计算按升序排列然后降序排列的数组中的反转次数

Question

Given an array that is sorted in increasing order then decreasing order eg: {1, 3, 5, 4, 2}, I want to find the number of inversions required to sort this array in O(n). This is my code which sorts such arrays in O(n):

void sort(long* a, long* result, long n)
 {
      long* start = &a[0];
      long* end = &a[n - 1];
      long count = 0;
      while (start <= end) {
        if (*start < *end) {
            result[count] = *start;
            start += 1;
        }
        else {
           result[count] = *end;
           end -= 1;
       }
      }
  }

In my code, I used 2 pointers from the start and end and stored the sorted array into the result array. How do I count the number of inversions using a similar algorithm?

Answer 1

You could use this algorithm from "https://www.geeksforgeeks.org/counting-inversions/":
Algorithm:

1. Traverse through the array from start to end.
   
2. For every element find the count of elements smaller than the current number upto that index using another loop.
   
3. Sum up the count of inversion for every index.
   
4. Print the count of inversions.
   
   You can also find a code example on their website.

Moreover, for your function it seems that you JUST need a new variable to store the number of oinversions you do.

Hope it helped!