Reputation: 31
Is the time complexity of the following piece of code O(n^5)
? My reasoning is that the outer for loop is O(n)
, the middle for loop is O(n^2)
since the value of i is based on the value of n, and the inner for loop is O(n^2)
since the value of j is based on the value if i^2 which is based on the value of n^2.
int x = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < i * i; j++) {
for (int k = 0; k < j; k++) {
x++;
}
}
}
Upvotes: 1
Views: 402
Reputation: 15852
That is not that simple. To determine the complexity, one needs to calculate how many times the x
will increase.
The most inner loop runs `j` times.
The middle loop runs `i*i` times.
The outer loop runs n times.
Let's reduce:
The middle loop's complexity is:
1+2+3+...+(i-1)+i+(i+1)+...+(i-1)*(i-1) = (i^2-2i+1)*i*i/2=(i^4-2i^3+i^2)/2
And the outer loop runs n
times for each n from 0 to n-1
. It sums up to:
n^5/10 - n^4/2 + 5n^3/6 - n^2/2 + n/15
and it's actually O(n^5)
.
The mathematical notation is:
Upvotes: 3