+1 vote
136 views
for(i=1;i<=n;i++)
for(j=1;j<=i;j++)
for(k=1;k<=j;k++)
for(l=1;l<=k;l++)
printf("gate");

+1 vote
Time complexity of nested loops is equal to the number of times the innermost statement is executed.

I think its O(n2)

$T(n) = \sum_{i=1}^{n} \sum_{j=1}^{i} \sum_{k=1}^{j} \sum_{l=1}^{k} c$

( Taking cost of printf as $c$)

$= c\sum_{i=1}^{n} \sum_{j=1}^{i} \sum_{k=1}^{j} k$

$= c\sum_{i=1}^{n} \sum_{j=1}^{i} \frac{j.(j+1)}{2}$

$= d \sum_{i=1}^{n} \sum_{j=1}^{i} (j^2 + j)$

$(d = c/2)$

$= d \sum_{i=1}^{n} \frac{i.(i+1)(2i+1)}{6} + \frac{i.(i+1)}{2})$

$= e \sum_{i=1}^{n} 2i^3 + 3i^2 + i + 3i^2 + 3i$

$(e = d/6)$

$= e \sum_{i=1}^{n} 2i^3 + 6i^2 + 4i$

Well, sum of the cubes of first $n$ natural numbers is $\left({\frac{n.(n+1)}{2}}\right)^2 = \Theta\left (n^4\right)$

So, our answer will also be $\Theta\left( n^4\right)$ due to the $i^3$ term going from $i=1..n$