For Array[0.....N-1], an inversion is an element pair that satisfes i<j and A[i]>A[j]. For instance, in array (1,3,5,2,4,6), such pairs are (3,2) (5,2) (5,4), so number of inversions is 3. The challenge is to implement an algorithm that computes the number of inversions with the array given to you.

**Method1 (Brute Force Approach)**

For each element, count number of elements which are on right side of it and are smaller than it.

public static int getInversionCount(int arr[])

{

int invCount = 0;

int i, j;

int n = arr.length;

for(i = 0; i < n - 1; i++)

{

for(j = i+1; j < n; j++)

{

if(arr[i] > arr[j])

invCount++;

}

}

return invCount;

}

**Method2 ( Merge Sort Approach)**

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().

**How to get 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]

/* An auxiliary recursive function that sorts the input array and

returns the number of inversions in the array. */

public static int mergeSort(int arr[], int temp[], int left, int right)

{

int mid, invCount = 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

invCount = mergeSort(arr, temp, left, mid);

invCount += mergeSort(arr, temp, mid+1, right);

//Merge the two parts

invCount += merge(arr, temp, left, mid+1, right);

}

return invCount;

}

// This function merges two sorted arrays and returns inversion count in

// the arrays.

public 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; // i is index for right subarray

k = left; // i 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++];

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;

}

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