Better solution using BIT of size Θ(n) using C#

Software Development
#1

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 BIT. Sum up the counts for all indexes in the array and print the sum. The approach remains the same but the problem with the previous approach is that it doesn’t work for negative numbers as the index cannot be negative. The idea is to convert the given array to an array with values from 1 to n and the relative order of smaller and greater elements remains the same.

Example:-

arr[] = {7, -90, 100, 1} It gets converted to, arr[] = {3, 1, 4 ,2 } as -90 < 1 < 7 < 100. Explanation: Make a BIT array of a number of elements instead of a maximum element. Changing element will not have any change in the answer as the greater elements remain greater and at the same position.

  • Algorithm:

    1. Create a BIT, to find the count of the smaller elements in the BIT for a given number and also a variable result = 0.
    2. The previous solution does not work for arrays containing negative elements. So, convert the array into an array containing relative numbering of elements,i.e make a copy of the original array and then sort the copy of the array and replace the elements in the original array with the indices of the same elements in the sorted array.
      For example, if the array is {-3, 2, 0} then the array gets converted to {1, 3, 2}
    3. Traverse the array from end to start.
    4. For every index check how many numbers less than the current element are present in BIT and add it to the result

  • Implementation:

// C# program to count inversions

// using Binary Indexed Tree

using System;

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];

Array.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;

Console.Write( "Number of inversions are : " +

getInvCount(arr, n));

}

}

  • Output:

Number of inversions are : 6

  • Complexity Analysis:
    • Time Complexity: The update function and getSum function runs for O(log(n)). The getSum function has to be run for every element in the array. So overall time complexity is O(nlog(n)).
    • Auxiliary Space: O(n).
      Space required for the BIT is an array of the size n.