Menu
Started studying about complexity, I'm struggling with this one:
void what(int n) { int i; for (i = 1; i <= n; i++) { int x = n; while (x > 0) x -= i; } } Well, the first for loop is clearly O(n). The first iteration is O(n), second one is O(n/2).. and like that log(n) times I guess? Which means O(n) * O(log(n)) = O(n * log(n)) complexity. Did I get this right?
Edit: (not a duplicate) I know what Big O is. I've asked the correct evaluation in a specific case.
965
Sorting approach: Sort the array, this will bring all the duplicates together if present. Now navigate the array and check the adjacent elements to check for duplicates. Time Complexity : O(nlogn) Space Complexity: O(n) by using merge sort. Time Complexity : O(n) and Space Complexity: O(n).
The time complexity of a function (or set of statements) is considered as O(1) if it doesn't contain a loop, recursion, and call to any other non-constant time function. Example: swap() function has O(1) time complexity. A loop or recursion that runs a constant number of times is also considered O(1).
The outer loop runs n times.
For each iteration, the inner loops runs n / i times.
The total number of runs is:
n + n/2 + n/3 + ... + n/n Asymptotically (ignoring integer arithmetic rounding), this simplifies as
n * (1 + 1/2 + 1/3 + ... + 1/n) This series loosely converges towards n * log(n).
Hence the complexity is O(N.log(N)) as you expected.
52
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
