Time complexity of nested for-loop
Big OComplexity TheoryTime ComplexityBig O Problem Overview
I need to calculate the time complexity of the following code:
for (i = 1; i <= n; i++)
{
for(j = 1; j <= i; j++)
{
// Some code
}
}
Is it O(n^2)?
Big O Solutions
Solution 1 - Big O
Yes, nested loops are one way to quickly get a big O notation.
Typically (but not always) one loop nested in another will cause O(n²).
Think about it, the inner loop is executed i times, for each value of i. The outer loop is executed n times.
thus you see a pattern of execution like this: 1 + 2 + 3 + 4 + ... + n times
Therefore, we can bound the number of code executions by saying it obviously executes more than n times (lower bound), but in terms of n how many times are we executing the code?
Well, mathematically we can say that it will execute no more than n² times, giving us a worst case scenario and therefore our Big-Oh bound of O(n²). (For more information on how we can mathematically say this look at the Power Series)
Big-Oh doesn't always measure exactly how much work is being done, but usually gives a reliable approximation of worst case scenario.
4 yrs later Edit: Because this post seems to get a fair amount of traffic. I want to more fully explain how we bound the execution to O(n²) using the power series
From the website: 1+2+3+4...+n = (n² + n)/2 = n²/2 + n/2. How, then are we turning this into O(n²)? What we're (basically) saying is that n² >= n²/2 + n/2. Is this true? Let's do some simple algebra.
- Multiply both sides by 2 to get: 2n² >= n² + n?
- Expand 2n² to get:n² + n² >= n² + n?
- Subtract n² from both sides to get: n² >= n?
It should be clear that n² >= n (not strictly greater than, because of the case where n=0 or 1), assuming that n is always an integer.
Actual Big O complexity is slightly different than what I just said, but this is the gist of it. In actuality, Big O complexity asks if there is a constant we can apply to one function such that it's larger than the other, for sufficiently large input (See the wikipedia page)
Solution 2 - Big O
A quick way to explain this is to visualize it.
if both i and j are from 0 to N, it's easy to see O(N^2)
O O O O O O O O
O O O O O O O O
O O O O O O O O
O O O O O O O O
O O O O O O O O
O O O O O O O O
O O O O O O O O
O O O O O O O O
in this case, it's:
O
O O
O O O
O O O O
O O O O O
O O O O O O
O O O O O O O
O O O O O O O O
This comes out to be 1/2 of N^2, which is still O(N^2)
Solution 3 - Big O
Indeed, it is O(n^2). See also a very similar example with the same runtime here.
Solution 4 - Big O
Let us trace the number of times each loop executes in each iteration.
for (int i = 1; i <= n; i++){ // outer loop
for (int j = 1; j <= i; j++){ // inner loop
// some code
}
}
In the first iteration of the outer loop (i = 1), the inner loop executes once
.
In the second iteration of the outer loop (i = 2), the inner loop executes twice
.
In the third iteration of the outer loop (i = 3), the inner loop executes thrice
.
So, in the last iteration of the outer loop (i = n), the inner loop executes n times
.
Therefore, the total number of times this code executes is
1 + 2 + 3 + … + n
= (n(n + 1) / 2)
(Sum of Natural Numbers Formula)
= (((n^2) + n) / 2)
= O(n^2)
——————
Also, do take a look at these
Solution 5 - Big O
On the 1st iteration of the outer loop (i = 1), the inner loop will iterate 1 times
On the 2nd iteration of the outer loop (i = 2), the inner loop will iterate 2 time
On the 3rd iteration of the outer loop (i = 3), the inner loop will iterate 3 times
.
.
On the FINAL iteration of the outer loop (i = n), the inner loop will
iterate n times
So, the total number of times the statements in the inner loop will be executed will be equal to the sum of the integers from 1 to n, which is:
((n)*n) / 2 = (n^2)/2 = O(n^2) times
Solution 6 - Big O
Yes, the time complexity of this is O(n^2).
Solution 7 - Big O
I think the easiest way to think about it is like this:
The outer loop runs n times, and for at least n/2 of those iterations, the inner loop runs at least n/2 times. The total number of inner loop iterations is therefore at least n2/4. That's O(n2)
Similarly, the outer loop runs n times, and in every iteration, the inner loop runs at most n times. The total number of inner loop iterations, therefore, is at most n2. That's also in O(n2).
Solution 8 - Big O
The inner loop depends on outer loops and the inner loop runs I times which gives me
for n = 5 if i = 1 inner loops runs 1 times 1 = 1
if i = 2 inner loops runs 2 times 1 + 2 = 3
if i = 3 inner loops runs 3 times 1 + 2 + 3 = 6
if i = 4 inner loops runs 4 times 1 + 2 + 3 + 4 = 10
if i = 5 inner loops runs 5 times 1 + 2 + 3 + 4 + 5 = 15
From above, we can know that n (n + 1) / 2
So O(n *(n+1))/2 = O(n2/2 + n/2) = O(n2/2) + O(n/2)
I am not great at algorithm analysis so please feel free to correct my answer.
Solution 9 - Big O
First we'll consider loops where the number of iterations of the inner loop is independent of the value of the outer loop's index. For example:
for (i = 0; i < N; i++) {
for (j = 0; j < M; j++) {
sequence of statements
}
}
The outer loop executes N times. Every time the outer loop executes, the inner loop executes M times. As a result, the statements in the inner loop execute a total of N * M times. Thus, the total complexity for the two loops is O(N2).