CODEWITHSUNDEEP
javaalgorithmbig-o
DeADLY
DeADLY

Reputation: 79

What will be the time complexity of this algorithm

I am very new to competitive programming and to Big O notation.

public void function(int n){
   for(int i = n; i > 0; i/=3){
       for(int j = 0; j < i; j++){
           System.out.println("Hello");
       }
   }
}

This is the algorithm. As far as i know about time complexity.It defines how run time gets affected from number of inputs.

So here if we take a example if 'n' is 10. The outer loop runs log n times and the inner loop runs 'i' times.

the inner loop runs relatively to 'i' not 'n'. So im a bit confused here as to how the time complexity is calculated. I think it is O(log n).Please correct me if i am wrong.

Will it be O(log n) or O (n log n) or (n^2). Please help me out with this. Thank you.

Upvotes: 3

Views: 4518

Answers (5)

Will Davitt
Will Davitt

Reputation: 11

So this is a great question! It's a tricky one that takes a little more thinking to analyse.

As correctly stated in some of the other answers, the outer loop:

for(int i = n; i > 0; i/=3)

Will run log(n) times. Specifically log_3(n) times but in big O notation we don't often worry about the base so log(n) will be fine.

Now the nested loop is a bit trickier:

for(int j = 0; j < i; j++){

On first glance you may think this is a simple log(n) loop but lets look a little further. So on the first iteration this will run N times since the value of i will be n. Next iteration it will be run n/3 times. Then n/9, n/27, n/81 etc....

If we sum this series, it is clear to see it will total less than 2n. Therefore we can conclude this algorithm has a complexity of O(n).

Upvotes: 1

bhuwan
bhuwan

Reputation: 116

for your code :

for(int i = n; i > 0; i/=3){
   for(int j = 0; j < i; j++){
       System.out.println("Hello");
   }

}

Inner loop variable j is dependent on outer loop variable i, so your inner loop will be the one which will decide the complexity for your algorithm. since j will run 'n' times in first run, 'n/3' times in second run and so on.. therefore your total complexity can be calculated as

n + n/3 + n/9 + n/27 + .......

resulting in O(n)

Upvotes: 1

Dhruv garg
Dhruv garg

Reputation: 844

I will try to explain it in the simplest term possible

The outer loop will simply run log(n) with base 3 times.

Since, i is decreasing by factor of 3 every time. The total work done is equal to :

n + n/3 + n/9 + n/27 + .... n/(3^log(n))

since, n/3 + ... + n/(3^log(n)) will always be less than n

for e.g. let n = 100 then, 100 + 100/3 + 100/9 + 100/27 + ... = 100 + (33.3 + 11.11 + 3.7 + ...)

we can clearly see the terms in the bracket will always be less than 100

The total time complexity of the overall solution will be O(n).

Upvotes: 5

Photon
Photon

Reputation: 2777

Actually it will never terminate cause i=0 and update is i *= 3 so i will stay 0 so we can say O(+oo)

assuming you meant for(int i =1... instead, then its O(n):

  • Outer loop is clearly O(log_3 n) cause we keep multiplying by 3
  • Inner loop will get executed O(log_3 n) times with iteration count of (1 + 3 + 9 + 27 + ... + 3^log_3(n)) which is clearly a geometric progression, solving which gives us approx 3^log_3(n)) which according to log rules gives n so this loop takes O(n) for all iterations, so total complexity is O(n)

Upvotes: 1

Tim Biegeleisen
Tim Biegeleisen

Reputation: 522654

In your code snippet:

for (int i=0; i < n; i*=3) { 
    for (int j=0; j < i; j++) {
        System.out.println("Hello");
    }
}

The outer loop in i is O(log_3(n)), because each increment of the loop decreases the amount of ground needed for i to reach n by a factor of 3. This is logarithmic behavior (log_3 in this case). The inner loop in j simply iterates the same number of times as whatever the outer value of i might be, so we can just square the outer complexity, to arrive at:

O(log_3(n)^2)

Upvotes: 0

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