for(int i = 1;i<=n;i++){
      for(int j = i+1;j<=n-i;j++){

The first loop has time complexity of n and second for loop is dependent on i. That means for every i, the number of time inner for loop gets executed is: $\sum_{k=1}^{n/2}(n-2k)$ After doing some algebra, I get: $\frac{n^2}{2}-\frac{n^2}{4}-\frac{n}{2}$. This is the time complexity for inner for loop. Now we multiplying outer loop $n*(\frac{n^2}{2}-\frac{n^2}{4}-\frac{n}{2})$. Therefore, time complexity is $\Theta (n^3)$. But this can't be right? It should max be $\Theta n^2$. Cause inner loop doesnt get executed for every i value.$\forall i \; \; \;i>n/2$ inner loop doesnt get executed at all.

  • 1
    $\begingroup$ The sum $\sum\limits_{k=1}^{n/2}(n-2k)$ is already the total number of print done by your execution, not the number of times the inner loop gets executed. So the complexity is indead $\Theta(n^2)$. $\endgroup$
    – Nathaniel
    Mar 18, 2021 at 9:35
  • $\begingroup$ hmmm. Interesting. What if print is outside the inner for loop? What is the time complexity then? $\endgroup$
    – Jonathen
    Mar 18, 2021 at 9:41
  • $\begingroup$ If you mean for i … {print("hello"); for j… {}}, then the time complexity stays the same (there are $n$ total print, but the inner loop is still executed). $\endgroup$
    – Nathaniel
    Mar 18, 2021 at 9:52

1 Answer 1


Suppose $i=1$ then inner loop get executed $n-2$ times because of $2\leq j\leq n-1$ and $j $increment each step by one, As a result $n-1-2+1=n-2$ times inner loop executed.

If $i=2$ then inner loop executed $n-4$ times ... until $i=\frac{n}{2}$ inner loop executed 1 time.

Value of $i$ #inner loop executed
$i=1$ $n-2$
$i=2$ $n-4$
$i=3$ $ n-6$
... ...
$i=\frac{n}{2}$ $n-(n-1)$

It's equal to following summation:

$\frac{n}{2}\times n-2\sum_{i=1}^{\frac{n}{2}}i=\frac{n^2}{2}-\frac{n\times(\frac{n}{2}+1)}{2}=O(n^2)$


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