算法复杂度分析

Question

What's the big-O time complexity of the two methods listed below?

Is method 1 O(log n²) or O(log n) ?

Is method 2 O(n²) or something else?

public static void method1(int n) {
    int k = 0;      
    for (int i = 1; i <= n; i = i*2)
       for (int j = n; j > 1; j = j/2)
           k++;     
}               

public static void method2(int n) {             
    for (int i = 1; i <= n; i = i + 1)
       for (int j = i; j <= n; j++)
          method1(n);
}

Answer 1

In the first example, the outer loop executes logn times, and the second loop also executes log n times. The complexity is O(log² n).

In the second example, the first loop executes n times and the second loop executes i times where i is the current index of the first loop. So, it executes 1 + 2 + 3 + ... + n times, summing up to n ⋅ (n - 1) / 2. The resulting complexity is O( (n² - n)/2 ) = O(n²). Read more here 1 + 2 + 3 ... series

So, to wrap it up, method2 executes method1 n² times, giving the total complexity of O(n² ⋅ log² n)

Answer 2

Well in the first method you have 2 nested loops where each one is exactly log(n) so yes that would turns them into a O((logn)^2)

For the second method, if we focus on the second loop we can see that for

• i = 1 => runs n times
• i = 2 => runs n-1 times
• ...
• i = n => runs 1 times

This is an arithmetic progression and its sum is equal to n * (n+1) / 2 which turns it into an O((nlogn)^2) since we are calling method1 about n^2 times.

Answer 3

The complexity of method1 is O((log n)²) since we have a nested loop where each loop runs O(log n) times.

In Method2 we execute Method1 a triangluar number of times, ie, we execute it O(n²) times. Since we execute an O((log n)²) function O(n²) times, the resulting complexity for Method2 is O(n² ⋅ (log n)²).

Answer 4

Methodically and formally, you can use Sigma notation like this: