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The cube root of a natural number $n$ is defined as the largest natural number $m$ such that $(m^3 \leq n)$ . The complexity of computing the cube root of $n$ ($n$ is represented by binary notation) is 

  1. $O(n)$ but not $O(n^{0.5})$
  2. $O(n^{0.5})$ but not $O((\log n)^k)$ for any constant $k>0$
  3. $O((\log n)^k)$ for some constant $k>0$, but not $O( (\log \log n)^m)$ for any constant $m>0$
  4. $O( (\log \log n)^k )$ for some constant $k > 0.5$, but not $O( (\log \log n)^{0.5} )$
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The best answer by @gatecse talks about an array. Don’t be confused there is no array in question the reference is used only to make the Visualization of the concept easier.

Eg. finding cube root of 1000, there is no array. So lets try to find out the cube root manually starting from 1, 2, 3, 4……. this will take O(n) time. but since we know the beginning(1) and end(1000) and also logically they are sorted we can make use of binary search concept to get to our answer.

Even if we search for cube root linearly, it should not take more than O(n^0.33) which eliminates option a and b?

7 Answers

98 votes
Best answer
We can simply do a binary search in the array of natural numbers from $1..n$ and check if the cube of the number matches $n$ (i.e., check if $a[i] * a[i] * a[i] == n$). This check takes $O(\log n)$ time and in the worst case we need to do the search $O(\log n)$ times. So, in this way we can find the cube root in $O(\log^2 n)$. So, options (A) and (B) are wrong.

Now, a number is represented in binary using $\log n$ bit. Since each bit is important in finding the cube root, any cube root finding algorithm must examine each bit at least once. This ensures that complexity of cube root finding algorithm cannot be lower than $\log n$. (It must be $\Omega \left( \log n \right)$). So,  (D) is also false and (C) is the correct answer.
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what if the number n is converted to decimal to find cube root and then find its complexity.will it be wrong to convert the number to decimal?
"We can simply do a binary search in the array of natural numbers from 1..n"
@Arjun sir from where this array comes?

@Ahwan There's no need to store the numbers in array. He just means perform a binary search on the range [1, n] for the cube root of n. This is because the cube root of n will always be in the range [1,n].

" the worst case we need to do the search O(logn) times".
I am not getting this. In O(logn) time, we get the i for which i^3<=n using binary search.  What will be worst case? Please help
@sks24 It's given that the number n is given in binary, so there are $logn$ bits, naive multiplication will be done in $log^2n$, then $logn$ for the binary search, so I think final complexity should be $O(log^3n)$

EDIT: If we convert the binary to decimal in the beginning, then it will be $O(logn)$
Thanks. Got it.
@Arjun sir. How is it logn^2. It should be logn only? 1...n we do binary search and each time we check a(i)*a(i)*a(i)=n or not and this is a constant time and it will be done logn times at max.Can you tell please how is it logn^2?


 I think final complexity should be O(log3n)O(log3n)

EDIT: If we convert the binary to decimal in the beginning, then it will be O(logn)

Please explain how it becomes O(logn)?
a == b? Usually we consider this as an $O(1)$ operation. But when $a$ is allowed to be arbitrarily large, this can take $O(\log a)$ as each bit needs to be checked. That is why the check is taking $O(\log n)$ in the given question.
Sir in asymptotic analysis we dont bother about how long arithmetic operations take.It depends on h/w .

even when we check a==b in constant as you mentioned,there also a can be any large or small number and we need to check all bit,but we dont bother about that.I still did not get:(
@Arjun sir, why do the binary search on the whole of n elements? Isn't it sufficient to do it only for the 1st n^1/3 elements? If so, then the time taken is also reduced to O(log n^1/3).

cant we do like this?

1) convert given binary representation of $n$ into decimal----$O(logn)$

2) then find $\left \lfloor n^{0.333} \right \rfloor$ --------------$O(1)$

total complexity $O(logn)$

@Arjun sir I have same doubt as above @rahul said.

Do we consider the underlying hardware complexity in asymptotic analysis of our algorithm?

i have the same doubt did u get the answer?
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How did you get log^2(n) ? I got it we need log(n) for binary search, but why did you square it ?
If  'n'  is not binary number then what is complexity?
If you are wondering how O$(log n)^{2}$ came then see Arjun Sir comment above dated Nov 11, 2017.

and remember we are not converting binary to decimal here so in this method answer would be O$(log n)^{2}$

But if we convert n into decimal first then the time complexity will be reduced to O$(log n)^{}$ which is only the binary search time. But I don't know converting would be appropriate or not.
The number 'n' is represented in 'b' bits. So the cube root will lie in the range of 0 to 2^(ceil(b/3)). So we can do a binary search in this range.
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Here  the elements are from 1..n  so for every increment in i we have to check that the element present at position i that is a[i] having its cube as n or not.

so there are two ways to do it either we can do it by sequential search or by binary search(if the mid element is having its cube > n then move to left otherwise move to right)

[Note: here we can move right in binary search because we are not sure that we have n contiguous natural numbers so any number can be possible]

56 votes

Consider the below function:

int cuberoot(int n)
    int crt;
    for(int i =1; i<=n;i++)
    if (i*i*i<=n)  crt=i;
    else return(crt);

The above function will return the cube root of value 'n' as defined in question.

Now if n = 64, the for loop will run 5 times(i=1,2,3,4,5) O(logn).

But if we take larger value say n= 2^30 then the for loop will run (2^10 +1=1025)  times (since 2^10 * 2^10 * 2^10 =2^30) which is  (logn)^k , where k=2.038.

