Inversion count in Array using BIT

// Java program to count inversions 
// using Binary Indexed Tree

class GFG
{
    
// Returns sum of arr[0..index]. 
// This function assumes that the 
// array is preprocessed and partial
// sums of array elements are stored
// in BITree[].
static int getSum(int[] BITree, int index)
{
    int sum = 0; // Initialize result

    // Traverse ancestors of BITree[index]
    while (index > 0)
    {
        // Add current element of BITree to sum
        sum += BITree[index];

        // Move index to parent node in getSum View
        index -= index & (-index);
    }
    return sum;
}

// Updates a node in Binary Index
// Tree (BITree) at given index
// in BITree. The given value 'val' 
// is added to BITree[i] and all
// of its ancestors in tree.
static void updateBIT(int[] BITree, int n,
                        int index, int val)
{
    // Traverse all ancestors and add 'val'
    while (index <= n)
    {
        // Add 'val' to current node of BI Tree
        BITree[index] += val;

        // Update index to that of parent in update View
        index += index & (-index);
    }
}

// Returns inversion count arr[0..n-1]
static int getInvCount(int[] arr, int n)
{
    int invcount = 0; // Initialize result

    // Find maximum element in arr[]
    int maxElement = 0;
    for (int i = 0; i < n; i++)
        if (maxElement < arr[i])
            maxElement = arr[i];

    // Create a BIT with size equal to 
    // maxElement+1 (Extra one is used so 
    // that elements can be directly be
    // used as index)
    int[] BIT = new int[maxElement + 1];
    for (int i = 1; i <= maxElement; i++)
        BIT[i] = 0;

    // Traverse all elements from right.
    for (int i = n - 1; i >= 0; i--)
    {
        // Get count of elements smaller than arr[i]
        invcount += getSum(BIT, arr[i] - 1);

        // Add current element to BIT
        updateBIT(BIT, maxElement, arr[i], 1);
    }

    return invcount;
}

// Driver code
public static void main (String[] args)
{
    int []arr = {8, 4, 2, 1};
    int n = arr.length;
    System.out.println("Number of inversions are : " +
                                getInvCount(arr,n));
}
}

// This code is contributed by mits

// Java program to count inversions 
// using Binary Indexed Tree
import java.util.*;
class GFG{

// Returns sum of arr[0..index]. 
// This function assumes that the 
// array is preprocessed and partial 
// sums of array elements are stored 
// in BITree[].
static int getSum(int BITree[],
                  int index)
{
  // Initialize result
  int sum = 0; 

  // Traverse ancestors of
  // BITree[index]
  while (index > 0)
  {
    // Add current element of
    // BITree to sum
    sum += BITree[index];

    // Move index to parent node 
    // in getSum View
    index -= index & (-index);
  }
  return sum;
}

// Updates a node in Binary Index Tree 
// (BITree) at given index in BITree.  
// The given value 'val' is added to 
// BITree[i] and all of its ancestors 
// in tree.
static void updateBIT(int BITree[], int n, 
                      int index, int val)
{
  // Traverse all ancestors 
  // and add 'val'
  while (index <= n)
  {
    // Add 'val' to current 
    // node of BI Tree
    BITree[index] += val;

    // Update index to that of 
    // parent in update View
    index += index & (-index);
  }
}

// Converts an array to an array 
// with values from 1 to n and 
// relative order of smaller and 
// greater elements remains same.  
// For example, {7, -90, 100, 1} 
// is converted to {3, 1, 4 ,2 }
static void convert(int arr[], 
                    int n)
{
  // Create a copy of arrp[] in temp 
  // and sort the temp array in 
  // increasing order
  int []temp = new int[n];
  
  for (int i = 0; i < n; i++)
    temp[i] = arr[i];
  Arrays.sort(temp);

  // Traverse all array elements
  for (int i = 0; i < n; i++)
  {
    // lower_bound() Returns pointer 
    // to the first element greater 
    // than or equal to arr[i]
    arr[i] =lower_bound(temp,0, 
                        n, arr[i]) + 1;
  }
}

static int lower_bound(int[] a, int low, 
                       int high, int element)
{
  while(low < high)
  {
    int middle = low + 
                (high - low) / 2;
    if(element > a[middle])
      low = middle + 1;
    else 
      high = middle;
  }
  return low;
}

// Returns inversion count 
// arr[0..n-1]
static int getInvCount(int arr[], 
                       int n)
{
  // Initialize result
  int invcount = 0; 

  // Convert arr[] to an array 
  // with values from 1 to n and
  // relative order of smaller 
  // and greater elements remains
  // same.  For example, {7, -90, 
  // 100, 1} is converted to
  //  {3, 1, 4 ,2 }
  convert(arr, n);

  // Create a BIT with size equal 
  // to maxElement+1 (Extra one is 
  // used so that elements can be 
  // directly be used as index)
  int []BIT = new int[n + 1];
  
  for (int i = 1; i <= n; i++)
    BIT[i] = 0;

  // Traverse all elements 
  // from right.
  for (int i = n - 1; i >= 0; i--)
  {
    // Get count of elements 
    // smaller than arr[i]
    invcount += getSum(BIT, 
                       arr[i] - 1);

    // Add current element to BIT
    updateBIT(BIT, n, arr[i], 1);
  }

  return invcount;
}

// Driver code
public static void main(String[] args)
{
  int arr[] = {8, 4, 2, 1};
  int n = arr.length;
  System.out.print("Number of inversions are : " +  
                    getInvCount(arr, n));
}
}

// This code is contributed by Amit Katiyar