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Count pairs in an array that hold i*arr[i] > j*arr[j]

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Given an array of integers arr[0..n-1], count all pairs (arr[i], arr[j]) in the such that i*arr[i] > j*arr[j], 0 =< i < j < n.

Examples : 

Input : arr[] = {5 , 0, 10, 2, 4, 1, 6}
Output: 5
 Pairs which hold condition i*arr[i] > j*arr[j]
 are (10, 2) (10, 4) (10, 1) (2, 1) (4, 1)

Input  : arr[] = {8, 4, 2, 1}
Output : 2
Recommended Practice

A Simple solution is to run two loops. Pick each element of the array one by one and for each element find an element on the right side of the array that holds the condition, then increment the counter, and last return the counter value. 

Below is the implementation of the above idea: 

C++




// C++ program to count all pair that
// hold condition i*arr[i] > j*arr[j]
#include<iostream>
using namespace std;
 
// Return count of pair in given array
// such that  i*arr[i] > j*arr[j]
int CountPair(int arr[] , int n )
{
    int result = 0; // Initialize result
 
    for (int i=0; i<n; i++)
    {
        // Generate all pair and increment
        // counter if the hold given condition
        for (int j = i + 1; j < n; j++)
            if (i*arr[i] > j*arr[j] )
                result ++;
    }
    return result;
}
 
// Driver code
int main()
{
    int arr[] = {5 , 0, 10, 2, 4, 1, 6} ;
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << "Count of Pairs : "
         << CountPair(arr, n);
    return 0;
}


Java




// Java Code for Count pairs in an
// array that hold i*arr[i] > j*arr[j]
class GFG {
     
    // Return count of pair in given array
    // such that  i*arr[i] > j*arr[j]
    public static int CountPair(int arr[] , int n )
    {
        int result = 0; // Initialize result
      
        for (int i = 0; i < n; i++)
        {
            // Generate all pair and increment
            // counter if the hold given condition
            for (int j = i + 1; j < n; j++)
                if (i*arr[i] > j*arr[j] )
                    result ++;
        }
        return result;
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int arr[] = {5 , 0, 10, 2, 4, 1, 6} ;
        int n = arr.length;
        System.out.println("Count of Pairs : " +
                            CountPair(arr, n));
    }
  }
// This code is contributed by Arnav Kr. Mandal.


Python3




# Python3 Code to Count pairs in an
# array that hold i*arr[i] > j*arr[j]
 
# Return count of pair in given array
# such that i*arr[i] > j*arr[j]
def CountPair(arr , n ):
     
    # Initialize result
    result = 0;
     
    for i in range (0, n):
         
        # Generate all pair and increment
        # counter if the hold given condition
        j = i + 1
        while(j < n):
            if (i * arr[i] > j * arr[j] ):
                result = result +1
            j = j + 1
    return result;
     
# Driver program to test above function */
     
arr = [5, 0, 10, 2, 4, 1, 6]
n = len(arr)
print("Count of Pairs : " , CountPair(arr, n))
 
# This code is contributed by Sam007.


C#




// C# Code to Count pairs in an
// array that hold i*arr[i] > j*arr[j]
using System;
 
class GFG
{
    // Return count of pair in given array
    // such that i*arr[i] > j*arr[j]
    public static int CountPair(int []arr , int n )
    {
        // Initialize result
        int result = 0;
     
        for (int i = 0; i < n; i++)
        {
            // Generate all pair and increment
            // counter if the hold given condition
            for (int j = i + 1; j < n; j++)
                if (i*arr[i] > j*arr[j] )
                    result ++;
        }
        return result;
    }
     
    /* Driver program to test above function */
    public static void Main()
    {
        int []arr = {5, 0, 10, 2, 4, 1, 6};
        int n = arr.Length;
        Console.WriteLine("Count of Pairs : " +
                           CountPair(arr, n));
    }
}
 
// This code is contributed by Sam007


PHP




<?php
// PHP program to count all pair that
// hold condition i*arr[i] > j*arr[j]
 
// Return count of pair in given array
// such that i*arr[i] > j*arr[j]
function CountPair($arr , $n )
{
     
    // Initialize result
    $result = 0;
     
 
    for($i = 0; $i < $n; $i++)
    {
         
        // Generate all pair and increment
        // counter if the hold given condition
        for ($j = $i + 1; $j < $n; $j++)
            if ($i $arr[$i] > $j * $arr[$j] )
                $result ++;
    }
    return $result;
}
 
    // Driver code
    $arr = array(5, 0, 10, 2, 4, 1, 6) ;
    $n = sizeof($arr);
    echo "Count of Pairs : ",
    CountPair($arr, $n);
     