Therefore we can say that the complexity of computing cuberoot  will O((logn)^k) but not O(loglogn).

Hence Answer is (C)

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1 comment

Your answer is better  than previous one
5 votes

Option C

Using mcLaurin series

1 comment

Good one! So it would take $O((log n)^{k}$) for square root also? (plus extra logn time for binary to decimal conversion)
4 votes


See, if we have all the numbers in a sorted fashion in an array, we can use binary search which takes $O(logn)$ time.

So, the time complexities mentioned in Options A and B hold. The "but not" part makes these options wrong.

The question says that 

n is represented by binary notation

Hence the size of each number is $O(logn)$.

When performing the binary search, all the $O(logn)$ bits of a number would be potentially checked to find a match. Hence, at the minimum this algorithm would take $O(logn*logn)$ time = $O(logn)^2$

Since this would be the minimum time, we can't get $O(loglogn)$, hence Option D is incorrect, too.


Option C

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edited by
When n shrinks by sqrt time complexity would be O(Log.Log n) and same for the others. Pls check once again.
edited by

; Thanks for pointing it out, this answer is old and stupid as hell.

I'll edit to provide the simplified solution in a while :)

PS: When n shrinks by sqrt, time complexity would be $O(loglogn)$ and not $O(√n)$

1 vote

Here we  need to find upper and perhaps lower bounds on the complexity of finding an integer cube root m of n. At least one upper bound is trivial, and rules out answers A and B: m can be found in O(log n) time using binary search.

Also note that the input size is O(log n) because the minimum number of bits needed to represent an arbitrary n in binary notation is proportional to log n. Because all bits of the number must be processed to solve the problem, θ(log n) is a lower bound on the time to solve the problem, and therefore the problem cannot be solved in time O((log log n)^w) [where w is some constant > 0] because that isn't O(log n). Thus, answer C applies.


0 votes

Consider the array of natural numbers .

 00   01   10   11  100  101  110

Suppose we want to find the cube root of $ 6$ , $(n=6)$ to store in binary it takes 3-bits ($ \left \lceil \log_2 n \right \rceil$)

take $m$ as middle element of the array.  (here $m=3$)

step 1 : Find middle element $m$ [ will take $O(1)$ time ]
step 2 : Find cube of m  [ will take $O(1)$ time ]
step 3 : Compare $m^{3} < n$ [ will take $\left \lceil \log_2 n \right \rceil$ time .  We need to compare $\left \lceil \log_2 n \right \rceil$ bits because of binary representation ]
step 4 : If there is match or nothing in array is left we return m else if compared value is grater than $m$ we go left of $m$ else we go right of $m$ loop from step 1

Running Time Complexity Of the Algorithm

  • $\log n$  -> Implementing Binary Search
  • $( \log n )^{k}$ -> Comparing Each bit of the number represented in binary notation where k is the number of loop iterations 
  • So on best case it will take $O(\log n)$ and worst case $O(\log^{k+1} n)$


Iteration 1

  1. $m=3$
  2. $3^{3}=9$
  3. $9 \leq 6$ False we go LEFT of the array [this is compared as $1001\leq 110 $] 

Iteration 2

  1. $m=1$
  2. $1^{3}=1$
  3. $1\leq 6$ True we go RIGHT of the array [this is compared as $01\leq 110 $] 

Iteration 3

  1. $m=2$
  2. $2^{3}=8$
  3. $8\leq 6$ False we go LEFT of the array [this is compared as $1000\leq 110 $] 
  4. Since sonthing in LEFT of the array we return $2$ as it is the nearest cube root

Here we took $O(\log ^{4} n)$

[ @arjun sir can u pls check this one ]

edited by


$k = \log n$ rt?
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@arjun sir ,

Actually What i Thought about upperbound complexity $O(\log^{k+1} n)$ is that 

  • Outer Loop  for implementing binary search will take $O( \log n)$ time
  • inner loop for comparing each bit of $n$ for each iteration takes $O( \log n)$ each

Isn't the example correct ?

  1.  $O(\log n) $ for implementing binary search
  2. $O(\log n) $ for comparing bits of 9 and 6
  3. $O(\log n) $ for comparing bits of 1 and 6
  4. $O(\log n) $ for comparing bits of 8 and 6

Here k = number of iterarions . The loop executes for $\log n $ time hence $k=\log n$

Correct me if wrong...

You can count k rt? Also why you taking power and not multiply for getting the time complexity?
Multiplying a number 3 time , is const operation  That is why i took it as $O(1)$
But it varies with the input- not a constant.
See my answer now, there was a mistake which I corrected now.
Got my mistake :)
This is wrong .
@arjun sir complexity will be always $O(\log^{2} n)$  ?  
How does k counted ? will it affect the complexity ?
yes, though I did not prove it. Complexity of multiplying 3 numbers of $\log n$ bits each. Normally we take this as $O(1)$ assuming CPU has a MUL instruction. For this question, we can assume such an instruction is not there.
Such a clear analysis. Thanx a lot sir.
While changing decimal to binary(in 1st case like 9 we will get so we r changing it to binary then comparing with desired binary no .) will not be considered in time complexity??

Just this doubt coming.
In question it is mentioned that (n is represented in binary notation).
0 votes



{ for(i=1 to n^3)

 if(i^3>n) return(i-1)


it takes n^(1/3) time


(1)^3   (2)^3  3  (4)^3  5  6  7  (8)^3  9  10  11  12  13  14  15  (16)^3 ..........

it takes logn time

so c is answer


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