Count inversions in a sequence generated by appending given array K times

Given an array arr[], the task is to append the given array exactly K - 1 times to its end and print the total number of inversions in the resulting array.

Examples:

Input: arr[]= {2, 1, 3}, K = 3
Output: 12
Explanation:
Appending 2 copies of array arr[] modifies arr[] to {2, 1, 3, 2, 1, 3, 2, 1, 3}
The pairs (arr[i], arr[j]), where i < j and arr[i] > arr[j] are (2, 1), (2, 1), (2, 1), (3, 2), (3, 1), (3, 2), (3, 1), (2, 1), (2, 1), (3, 2), (3, 1), (2, 1)
Therefore, the total number of inversions are 12.

Input: arr[]= {6, 2}, K = 2
Output: 3
Explanation:
Appending 2 copies of array arr[] = {6, 2, 6, 2}
The pairs (arr[i], arr[j]), where i < j and arr[i] > arr[j] are (6, 2), (6, 2), (6, 2)
Therefore, the total number of inversions are 3.

Naive Approach: The simplest approach is to store K copies of the given array in a vector and then, find the count of inversions of the resulting vector. 

Time Complexity: O(N²)
Auxiliary Space: O(K * N)

Efficient Approach: The idea to solve this problem is to first find the total number of inversions in the given array, say inv. Then, count pairs of distinct elements in a single copy, say X. Now, calculate the total number of inversions after appending K copies of the array by the equation: 

(inv*K + ((K*(K-1))/2)*X).

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to count the number of
// inversions in K copies of given array
void totalInversions(int arr[],
                     int K, int N)
{
    // Stores count of inversions
    // in the given array
    int inv = 0;

    // Stores the count of pairs
    // of distinct array elements
    int X = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // Generate each pair
        for (int j = 0; j < N; j++) {

            // Check for each pair, if the
            // condition is satisfied or not
            if (arr[i] > arr[j] and i < j)
                inv++;

            // If pairs consist of
            // distinct elements
            if (arr[i] > arr[j])
                X++;
        }
    }

    // Count inversion in the sequence
    int totalInv = X * K * (K - 1) / 2
                   + inv * K;

    // Print the answer
    cout << totalInv << endl;
}

// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 1, 3 };

    // Given K
    int K = 3;

    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);

    totalInversions(arr, K, N);
}

// Java program for the above approach
import java.util.*;
class GFG
{
 
 // Function to count the number of
// inversions in K copies of given array
static void totalInversions(int arr[],
                     int K, int N)
{
  
    // Stores count of inversions
    // in the given array
    int inv = 0;

    // Stores the count of pairs
    // of distinct array elements
    int X = 0;
    int i, j;

    // Traverse the array
    for (i = 0; i < N; i++)
    {

        // Generate each pair
        for (j = 0; j < N; j++)
        {

            // Check for each pair, if the
            // condition is satisfied or not
            if(arr[i] > arr[j] && i < j)
                inv++;

            // If pairs consist of
            // distinct elements
            if(arr[i] > arr[j])
                X++;
        }
    }

    // Count inversion in the sequence
    int totalInv = X * K * (K - 1) / 2
                   + inv * K;

    // Print the answer
    System.out.println(totalInv);
}

// Driver Code
public static void main(String args[])
{
  
    // Given array
    int arr[] = { 2, 1, 3 };

    // Given K
    int K = 3;

    // Size of the array
    int N = arr.length;
    totalInversions(arr, K, N);
}
}

// This code is contributed by bgangwar59.

# Python program of the above approach

# Function to count the number of
# inversions in K copies of given array
def totalInversions(arr, K, N) :
                         
    # Stores count of inversions
    # in the given array
    inv = 0

    # Stores the count of pairs
    # of distinct array elements
    X = 0

    # Traverse the array
    for i in range(N):

        # Generate each pair
        for j in range(N):

            # Check for each pair, if the
            # condition is satisfied or not
            if (arr[i] > arr[j] and i < j) :
                inv += 1

            # If pairs consist of
            # distinct elements
            if (arr[i] > arr[j]) :
                X += 1
        
    # Count inversion in the sequence
    totalInv = X * K * (K - 1) // 2 + inv * K

    # Print the answer
    print(totalInv)

# Driver Code

# Given array
arr = [ 2, 1, 3 ] 

# Given K
K = 3

# Size of the array
N = len(arr) 
totalInversions(arr, K, N)

# This code is contributed by susmitakundugoaldanga

// C# program to implement
// the above approach
using System;

class GFG
{

  // Function to count the number of
  // inversions in K copies of given array
  static void totalInversions(int []arr,
                              int K, int N)
  {

    // Stores count of inversions
    // in the given array
    int inv = 0;

    // Stores the count of pairs
    // of distinct array elements
    int X = 0;
    int i, j;

    // Traverse the array
    for (i = 0; i < N; i++)
    {

      // Generate each pair
      for (j = 0; j < N; j++)
      {

        // Check for each pair, if the
        // condition is satisfied or not
        if(arr[i] > arr[j] && i < j)
          inv++;

        // If pairs consist of
        // distinct elements
        if(arr[i] > arr[j])
          X++;
      }
    }

    // Count inversion in the sequence
    int totalInv = X * K * (K - 1) / 2
      + inv * K;

    // Print the answer
    Console.WriteLine(totalInv);
  }

  // Driver Code
  public static void  Main()
  {

    // Given array
    int []arr = { 2, 1, 3 };

    // Given K
    int K = 3;

    // Size of the array
    int N = arr.Length;
    totalInversions(arr, K, N);
  }
}

// This code is contributed by jana_sayantan.

<script>

// JavaScript program for
// the above approach
 
 
 // Function to count the number of
// inversions in K copies of given array
function totalInversions(arr, K, N)
{
  
    // Stores count of inversions
    // in the given array
    let inv = 0;

    // Stores the count of pairs
    // of distinct array elements
    let X = 0;
    let i, j;

    // Traverse the array
    for (i = 0; i < N; i++)
    {

        // Generate each pair
        for (j = 0; j < N; j++)
        {

            // Check for each pair, if the
            // condition is satisfied or not
            if(arr[i] > arr[j] && i < j)
                inv++;

            // If pairs consist of
            // distinct elements
            if(arr[i] > arr[j])
                X++;
        }
    }

    // Count inversion in the sequence
    let totalInv = X * K * (K - 1) / 2
                   + inv * K;

    // Print the answer
    document.write(totalInv);
}
 
// Driver Code
 
    // Given array
    let arr = [ 2, 1, 3 ];

    // Given K
    let K = 3;

    // Size of the array
    let N = arr.length;
    totalInversions(arr, K, N);
 
</script>

12

Time Complexity: O(N²)
Auxiliary Space: O(1)