这个 python function 的时间复杂度是多少？

Question

def hasPairWithSum(arr,target):
     for i in range(len(arr)):
         if ((target-arr[i])  in arr[i+1:]):
             return True

    return False    

In python, is time complexity for this function O(n) or O(n^2), in other words 'if ((target-arr[i]) in arr[i+1:])' is this another for loop or no? Also what about following function, is it also O(n^2), if not why:

def hasPairWithSum2(arr,target):
   seen = set() 
   for num in arr:
      num2 = target - num
      if num2 in seen:
         return True
      seen.add(num)
   return False

Thanks!

Answer 1

The first version has a O(n²) time complexity:

Indeed the arr[i+1:] already creates a new list with n-1-i elements, which is not a constant time operation. The in operator will scan that new list, and in worst case visit each value in that new list.

If we count the number of elements copied into a new list (with arr[i+1 ), we can sum those counts for each iteration of the outer loop:

  n-1
+ n-2
+ n-3
+ ...
+ 1
+ 0

This is a triangular number , and equals n(n-1)/2 , which is O(n²) .

Second version

The second version, using a set , runs in O(n) average time complexity.

There is no list slicing here, and the in operator on a set has -- contrary to a list argument -- an average constant time complexity. So now every action within the loop has an (average) constant time complexity, giving the algorithm an average time complexity of O(n) .

According to the python docs , the in operator for a set can have an amortised worst time complexity of O(n) . So you would still get a worst time complexity for your algorithm of O(n²) .

Answer 2

It's O(n^2)

First loop will run n times and second will run (nm) for every n. So the whole thing will run n(nm) times that is n^2 - nm . Knowing that m<n we know that it has a complexity of O(n^2)

But if it's critical to something you might consult someone who is better at that.