此方法在数组 (python) 中查找反转次数的时间复杂度是多少？

Question

inv=0
for j in range(n):
     inv=inv+ sum((x<arr[j]) for x in arr[j:] )

For every element I am checking the number of elements smaller than it occurring after it in the array.( arr[j : ] )

Answer 1

It is O(n ² ) . Here is how you can compute this:

• for the 1 ^st element, you need to compare with the next n-1 elements.

• for the 2 ^nd element, you need to compare with the next n-2 elements.

  ...

• for the n ^th element, you need to compare with the next 0 elements.

Therefore, in total you are making (n-1) + (n-2) + ... + 1 + 0 = n(n-1) / 2 comparisons, which is quadratic in n.

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More efficient approaches do exist. For example, by using a divide and conquer based strategy, you can count them in O(n log(n)) . See this nice link !

Answer 2

inv=0
for j in range(n):
    inv=inv+ sum((x<arr[j]) for x in arr[j:] )

Let's break this code into three parts

1: inv = 0 This will take contant time operation sat T1

2: for j in range(n): here we are running a loop for variable n

total time required now is T1 + N * f(a) here f(a) is the time taken by the body of loop. For simplicity we can remove constant factor. So complexity is N * f(a)

Now here comes the tricky part. What is f(a)

3: inv = inv + sum((x<arr[j]) for x in arr[j:] ) concentrate on sum((x < arr[j] for x in arr[j:]) sum will add all the values that are below arr[j] in

loop for x in arr[j:]

so you are left with f(a) as N , N - 1 , N - 2 up to N - N

Combining this all together you get N * (N + N - 1 + N - 2 + ... + N - N) which is (N * N - 1) / 2 that is O(N^2)

Hope you get it.