Time complexity of nested for-loop

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)

Indeed, it is O(n^2). See also a very similar example with the same runtime here.

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

1. https://stackoverflow.com/a/71805214/17112163
2. https://stackoverflow.com/a/71537431/17112163
3. https://stackoverflow.com/a/69821878/17112163
4. https://stackoverflow.com/a/72046825/17112163
5. https://stackoverflow.com/a/72046933/17112163

Yes, the time complexity of this is O(n^2).

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).

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.