Big O Theory- triple nested loop

Question

If I have a function of:

for(i=0;i<n;i++)
   for(j=0;j<i*i;j++)
      for(k=0;k<j;k++)
         System.out.println(k);

Would the big O of this function be n^5 from having: n*((n-1)^2)*((n-1)^2)-1?

possible duplicate of [Big O analysis for this for loop](http://stackoverflow.com/questions/10071692/big-o-analysis-for-this-for-loop) — Mohamed Ennahdi El Idrissi, Mar 14 '14 at 22:12

fastcodejava · Answer 1 · 2014-01-29T00:44:20.267

3

Your function is O(1) because it returns the first k, the loop ends at first iteration. Assuming it don't return immediately, it is n^5 as you thought.

For each i the second loop is looping i^2 times, and the third loop is going j times. So for each i it is looping i^4 times. So the total is Sum(i^4) (1..n) which is O(n^5).

edited Jan 29 '14 at 00:44

answered Jan 28 '14 at 03:58

fastcodejava

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Sorry, I edited and fixed. So it is n^5 based on what I wrote, correct? – Danny2036 Jan 28 '14 at 04:17
@user3242955 Yeah it is. – fastcodejava Jan 28 '14 at 17:23

score 1 · Answer 2 · answered Mar 17 '14 at 22:33

1

Formally, using Sigma Notation, you can infer the order of growth through the following:

enter image description here

answered Mar 17 '14 at 22:33

Mohamed Ennahdi El Idrissi

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Big O Theory- triple nested loop

2 Answers2