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A Time Complexity Question

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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 << "GeeksforGeeks";
}
 
// 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("GeeksforGeeks");
}

                    

Java

static void fun()
{
    int i, j;
    for (i = 1; i <= n; i++)
        for (j = 1; j <= log(i); j++)
            System.out.printf("GeeksforGeeks");
}
 
// This code is contributed by umadevi9616

                    

Python3

import math
def fun():
    i = 0
    j = 0
    for i in range(1, n + 1):
        for j in range(1,math.log(i) + 1):
            print("GeeksforGeeks")
 
# 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("GeeksforGeeks");
}
 
// 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("GeeksforGeeks");
}
 
// This code is contributed by SHUBHAMSINGH10.

                    

Time Complexity of the above function can be written as ?(log 1) + ?(log 2) + ?(log 3) + . . . . + ?(log n) which is ?(log n!)
Order of growth of ‘log n!’ and ‘n log n’ is same for large values of n, i.e., ?(log n!) = ?(n log n). So time complexity of fun() is ?(n log n).
The expression ?(log n!) = ?(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)) 


Sources: 
http://en.wikipedia.org/wiki/Stirling%27s_approximation



Last Updated : 27 Dec, 2021
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