Inversion Count for an array indicates – how far (or close) the array is from being sorted. If the array is already sorted, then the inversion count is 0, but if the array is sorted in the reverse order, the inversion count is the maximum.
Formally speaking, two elements a[i] and a[j] form an inversion if a[i] > a[j] and i Example:
Input: arr[] = {8, 4, 2, 1}
Output: 6
Explanation: Given array has six inversions:
(8, 4), (4, 2), (8, 2), (8, 1), (4, 1), (2, 1).
Input: arr[] = {3, 1, 2}
Output: 2
Explanation: Given array has two inversions:
(3, 1), (3, 2) METHOD 1 (Simple):
Approach: Traverse through the array, and for every index, find the number of smaller elements on its right side of the array. This can be done using a nested loop. Sum up the counts for all index in the array and print the sum.
Algorithm:
- Traverse through the array from start to end
- For every element, find the count of elements smaller than the current number up to that index using another loop.
- Sum up the count of inversion for every index.
- Print the count of inversions.
Implementation:
// C# program to count inversions // in an array using System;
using System.Collections.Generic;
class GFG
{ static int[] arr =
new int[] {1, 20, 6, 4, 5};
static int getInvCount(int n)
{
int inv_count = 0;
for (int i = 0; i < n - 1; i++)
for (int j = i + 1; j < n; j++)
if (arr[i] > arr[j])
inv_count++;
return inv_count;
}
// Driver code
public static void Main()
{
Console.WriteLine("Number of " +
"inversions are " +
getInvCount(arr.Length));
}
} // This code is contributed by Sam007 |
Output:
Number of inversions are 5
Complexity Analysis:
- Time Complexity: O(n^2), Two nested loops are needed to traverse the array from start to end, so the Time complexity is O(n^2)
- Space Complexity:O(1), No extra space is required.
METHOD 2(Enhance Merge Sort):
Approach:
Suppose the number of inversions in the left half and right half of the array (let be inv1 and inv2); what kinds of inversions are not accounted for in Inv1 + Inv2? The answer is – the inversions that need to be counted during the merge step. Therefore, to get the total number of inversions that needs to be added are the number of inversions in the left subarray, right subarray, and merge().
How to get the number of inversions in merge()?
In merge process, let i is used for indexing left sub-array and j for right sub-array. At any step in merge(), if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j]
The complete picture:
Algorithm:
- The idea is similar to merge sort, divide the array into two equal or almost equal halves in each step until the base case is reached.
- Create a function merge that counts the number of inversions when two halves of the array are merged, create two indices i and j, i is the index for the first half, and j is an index of the second half. if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j].
- Create a recursive function to divide the array into halves and find the answer by summing the number of inversions is the first half, the number of inversion in the second half and the number of inversions by merging the two.
- The base case of recursion is when there is only one element in the given half.
- Print the answer
Implementation:
// C# implementation of counting the // inversion using merge sort using System;
public class Test
{ /* This method sorts the input array and
returns the number of inversions in
the array */
static int mergeSort(int[] arr,
int array_size)
{
int[] temp = new int[array_size];
return _mergeSort(arr, temp, 0,
array_size - 1);
}
/* An auxiliary recursive method that sorts
the input array and returns the number
of inversions in the array. */
static int _mergeSort(int[] arr, int[] temp,
int left, int right)
{
int mid, inv_count = 0;
if (right > left)
{
/* Divide the array into two parts and
call _mergeSortAndCountInv() for
each of the parts */
mid = (right + left) / 2;
/* Inversion count will be the sum of
inversions in left-part, right-part
and number of inversions in merging */
inv_count += _mergeSort(arr, temp,
left, mid);
inv_count += _mergeSort(arr, temp,
mid + 1, right);
// Merge the two parts
inv_count
+= merge(arr, temp, left,
mid + 1, right);
}
return inv_count;
}
/* This method merges two sorted arrays
and returns inversion count in the
arrays.*/
static int merge(int[] arr, int[] temp,
int left, int mid, int right)
{
int i, j, k;
int inv_count = 0;
/* i is index for left subarray*/
i = left;
/* j is index for right subarray*/ j = mid;
/* k is index for resultant merged
subarray*/
k = left;
while ((i <= mid - 1) &&
(j <= right))
{
if (arr[i] <= arr[j])
{
temp[k++] = arr[i++];
}
else {
temp[k++] = arr[j++];
/* this is tricky -- see above
explanation/diagram for merge()*/
inv_count = inv_count + (mid - i);
}
}
/* Copy the remaining elements of left
subarray (if there are any) to temp*/
while (i <= mid - 1)
temp[k++] = arr[i++];
/* Copy the remaining elements of right
subarray (if there are any) to temp*/
while (j <= right)
temp[k++] = arr[j++];
/* Copy back the merged elements to
original array*/
for (i = left; i <= right; i++)
arr[i] = temp[i];
return inv_count;
}
// Driver code
public static void Main()
{
int[] arr =
new int[] {1, 20, 6, 4, 5};
Console.Write("Number of inversions are " +
mergeSort(arr, 5));
}
} // This code is contributed by Rajput-Ji |
Output:
Number of inversions are 5
Complexity Analysis:
- Time Complexity: O(n log n), The algorithm used is divide and conquer, So in each level, one full array traversal is needed, and there are log n levels, so the time complexity is O(n log n).
- Space Complexity: O(n), Temporary array.
Note that the above code modifies (or sorts) the input array. If we want to count only inversions, we need to create a copy of the original array and call mergeSort() on the copy to preserve the original array’s order.