By looking at the constraints of a problem, we can often "guess" the solution.

**Common time complexities**

Let *n* be the main variable in the problem.

- If
*n*≤ 12, the time complexity can be O(n!). - If
*n*≤ 25, the time complexity can be O(2^{n}). - If
*n*≤ 100, the time complexity can be O(n^{4}). - If
*n*≤ 500, the time complexity can be O(n^{3}). - If
*n*≤ 10^{4}, the time complexity can be O(n^{2}). - If
*n*≤ 10^{6}, the time complexity can be O(n log n). - If
*n*≤ 10^{8}, the time complexity can be O(n). - If
*n*> 10^{8}, the time complexity can be O(log n) or O(1).

**Examples of each common time complexity**

- O(n!) [Factorial time]: Permutations of 1 ...
*n* - O(2
^{n}) [Exponential time]: Exhaust all subsets of an array of size*n* - O(n
^{3}) [Cubic time]: Exhaust all triangles with side length less than*n* - O(n
^{2}) [Quadratic time]: Slow comparison-based sorting (eg. Bubble Sort, Insertion Sort, Selection Sort) - O(n log n) [Linearithmic time]: Fast comparison-based sorting (eg. Merge Sort)
- O(n) [Linear time]: Linear Search (Finding maximum/minimum element in a 1D array), Counting Sort
- O(log n) [Logarithmic time]: Binary Search, finding GCD (Greatest Common Divisor) using Euclidean Algorithm
- O(1) [Constant time]: Calculation (eg. Solving linear equations in one unknown)

**Explanations based on Codeforces problems**

255D Mr. Bender and Square

Observe that 1 ≤*n, c*≤ 10^{9}. Referring to the information above, the program's time complexity should be either O(log n) or O(1). Since no O(1) solution exists, we conclude that binary search must be used.580B Kefa and Company

In this problem, 1 ≤*n*≤ 10^{5}, which suggests that the time complexity can be either O(n log n) or O(n). It is quite obvious that sorting is required. Therefore, O(n log n) is the correct solution of this problem.583B Robot's Task

Notice that*n*in very small (1 ≤*n*≤ 1000) in this problem. It means that a O(n^{2}) solution can solve it. We simply need to simulate the robot's moves.

I hope that this can help you figure out the solution of some problems quicker :)

Note: The above method may not always work in all problems. Some may require algorithms that have complex time complexities, while in some problems like 591B Rebranding, the range of *n* does not match the time complexity of the "optimal" solution. (1 ≤ *n, m* ≤ 200 000 suggests that the time complexity is O(n log n) or O(n) but the time complexity of the solution is actually O(1).)

10

^{8}for sounds too optimistic. I'd say, forn≤ 10^{6}has a chance to pass.Hey, why do you call mergesort non-comparison-based?

Why do you care about lower bounds? Only upper bounds matter (except maybe some special cases / undefined-ness).

If it's

n≤ 50, it can be anything polynomial, but there's probably a simple slower solution and a faster one that's harder to implement. Also, you have 75 minutes.What do you mean by "75 minutes"?

Old TC, of course.

Thanks for pointing out my mistakes. I have changed them.

Another good way to remember this is that whatever algorithm you chose, if you plug in

`n`

, you should get around`~10^8`

. This makes is easier to analyze where your complexity might depend on 2 things e.g.`O(MN log N)`

some problems have multiple test cases.As like 1<=t<=100000 and 1<=n<=1000000. At this type of problem how I can understand what algorithm will be needed?

These problems often say that the sum of all cases is, at most,

N.Should u consider value of each element in the constarint ?

Thanks for sharning.

When

`N <= 10^5`

We can use

`O(N√N)`

some kind of brute force or any square root algorithm like Mo's.Problem for this senario1147B - Chladni Figure

This problem can also be solved in

`O(NlogN)`

dont seconds matter here? I mean since n ≤ 10^8 -> time complexity can be O(n) Lets say now n = 10^9, is there any chance it is going to give a TLE for 1 second but might get accepted for 2 seconds? Can you also include the time paramenter as well, to check if the solutions time complexity can get accepted?

I have seen problems with a colossal time limit to kill a solution. For instance, N = 30000 and 10 seconds TL. It would be enough to get AC with the O(N^2) solution (intended solution for the problem), yet it would kill the O(N^3) heavily optimized algorithm (that used vectorization I think). But as far as I remember, I have only encountered such instances 2 or 3 times.

So following your example, maybe a writer wanted to kill an O(N log N) solution and increase the time limit to 20 seconds and N=2e9.

But yeah, these problems are not that common.

Anyway, if you are unsure if your program will fit into the time limit, try plugging the numbers in the calculator. I do that all the time. It works out pretty well.

PS: Sorry for necroposting ;-;

10 times data means 10 times processing time for O(n) solution. so n = 10^9 will be accepted in more than 10 seconds if n ≤ 10^8 takes 1 second

Sharing one of my related answers in Quora

What will be the ideal complexity for n<= 10^16