具有三个嵌套for循环的函数运行时间的渐近增长

Question

I have pseudo code for a function (below).
I understand that if each of i , j and k were 1 then worst case run time would be O(n^3) . I am struggling to understand the impact of n/2 though - if any - on run time. Any guidance would be great.

for i=n/2; i < n; increase i by 1
    for j=1; j < n/2; increase j by 1
        for k = 1; k < n; increase k by k*2
            Execute a Statement

Answer 1

Your understanding is not correct

k is increased by k*2 which leads to logarithmic time, the complexity is actually O(n^2 * log n)

The O(n/2) = O(n) , therefore the n/2 does not have any impact on asymptotic growth.

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If you are not sure, the general approach is to count it as precise as possible and then remove the constants.

for i will be execute n/2 times, the for j also n/2 times and k will be executed log n times.

n/2 * n/2 * log n = (n^2/4) * log n . You can remove constants, so O(n^2 * log)

Answer 2

The worst case time complexity is not O(N^3)

Check the innermost for loop. K increases by K * 2 That means, the innermost for loop will take O(lgN) time

Other two outer loops would take O(N) time each and N/2 would not have any effect on the asymptotic growth of run time.

So, the overall time complexity would be O(N^2 * lgN)