CareerCup

int main() {
    int a[20],n,k;
	scanf("%d",&n);
	for(int i =0 ;i<n;i++) {
		scanf("%d",&a[i]);
	}
	int c = 0;
	for ( int i = 0; i< n;i++) {
		for (int j = i+1;j<n;j++) {
			if(a[i]>a[j]) {
				printf("%d %d\n",a[i],a[j]);
				c++;
			}
		}

}

Modifed version of mergesort algorithm could give us the no of inversions in the array. I tested this on
(1), (1,1,1,1), (1,2,3,4,5,6), (3,5,7,2,8)

package arrays;

public class FindIInversionCount {

	public static int[] temp_data;
	public static int inversion_count = 0;

	public static void main(String[] args) {
		int[] a = {3,5,7,2,8};
		temp_data = new int[a.length];
		inversion_count = merge_and_count(a, 0, a.length - 1);

//		for (int i : temp_data) {
//			System.out.println(i);
//
//		}
		System.out.println("inversion count: " + inversion_count);
	}

	private static int merge_and_count(int[] a, int low, int high) {
		if (high - low > 1) {
			// more than two element

			int mid = (int) Math.floor((high + low) / 2);

			merge_and_count(a, low, mid);
			merge_and_count(a, mid + 1, high);

			// merge the data and count the inversions
			int i = low;
			int j = mid + 1;
			int count = low;
			while (i <= mid && j <= high) {
				if (a[i] <= a[j]) {
					temp_data[count++] = a[i];
					i++;
				} else if (a[i] > a[j]) {
					temp_data[count++] = a[j];
					j++;
					for (int k = i; k <= mid; k++) {
						inversion_count++;
					}
				}
			}

			if (j <= high) {
				while (j <= high) {
					temp_data[count++] = a[j];
					j++;
				}
			}
			if (i <= mid) {
				while (i <= mid) {
					temp_data[count++] = a[i];
					i++;
					inversion_count++;
				}
			}
		} else {
			// we have to deal with less than two elements
			if (high == low + 1) {
				if (a[low] > a[high]) {
					int tmp = a[low];
					a[low] = a[high];
					a[high] = tmp;
					inversion_count++;
				}
			} else if (high == low) {
				System.out.println("One element no sorting needed");
				temp_data[low] = a[low];
			}
		}
		return inversion_count;
	}
}

I think I have solution.
I am not sure about running time as I haven't calculated it yet.

Let's say we have an array of 11 elements (10,14,12,18,7,9,13,2,6,15,1)
Possible inversions :
(10,8) (10,7) (10,9) (10,2) (10,6) (10,1)
(14,12) (14,8) (14,7) (14,9) (14,13) (14,2) (14,6) (14,1)
(12,8) (12,7) (12,9) (12,2) (12,6) (12,1)
(8,7) (8,2) (8,6) (8,1)
(7,2) (7,6) (7,1)
(9,2) (9,6) (9,1)
(13,2) (13,6) (13,1)
(2,1)
(6,1)
(15,1)

Total : 36 Inversions

Now solution :

1. Let's define element with their positions.

10-1
14-2
12-3
8-4
7-5
9-6-
13-7
2-8
6-9
15-10
1-11

2. Find Minimum element , in the above case we have 1 at position 11 , so all above 1 will be greater than 1 -- 10 Inversion.

3. Find next minimum element and exclude 1 , in above case it's 2 at position 8 -- 7 Inversion.

4. Repeat step 3, next minimum element 6 at position 9 , Now exclude 2 (as it already been processed - Inversion (9 (position of 6 ) -1 (2 is already processed) - 1 (as element 6 itself )
--7 Inversion.

5. Repeat step 3 & 4 and in each iteration we will get number of inversions.

6. At last add all inversions--10+7+7+4+3+3+0+1+1+0+0 = 36 Inversion total.

Let me know if it doesn't apply to all case.
I will try to optimize it and write the running time.

Thanks,

Pankaj T.

Sorry, the above output was not mentioned correctly.
The Output should be 3. ( It has 3 inversions: 3>2,, 5>2, 7>2)

I forgot to mention the length of the array is unknown and your code should be optimized in less than O(n^2).

number of inversions = sum of distance of (current position of each element, its position in sorted array) / 2.
Prove by induction. n = 1 case is trial.
n = k to n = k+1: suppose the largest element is inserted to position i.
elements behind it increase the distance by 1, there are k+1-i of them;
itself has the distance of k+1-i.
This insertion created k+1-i new inversions.
So the conclusion is valid for n=k+1.

Back to the question, now it's easy to learn it has a solution of O(n*log(n)) with sorting and then counting distances.

Any idea on an O(n) solution?

divide and conquer, similar to merge sort. O(nlgn)

For each element in an array
--->Insert into a BST and count the number of successors in the BST.
the no if successors is the no of inversions for that element.
Since we need the count of the successors we can increase the efficiency by maintaining the number of nodes in right subtree at each node.
This should be a nlogn .

def find_highest(list) :
count = 0
print(list)
i = list[0]
print(i)
for i_max in list:
if i > i_max:
count += 1
print(f'count: {count}')
return count

def highest_count(list):
count = 0
for index,_ in enumerate(list):
count += find_highest(list[index:])

return count

# Test
print(highest_count([3,5,7,2,8]))

import java.util.*;
public class MyClass {
    public static int findHighest(List<Integer> numbers) {
        int count = 0;
        int max = numbers.get(0);
        for(int number : numbers) {
            if(max > number)
                count += 1;
        }
        return count;
    }
    
    public static int countHighest(List<Integer> numbers) {
       int count = 0;
        for(int i = 0; i<numbers.size(); i++) {
            count += findHighest(numbers.subList(i, numbers.size()));
        }
            
        return count;
    }
    
    public static void main(String[] args) {
        System.out.println(countHighest(Arrays.asList(3,5,7,2,8))); 
    }
}