# Count inversions of size k in a given array

Given an array of n distinct integers and an integer k. Find out the number of sub-sequences of a such that , and . In other words output the total number of inversions of length k.

Examples:

Input : a[] = {9, 3, 6, 2, 1}, k = 3
Output : 7
The seven inversions are {9, 3, 2}, {9, 3, 1},
{9, 6, 2}, {9, 6, 1}, {9, 2, 1}, {3, 2, 1} and
{6, 2, 1}.

Input : a[] = {5, 6, 4, 9, 2, 7, 1}, k = 4
Output : 2
The two inversions are {5, 4, 2, 1}, {6, 4, 2, 1}.


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have already discussed counting inversion of length three here. This post will generalise the approach using Binary Indexed Tree and Counting inversions using BIT. More on Binary Indexed Trees can be found from the actual paper that Peter M. Fenwick published, here.

1. To begin with, we first convert the given array to a permutation of elements (Note that this is always possible since the elements are distinct).
2. The approach is to maintain a set of k Fenwick Trees. Let it be denoted by where , keeps track of the number of – length sub-sequences that start with .
3. We iterate from the end of the converted array to the beginning. For every converted array element , we update the Fenwick tree set, as: For each , each sub-sequence of length that start with a number less than is also a part of sequences of length .
The final result is found by sum of occurences of .

The implementation is discussed below:

## C++

 // C++ program to count inversions of size k using  // Binary Indexed Tree  #include  using namespace std;     // It is beneficial to declare the 2D BIT globally   // since passing it into functions will create   // additional overhead  const int K = 51;  const int N = 100005;  int BIT[K][N] = { 0 };     // update function. "t" denotes the t'th Binary   // indexed tree  void updateBIT(int t, int i, int val, int n)  {      // Traversing the t'th BIT      while (i <= n) {          BIT[t][i] = BIT[t][i] + val;          i = i + (i & (-i));      }  }     // function to get the sum.  // "t" denotes the t'th Binary indexed tree  int getSum(int t, int i)  {      int res = 0;         // Traversing the t'th BIT      while (i > 0) {          res = res + BIT[t][i];          i = i - (i & (-i));      }      return res;  }     // 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 arr[] 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 count of inversions of size three  int getInvCount(int arr[], int n, int k)  {      // 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);         // iterating over the converted array in       // reverse order.      for (int i = n - 1; i >= 0; i--) {          int x = arr[i];             // update the BIT for l = 1          updateBIT(1, x, 1, n);             // update BIT for all other BITs          for (int l = 1; l < k; l++) {              updateBIT(l + 1, x, getSum(l, x - 1), n);          }      }         // final result      return getSum(k, n);  }     // Driver program to test above function  int main()  {      int arr[] = { 5, 6, 4, 9, 3, 7, 2, 1 };      int n = sizeof(arr) / sizeof(arr);      int k = 4;      cout << "Inversion Count : " << getInvCount(arr, n, k);      return 0;  }

## Java

 // Java program to count   // inversions of size k using  // Binary Indexed Tree  import java.io.*;  import java.util.Arrays;  import java.util.ArrayList;  import java.lang.*;  import java.util.Collections;     class GFG  {     // It is beneficial to declare  // the 2D BIT globally since   // passing it into functions   // will create additional overhead  static int K = 51;  static int N = 100005;  static int BIT[][] = new int[K][N];     // update function. "t" denotes   // the t'th Binary indexed tree  static void updateBIT(int t, int i,                        int val, int n)  {            // Traversing the t'th BIT      while (i <= n)      {          BIT[t][i] = BIT[t][i] + val;          i = i + (i & (-i));      }  }     // function to get the sum.  // "t" denotes the t'th   // Binary indexed tree  static int getSum(int t, int i)  {      int res = 0;         // Traversing the t'th BIT      while (i > 0)       {          res = res + BIT[t][i];          i = i - (i & (-i));      }      return res;  }     // 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 arr[] 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] = Arrays.binarySearch(temp,                                        arr[i]) + 1;      }  }     // Returns count of inversions  // of size three  static int getInvCount(int arr[],                          int n, int k)  {             // 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);         // iterating over the converted       // array in reverse order.      for (int i = n - 1; i >= 0; i--)       {          int x = arr[i];             // update the BIT for l = 1          updateBIT(1, x, 1, n);             // update BIT for all other BITs          for (int l = 1; l < k; l++)           {              updateBIT(l + 1, x,                         getSum(l, x - 1), n);          }      }         // final result      return getSum(k, n);  }     // Driver Code  public static void main(String[] args)  {             int arr[] = { 5, 6, 4, 9,                     3, 7, 2, 1 };      int n = arr.length;      int k = 4;      System.out.println("Inversion Count : " +                           getInvCount(arr, n, k));  }  }

