What is time complexity for the following code?

问题

It seems the complexity of the following code should be O(n^2) but it's O(n), how?

void fun(int n, int arr[])
{
    int i = 0, j = 0;
    for(; i < n; ++i)
        while(j < n && arr[i] < arr[j])
            j++;
}

回答1:

j is not reset to 0 with every iteration of the outer loop. As such, it runs to n-1 just once, same as i does. So you have two parallel/intermingled iterations from 0 to (at most) n-1.

In every step, the program increases i by one. The program terminates when i reaches n. The "outer loop" is run n times.

There is also an "inner loop" about j. But all it does is increase j until it reaches i (at most, sometimes it does less). j is never decreased. So that part is also run at most n times in total (not n times for each iteration of the "outer loop").

回答2:

In the first look, the time complexity seems to be O(n^2) due to two loops. But, please note that the variable j is not initialized for each value of variable i.

Hence, the inner j++ will be executed at most n times.

The i loop also runs n times.

So, the whole thing runs for O(n) times.

Please observe the difference between the function given in question and the below function:

void fun(int n, int arr[])
               {
int i = 0, j = 0;
for(; i < n; ++i)
{
    j = 0;
    while(j < n && arr[i] < arr[j])
        j++;
}

Still not convinced ?

Let's assume the array passed has its element in decreasing order. We will just dry run through the code :

               Iteration 1 : i = 0, j = 0. arr[0] < arr[0] is false. So, the 
                             inner while loop breaks.
               Iteration 2: i =1, j = 0. arr[1] < arr[0] is true. j becomes 
               Iteration 3 : i = 1, j = 1. Condition false. We break. Note 
                            that j will remain 1 and is not reset back to 0.
               Iteration 4 : i = 2, j = 1. arr[2] < arr[1]. True. j = 2.
               Iteration 5 : i = 2, j = 2. Condition false. Break.
               Iteration 6 : i = 3, j = 2. arr[3] < arr[2]. True. j = 3.
               Iteration 7 : i = 3, j = 3. Condition false. Break.

As you can see, the inner while loop only runs once in this case. So, total iterations is 2 * N.

回答3:

The answer is O(n) The outer loop runs 'n' times and the inner loop only runs to 'n' a single time in all the iterations combined as the value of j is never reset to 0. Therefore the answer is O(n+n)=O(n).

回答4:

Lets consider the worst case when the while loop is executed maximum no. of times. Initially: i=0 , j=0 => while loop does not gets executed since arr[0] = arr[0] i.e. j=0. Second iteration : i=1 , j=0 => while loop gets executed for the worst case i.e. j=1. Third iteration : i=2 , j=1 => while loop again gets executed for the worst case i.e. j=2. ... nth iteration : i=n-1 , j=n-2 => while loop again gets executed for the worst case i.e. j=n-1.

So, by doing this exercise we can observe that every time j = i-1 except i=0 and j=0 OR we can say that the while loop is just running in parallel to the for loop and thus the no. of executions of the while loop is equal to the no. of executions of the for loop. Hence Order = O(n);

来源：https://stackoverflow.com/questions/30073130/what-is-time-complexity-for-the-following-code

标签

time-complexity