What will be the time complexity of such kind of functions?

Question

void yash(){ 

    int i,j;
    for(i=1;i<=n;i++)
       for(j=1;j<=log(i);j++)
          printf("Hello !!");
}

what will be the complexity of this function?

In terms of big O annotation, O(n*log(n)) – Nilesh Sep 03 '18 at 15:32 — Nilesh, Sep 03 '18 at 15:32
What do **YOU** think? Share your thoughts! – MrSmith42 Sep 03 '18 at 15:42 — MrSmith42, Sep 03 '18 at 15:42
Depends on how costly the `log(i)`-call is. – MrSmith42 Sep 03 '18 at 15:43 — MrSmith42, Sep 03 '18 at 15:43

nice_dev · Answer 1 · 2018-09-03T15:50:31.730

2

Loop will iterate like this-

1 * log(1) time.
2 * log(2) times.
so on and so forth till n * log(n) times.

Among all these values, n * log(n) is the highest(upper bound). So, time complexity is n * log(n) asymptotically.

UPDATE:

If you are calculating log(i) for each iteration instead of storing it in a variable, then time complexity is O(n * log(n) ²), assuming calculating log(i) isn't O(1). If it's O(1), then time complexity is O(n log(n)).

edited Sep 03 '18 at 15:50

answered Sep 03 '18 at 15:34

nice_dev

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"assuming calculating log(i) isn't O(1)" What time complexity could log(i) have that would make the term log(n)^2? O(log n)? – David G Sep 03 '18 at 16:54
@0x499602D2 Yes, but not sure. Hence, I said `assuming`. – nice_dev Sep 03 '18 at 19:31

What will be the time complexity of such kind of functions?

1 Answers1