What is the time complexity of following function fun()? Assume that log(x) returns log value in base 2.
C++
void fun(){ int i, j; for (i = 1; i <= n; i++) for (j = 1; j <= log(i); j++) cout << "neveropen";}// This code is contributed by SHUBHAMSINGH10. |
C
void fun(){ int i, j; for (i = 1; i <= n; i++) for (j = 1; j <= log(i); j++) printf("neveropen");} |
Java
static void fun(){ int i, j; for (i = 1; i <= n; i++) for (j = 1; j <= log(i); j++) System.out.printf("neveropen");}// This code is contributed by umadevi9616 |
Python3
import mathdef fun(): i = 0 j = 0 for i in range(1, n + 1): for j in range(1,math.log(i) + 1): print("neveropen")# This code is contributed by SHUBHAMSINGH10. |
C#
static void fun(){ int i, j; for (i = 1; i <= n; i++) for (j = 1; j <= log(i); j++) Console.Write("neveropen");}// This code is contributed by umadevi9616 |
Javascript
const fun(){ let i, j; for (i = 1; i <= n; i++) for (j = 1; j <= Math.log(i); j++) document.write("neveropen");}// This code is contributed by SHUBHAMSINGH10. |
Time Complexity of the above function can be written as θ(log 1) + θ(log 2) + θ(log 3) + . . . . + θ(log n) which is θ(log n!)
Order of growth of ‘log n!’ and ‘n log n’ is same for large values of n, i.e., θ(log n!) = θ(n log n). So time complexity of fun() is θ(n log n).
The expression θ(log n!) = θ(n log n) can be easily derived from following Stirling’s approximation (or Stirling’s formula).
log n! = n*log n - n = O(n*log(n))
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Sources:
http://en.wikipedia.org/wiki/Stirling%27s_approximation
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