// This code is contributed by m_kit
?>


Javascript




<script>
 
// JavaScript program  to Count pairs in an
// array that hold i*arr[i] > j*arr[j]
 
// Return count of pair in given array
// such that  i*arr[i] > j*arr[j]
function CountPair(arr, n)
{
     
    // Initialize result
    let result = 0;
    
    for(let i = 0; i < n; i++)
    {
         
        // Generate all pair and increment
        // counter if the hold given condition
        for(let j = i + 1; j < n; j++)
            if (i * arr[i] > j * arr[j] )
                result ++;
    }
    return result;
}
   
// Driver Code
let arr = [5 , 0, 10, 2, 4, 1, 6] ;
let n = arr.length;
 
document.write("Count of Pairs : " +
               CountPair(arr, n));
                
// This code is contributed by souravghosh0416 
 
</script>


Output

Count of Pairs : 5

Time Complexity: O(n2)
Auxiliary Space: O(1)

An efficient solution to this problem takes O(n log n) time. The idea is based on an interesting fact about this problem after modifying the array such that every element is multiplied by its index, this problem converts into Count Inversions in an array

Algorithm : 

Given an array ‘arr’ and it’s size ‘n’

  1. First traversal array element, i goes from 0 to n-1
    1. Multiple each element with its index arr[i] = arr[i] * i 
  2. After that step 1. whole process is similar to Count Inversions in an array

Below is the implementation of the above idea  

C++




// C++ program to count all pair that
// hold condition i*arr[i] > j*arr[j]
#include <bits/stdc++.h>
using namespace std;
 
/* This function merges two sorted arrays and
   returns inversion count in the arrays.*/
int merge(int arr[], int temp[], int left,
                       int mid, int right)
{
    int inv_count = 0;
 
    int i = left; /* index for left subarray*/
    int j = mid;  /* index for right subarray*/
    int k = left; /* index for resultant subarray*/
    while ((i <= mid - 1) && (j <= right))
    {
        if (arr[i] <= arr[j])
            temp[k++] = arr[i++];
        else
        {
            temp[k++] = arr[j++];
 
            inv_count = inv_count + (mid - i);
        }
    }
 
    /* Copy the remaining elements of left
     subarray (if there are any) to temp*/
    while (i <= mid - 1)
        temp[k++] = arr[i++];
 
    /* Copy the remaining elements of right
     subarray (if there are any) to temp*/
    while (j <= right)
        temp[k++] = arr[j++];
 
    /* Copy back the merged elements to original
      array*/
    for (i=left; i <= right; i++)
        arr[i] = temp[i];
 
    return inv_count;
}
 
/* An auxiliary recursive function that sorts
   the input array and returns the number of
   inversions in the array. */
int _mergeSort(int arr[], int temp[], int left,
                                      int right)
{
    int mid, inv_count = 0;
    if (right > left)
    {
        /* Divide the array into two parts and call
          _mergeSortAndCountInv() for each of
          the parts */
        mid = (right + left)/2;
 
        /* Inversion count will be sum of inversions in
           left-part, right-part and number of inversions
           in merging */
        inv_count  = _mergeSort(arr, temp, left, mid);
        inv_count += _mergeSort(arr, temp, mid+1, right);
 
        /*Merge the two parts*/
        inv_count += merge(arr, temp, left, mid+1, right);
    }
 
    return inv_count;
}
 
/* This function sorts the input array and
   returns the number of inversions in the
   array */
int countPairs(int arr[], int n)
{
    // Modify the array so that problem reduces to
    // count inversion problem.
    for (int i=0; i<n; i++)
        arr[i] = i*arr[i];
 
    // Count inversions using same logic as
    // below post
    int temp[n];
    return _mergeSort(arr, temp, 0, n - 1);
}
 
// Driver code
int main()
{
    int arr[] = {5, 0, 10, 2, 4, 1, 6};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << "Count of Pairs : "
         << countPairs(arr, n);
    return 0;
}


Java




// Java program to count all pair that
// hold condition i*arr[i] > j*arr[j]
import java.io.*;
 
class GFG
{
    // This function merges two sorted arrays and
    // returns inversion count in the arrays.
    static int merge(int arr[], int temp[], int left,
                                   int mid, int right)
    {
        int inv_count = 0;
         
        /* index for left subarray*/
        int i = left;
         
        /* index for right subarray*/
        int j = mid;
        /* index for resultant subarray*/
        int k = left;
         
        while ((i <= mid - 1) && (j <= right))
        {
            if (arr[i] <= arr[j])
                temp[k++] = arr[i++];
            else
            {
                temp[k++] = arr[j++];
     
                inv_count = inv_count + (mid - i);
            }
        }
     
        /* Copy the remaining elements of left
        subarray (if there are any) to temp*/
        while (i <= mid - 1)
            temp[k++] = arr[i++];
     
