this will help Intuitively it is O(1) because as n increases the runtime does not increase after a certain point. However, this is an edge case, as were n bounded by a much higher number, say the maximum value of an int, it would seem to be no different than if n was not bounded at all. However, when considering runtime using complexity theory we usually ignore things like that maximum size of an int. Another way to think of this is that the number of iterations grows linearly with n for n in (0,100) and is constant otherwise. When considering n can be any value, however, the algorithm is definitely O(1) code :
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Time complexity analysis of function with recursion inside loop
By : user6658026
Date : March 29 2020, 07:55 AM
wish helps you The way to analyze this is by developing a recursion relationship based on the length of the input and the (unknown) probability that a prefix is in s. Let's assume that the probability of a prefix being in s is given by some function pr(L) of the length L of the prefix. Let the complexity (number of operations) be given by T(len). If len == 0 (word is the empty string), then T = 1. (The function is missing a final return statement after the loop, but we're assuming that the actual code is only a sketch of the idea, not what's actually executing). code :
len T(len)
0 1
1 1 + 1 = 2
2 2 + 2 + 1 = 5
3 3 + (4 + 2 + 1) = 11
4 4 + (11 + 5 + 2 + 1) = 23
5 5 + (23 + 11 + 5 + 2 + 1) = 47

Time complexity analysis for a while loop with a cubed iterator
By : Genius Racer Officia
Date : March 29 2020, 07:55 AM
This might help you The algorithm will iterate as long as the condition is fulfilled: j <= n. So it stops when j > n. So it will iterate with j = 3 27 (3^3) 19683 (27^3 = 3^(3^3)) ...

Time complexity analysis. while loop with inner for loop
By : Ming Tan
Date : March 29 2020, 07:55 AM
this will help No, your analysis is incorrect. Note that each iteration of the inner loop does O(1) work, so the total time complexity can be found by multiplying the number of loop iterations by O(1). In this case, the loop runs for O(log n) iterations, since i can only double O(log n) times before it exceeds n. Therefore, the total time complexity is O(log n).

Algorithm Time Complexity Analysis (for loop with inner while loop)
By : user2710498
Date : March 29 2020, 07:55 AM
To fix this issue Here, the complexity is measured by studying the number of times the program will run the operations of the inner loop. Each time Condition is triggered, the inner loop runs x times. Thus the inner loop complexity is O(x).

Misunderstanding small details w/ nested forloop time complexity analysis... How to tell O(n) and O(n²) apart
By : user3032716
Date : March 29 2020, 07:55 AM
like below fixes the issue You have a line that says if (j < i) j = j + n; which essentially breaks out of the loop (when j < i), and since the inner loop starts at 0, this will trigger on the first iteration every time (except the first time), making it run in constant time. You essentially only have one loop here. The code can be rewritten as code :
int x = 0;
for (int i = 0; i < n; i++) {
x = x + 1;
}

