time complexity of O(n).
In the case of your problem, i is initally equal to N and decrements by half per iteration. This would mean that when (N/pow(2, m) > 0 the loop terminates. So the loop runs at most m times which log(n).
log(N) = log(pow(2,m)) ==> log(N) = m
function(n):
{
for (i = 1; i <= n; i++)
{
for (j = 1; j <= n / 2; j++)
output("")
}
}
Now I have calculated the time complexity for the first for loop which is O(n). Now the second for loop shows j <= n / 2 so any given n I put, for example, a range of [1,2,...,10] will give me O(log(n)) since it will continuously give me a series of n,n/2,n/4,n/8 .... K.
So if we wanted to compare the relationship it must look something like this 2^n = k.
My question is will it give me O(log(n))?
The correct summation according to the code is:
So, it's not O(log n). It's O(n^2).
No, it does not give you o(logn).
The first for loop is O(n). The second loop is O(n) as well, as the number of iterations grows as a function of n (the growth rate is linear).
It would be the same even by changing the second loop to something like
for (j=1; j<=n/2000; j++)
or in general if you replace the denominator with any constant k.
To conclude, the time compexity is quadratic, i.e., 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)).
Here is the code given in the book "Cracking the Coding Interview" by Gayle Laakmann. Here time complexity of the code to find:-
int sumDigits(int n)
{ int sum=0;
while(n >0)
{
sum+=n%10;
n/=10
}
return sum ;
}
I know the time complexity should be the number of digits in n.
According to the book, its run time complexity is O(log n). Book provided brief description but I don't understand.
while(n > 0)
{
sum += n % 10;
n /= 10;
}
so, how much steps will this while loop take so that n comes to 0? What you do in each step, you divide n with 10. So, you need to do it k times in order to come to 0. Note, k is the number of digits in n.
Lets go step by step:
First step is when n > 0, you divide n by 10. If n is still positive, you will divide it by 10. What you get there is n/10/10 or n / (10^2). After third time, its n / (10^3). And after k times, its n/(10^k) = 0. And loop will end. But this is not 0 in mathematical sense, its 0 because we deal with integers. What you really have is |n|/(10^k) < 1, where k∈N.
So, we have this now:
n/(10^k) < 1
n < 10^k
logn < k
Btw. its also n/(10^(k-1)) > 1, so its:
k-1 < logn < k. (btw. don't forget, this is base 10).
So, you need to do logn + 1 steps in order to finish the loop, and thats why its O(log(n)).
The number of times the logic would run is log(n) to the base of 10 which is same as (log(n) to the base 2)/ (log (10) to the base of 2). In terms of time complexity, this would simply be O(log (n)). Note that log(n) to the base of 10 is how you would represent the number of digits in n.
What is the time complexity of
x = 1
while( x < SomeValue )
{
x *= 2;
}
I believe it is O(N), as the loop will continue for a fixed number of iteration.
Is my assumption correct?
The time complexity would be O(log(n)) because x is increasing exponentially.
The loop will execute in O(log n) time. Hopefully the math makes the reasoning more clear. Each iteration is constant time. To find the number of iterations relative to SomeValue, which we'll call t, you can see that after the nth iteration, x will have the value 2ⁿ. The loop ends once x≥t. So to find the number of iterations needed to meet or exceed t, plug in 2ⁿ for x and solve for t. You get t=log₂(n). Hence, O(log n) time.