        /* Copy the remaining elements of right
        subarray (if there are any) to temp*/
        while (j <= right)
            temp[k++] = arr[j++];
     
        // Copy back the merged elements
        // to original array
        for (i = left; i <= right; i++)
            arr[i] = temp[i];
     
        return inv_count;
    }
     
    /* An auxiliary recursive function
    that sorts the input array and
    returns the number of inversions
    in the array. */
    static int _mergeSort(int arr[], int temp[],
                               int left,int right)
    {
        int mid, inv_count = 0;
        if (right > left)
        {
            /* Divide the array into two parts and call
            _mergeSortAndCountInv() for each of
            the parts */
            mid = (right + left) / 2;
     
            // Inversion count will be sum of inversions in
            // left-part, right-part and number of inversions
            // in merging
            inv_count = _mergeSort(arr, temp, left, mid);
            inv_count += _mergeSort(arr, temp, mid+1, right);
     
            /*Merge the two parts*/
            inv_count += merge(arr, temp, left, mid+1, right);
        }
     
        return inv_count;
    }
     
    // This function sorts the input array and
    // returns the number of inversions in the
    // array
    static int countPairs(int arr[], int n)
    {
        // Modify the array so that problem reduces to
        // count inversion problem.
        for (int i = 0; i < n; i++)
            arr[i] = i * arr[i];
     
        // Count inversions using same logic as
        // below post
        int temp[] = new int [n];
        return _mergeSort(arr, temp, 0, n - 1);
    }
     
    // Driver code
 
    public static void main (String[] args)
    {
        int arr[] = {5, 0, 10, 2, 4, 1, 6};
        int n = arr.length;
        System.out.print( "Count of Pairs : "
                          + countPairs(arr, n)); 
         
    }
}
 
// This code is contributed by vt_m


Python3




# Python 3 program to count all
# pair that hold condition
# i*arr[i] > j*arr[j]
 
# This function merges two
# sorted arrays and returns
# inversion count in the arrays.
def merge(arr, temp, left, mid, right):
 
    inv_count = 0
 
    i = left # index for left subarray
    j = mid # index for right subarray
    k = left # index for resultant subarray
    while ((i <= mid - 1) and (j <= right)):
     
        if (arr[i] <= arr[j]):
            temp[k] = arr[i]
            i += 1
            k += 1
        else:
         
            temp[k] = arr[j]
            k += 1
            j += 1
 
            inv_count = inv_count + (mid - i)
 
    # Copy the remaining elements of left
    # subarray (if there are any) to temp
    while (i <= mid - 1):
        temp[k] = arr[i]
        i += 1
        k += 1
 
    # Copy the remaining elements of right
    # subarray (if there are any) to temp
    while (j <= right):
        temp[k] = arr[j]
        k += 1
        j += 1
 
    # Copy back the merged elements
    # to original array
    for i in range(left, right + 1):
        arr[i] = temp[i]
 
    return inv_count
 
# An auxiliary recursive function
# that sorts the input array and
# returns the number of inversions
# in the array.
def _mergeSort(arr, temp, left, right):
 
    inv_count = 0
    if (right > left):
     
        # Divide the array into two parts
        # and call _mergeSortAndCountInv()
        # for each of the parts
        mid = (right + left) // 2
 
        # Inversion count will be sum of
        # inversions in left-part, right-part x
        # and number of inversions in merging
        inv_count = _mergeSort(arr, temp, left, mid)
        inv_count += _mergeSort(arr, temp,
                                mid + 1, right)
 
        # Merge the two parts
        inv_count += merge(arr, temp, left,    
                           mid + 1, right)
 
    return inv_count
 
# This function sorts the input
# array and returns the number
# of inversions in the array
def countPairs(arr, n):
     
    # Modify the array so that problem
    # reduces to count inversion problem.
    for i in range(n):
        arr[i] = i * arr[i]
 
    # Count inversions using same
    # logic as below post
    temp = [0] * n
    return _mergeSort(arr, temp, 0, n - 1)
 
# Driver code
if __name__ == "__main__":
    arr = [5, 0, 10, 2, 4, 1, 6]
    n = len(arr)
    print("Count of Pairs : ",
           countPairs(arr, n))
            
# This code is contributed
# by ChitraNayal


C#




// C# program to count all pair that
// hold condition i*arr[i] > j*arr[j]
using System;
 
class GFG
{
    // This function merges two sorted arrays and
    // returns inversion count in the arrays.
    static int merge(int []arr, int []temp, int left,
                                int mid, int right)
    {
        int inv_count = 0;
         
