What is time complexity for the following code?

Question

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; ++         


        

Answer 1

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.