## Python3

 # Python3 program to count inversions   # of size k using Binary Indexed Tree      # It is beneficial to declare the 2D BIT   # globally since passing it o functions   # will create additional overhead   K = 51 N = 100005 BIT = [[0 for x in range(N)]             for y in range(K)]      # update function. "t" denotes  # the t'th Binary indexed tree   def updateBIT(t, i, val, n):          # Traversing the t'th BIT       while (i <= n):           BIT[t][i] = BIT[t][i] + val           i = i + (i & (-i))      # function to get the sum. "t" denotes   # the t'th Binary indexed tree   def getSum(t, i):         res = 0        # Traversing the t'th BIT       while (i > 0):           res = res + BIT[t][i]           i = i - (i & (-i))          return res      # 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   def convert( arr, n):          # Create a copy of arr[] in temp and sort       # the temp array in increasing order       temp =  * n       for i in range(n):           temp[i] = arr[i]       temp = sorted(temp)      j = 1     for i in temp:          arr[arr.index(i)] = j          j += 1    # Returns count of inversions   # of size three   def getInvCount(arr, n, k) :             # 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)          # iterating over the converted array       # in reverse order.       for i in range(n - 1, -1, -1):           x = arr[i]              # update the BIT for l = 1           updateBIT(1, x, 1, n)              # update BIT for all other BITs           for l in range(1, k):               updateBIT(l + 1, x, getSum(l, x - 1), n)          # final result       return getSum(k, n)      # Driver code   if __name__ =="__main__":      arr = [5, 6, 4, 9, 3, 7, 2, 1]       n = 8     k = 4     print("Inversion Count :",              getInvCount(arr, n, k))         # This code is contributed by  # Shubham Singh(SHUBHAMSINGH10)

## C#

 // C# program to count   // inversions of size k using   // Binary Indexed Tree   using System;  using System.Linq;     class GFG   {          // It is beneficial to declare       // the 2D BIT globally since       // passing it into functions       // will create additional overhead       static int K = 51;       static int N = 100005;       static int [,]BIT = new int[K, N];          // update function. "t" denotes       // the t'th Binary indexed tree       static void updateBIT(int t, int i,                           int val, int n)       {              // Traversing the t'th BIT           while (i <= n)           {               BIT[t, i] = BIT[t, i] + val;               i = i + (i & (-i));           }       }          // function to get the sum.       // "t" denotes the t'th       // Binary indexed tree       static int getSum(int t, int i)       {           int res = 0;              // Traversing the t'th BIT           while (i > 0)           {               res = res + BIT[t, i];               i = i - (i & (-i));           }           return res;       }          // 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 arr[] 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] = Array.BinarySearch(temp,                                           arr[i]) + 1;           }       }          // Returns count of inversions       // of size three       static int getInvCount(int []arr,                           int n, int k)       {              // 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);              // iterating over the converted           // array in reverse order.           for (int i = n - 1; i >= 0; i--)           {               int x = arr[i];                  // update the BIT for l = 1               updateBIT(1, x, 1, n);                  // update BIT for all other BITs               for (int l = 1; l < k; l++)               {                   updateBIT(l + 1, x,                           getSum(l, x - 1), n);               }           }              // final result           return getSum(k, n);       }          // Driver Code       public static void Main(String[] args)       {              int []arr = { 5, 6, 4, 9,                       3, 7, 2, 1 };           int n = arr.Length;           int k = 4;           Console.WriteLine("Inversion Count : " +                               getInvCount(arr, n, k));       }   }      // This code is contributed by PrinciRaj1992

Output:

Inversion Count : 11


Time Complexity Auxiliary Space It should be noted that this is not the only approach to solve the problem of finding k-inversions. Obviously, any problem solvable by BIT is also solvable by Segment Tree. Besides, we can use Merge-Sort based algorithm, and C++ policy based data structure too. Also, at the expense of higher time complexity, Dynamic Programming approach can also be used.

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