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Java Program to Count Inversions in an array | Set 1 (Using Merge Sort)
• Last Updated : 11 Dec, 2018

Inversion Count for an array indicates – how far (or close) the array is from being sorted. If array is already sorted then inversion count is 0. If array is sorted in reverse order that inversion count is the maximum.
Formally speaking, two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j.

Example:
The sequence 2, 4, 1, 3, 5 has three inversions (2, 1), (4, 1), (4, 3).

## Java

 `// Java program to count``// inversions in an array``class` `Test {``    ``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 method to test the above function``    ``public` `static` `void` `main(String[] args)``    ``{``        ``System.out.println(``"Number of inversions are "``                           ``+ getInvCount(arr.length));``    ``}``}`
Output:
```Number of inversions are 5
```

METHOD 2(Enhance Merge Sort)
Suppose we know 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 we have to count during the merge step. Therefore, to get number of inversions, we need to add number of inversions in left subarray, right subarray and merge().

## Java

 `// Java implementation of counting the``// inversion using merge sort`` ` `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 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 = left; ``/* i is index for left subarray*/``        ``j = mid; ``/* j is index for right subarray*/``        ``k = left; ``/* k is index for resultant merged subarray*/``        ``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 method to test the above function``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = ``new` `int``[] { ``1``, ``20``, ``6``, ``4``, ``5` `};``        ``System.out.println(``"Number of inversions are "` `+ mergeSort(arr, ``5``));``    ``}``}`
Output:
```Number of inversions are 5
```

Please refer complete article on Count Inversions in an array | Set 1 (Using Merge Sort) for more details!

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