Counting inversions in all subarrays of given size

Given an array and an integer k, count all inversions in all subarrays of size k.

Example:

Input : a[] = {7, 3, 2, 4, 1}, 
        k = 3;
Output : 6
Explanation: subarrays of size 3 are - 
{7, 3, 2}
{3, 2, 4}
{2, 4, 1} 
and there inversion count are 3, 1, 2 
respectively. So, total number of 
inversions are  6.


It is strongly recommended to refer Inversion count in an array using BIT

The process of counting inversion is same as in the above linked article. First, we calculate the inversion in first k (k is given size of subarray) elements in the array using BIT. Now after this for every next subarray we subtract the inversion of first element of previous subarray and add the inversions of the next element not included in the previous subarray. This process is repeated until the last subarray is processed.On the above example this algorithm works like this-

a[] = {7, 3, 2, 4, 1}, 
  k = 3;

Inversion are counted for first subarray 
A = {7, 3, 2} Let this be equal to invcount_A.

For counting the inversion of subarray B we 
subtract the inversion of first element of A,
from invcount_A and add inversions of 4 (last 
element of B) in the subarray B.
So, invcount_B = invcount_A - 2 + 0
               = 3 - 2 + 0
               = 1

Same process is repeated for next subarray
and sum of all inversion count is the answer.  
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// C++ program to count inversions in all sub arrays
// of size k using Binary Indexed Tree
#include <bits/stdc++.h>
using namespace std;
  
// Returns sum of arr[0..index]. This function assumes
// that the array is preprocessed and partial sums of
// array elements are stored in BITree[].
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.
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 }
void convert(int arr[], int n)
{
    // Create a copy of arrp[] in temp and sort 
    // the temp array in increasing order
    int temp[n];
    for (int i = 0; i < n; i++)
        temp[i] = arr[i];
    sort(temp, temp + n);
  
    // 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, temp + n, 
                         arr[i]) - temp + 1;
    }
}
  
// Returns inversion count of all subarray 
// of size k in arr[0..n-1]
int getInvCount(int arr[], int k, int n)
{
    int invcount = 0; // Initialize result
  
    // 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[n + 1];
    for (int i = 1; i <= n; i++)
        BIT[i] = 0;
  
    // Get inversion count of first subarray
    for (int i = k - 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);
    }
  
    // now calculate the inversion of all other subarrays
    int ans = invcount;
    int i = 0, j = k, icnt = 0, jcnt = 0;
    while (j <= n - 1) {
  
        // calculate value of inversion count of first element
        // in previous subarray
        icnt = getSum(BIT, arr[i] - 1);
        updateBIT(BIT, n, arr[i], -1);
  
        // calculating inversion count of last element in the
        // current subarray
        jcnt = getSum(BIT, n) - getSum(BIT, arr[j]);
        updateBIT(BIT, n, arr[j], 1);
  
        // calculating inversion count of current subarray
        invcount = invcount - icnt + jcnt;
  
        // adding current inversion to the answer
        ans = ans + invcount;
        i++, j++;
    }
    return ans;
}
int main()
{
    int arr[] = { 7, 3, 2, 4, 1 };
    int k = 3;
    int n = sizeof(arr) / sizeof(int);
    cout << "Number of inversions in all "
         "subarrays of size " << k <<" are : "
         << getInvCount(arr, k, n);
    return 0;
}

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Output:

Number of inversions in all subarrays of size 3 are : 6

Time Complexity :- inversion count of first subarray takes O(k log(n)) time and for all other subarray it takes O((n-k)log(n)). So overall time complexity is : O(n log n).
Auxiliary space:- O(n)

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