        /* index for left subarray*/
        int i = left;
         
        /* index for right subarray*/
        int j = mid;
        /* index for resultant subarray*/
        int k = left;
         
        while ((i <= mid - 1) && (j <= right))
        {
            if (arr[i] <= arr[j])
                temp[k++] = arr[i++];
            else
            {
                temp[k++] = arr[j++];
     
                inv_count = inv_count + (mid - i);
            }
        }
     
        /* Copy the remaining elements of left
        subarray (if there are any) to temp*/
        while (i <= mid - 1)
            temp[k++] = arr[i++];
     
        /* Copy the remaining elements of right
        subarray (if there are any) to temp*/
        while (j <= right)
            temp[k++] = arr[j++];
     
        // Copy back the merged elements
        // to original array
        for (i = left; i <= right; i++)
            arr[i] = temp[i];
     
        return inv_count;
    }
     
    /* An auxiliary recursive function
    that sorts the input array and
    returns the number of inversions
    in the array. */
    static int _mergeSort(int []arr, int []temp,
                            int left,int right)
    {
        int mid, inv_count = 0;
        if (right > left)
        {
            /* Divide the array into two parts and call
            _mergeSortAndCountInv() for each of
            the parts */
            mid = (right + left) / 2;
     
            // Inversion count will be sum of inversions in
            // left-part, right-part and number of inversions
            // in merging
            inv_count = _mergeSort(arr, temp, left, mid);
            inv_count += _mergeSort(arr, temp, mid+1, right);
     
            /*Merge the two parts*/
            inv_count += merge(arr, temp, left, mid+1, right);
        }
     
        return inv_count;
    }
     
    // This function sorts the input array and
    // returns the number of inversions in the
    // array
    static int countPairs(int []arr, int n)
    {
        // Modify the array so that problem reduces to
        // count inversion problem.
        for (int i = 0; i < n; i++)
            arr[i] = i * arr[i];
     
        // Count inversions using same logic as
        // below post
        int []temp = new int [n];
        return _mergeSort(arr, temp, 0, n - 1);
    }
     
    // Driver code
 
    public static void Main ()
    {
        int []arr = {5, 0, 10, 2, 4, 1, 6};
        int n = arr.Length;
        Console.WriteLine( "Count of Pairs : "
                        + countPairs(arr, n));
         
    }
}
 
// This code is contributed by anuj_67.


Javascript




<script>
// Javascript program to count all pair that
// hold condition i*arr[i] > j*arr[j]
     
    // This function merges two sorted arrays and
    // returns inversion count in the arrays.
    function merge(arr,temp,left,mid,right)
    {
        let inv_count = 0;
          
        /* index for left subarray*/
        let i = left;
          
        /* index for right subarray*/
        let j = mid;
        /* index for resultant subarray*/
        let k = left;
          
        while ((i <= mid - 1) && (j <= right))
        {
            if (arr[i] <= arr[j])
                temp[k++] = arr[i++];
            else
            {
                temp[k++] = arr[j++];
      
                inv_count = inv_count + (mid - i);
            }
        }
      
        /* Copy the remaining elements of left
        subarray (if there are any) to temp*/
        while (i <= mid - 1)
            temp[k++] = arr[i++];
      
        /* Copy the remaining elements of right
        subarray (if there are any) to temp*/
        while (j <= right)
            temp[k++] = arr[j++];
      
        // Copy back the merged elements
        // to original array
        for (i = left; i <= right; i++)
            arr[i] = temp[i];
      
        return inv_count;
    }
     
    /* An auxiliary recursive function
    that sorts the input array and
    returns the number of inversions
    in the array. */
    function _mergeSort(arr,temp,left,right)
    {
        let mid, inv_count = 0;
        if (right > left)
        {
            /* Divide the array into two parts and call
            _mergeSortAndCountInv() for each of
            the parts */
            mid = Math.floor((right + left) / 2);
      
            // Inversion count will be sum of inversions in
            // left-part, right-part and number of inversions
            // in merging
            inv_count = _mergeSort(arr, temp, left, mid);
            inv_count += _mergeSort(arr, temp, mid+1, right);
      
            /*Merge the two parts*/
            inv_count += merge(arr, temp, left, mid+1, right);
        }
      
        return inv_count;
    }
     
    // This function sorts the input array and
    // returns the number of inversions in the
    // array
    function countPairs(arr,n)
    {
        // Modify the array so that problem reduces to
        // count inversion problem.
        for (let i = 0; i < n; i++)
            arr[i] = i * arr[i];
      
        // Count inversions using same logic as
        // below post
        let temp = new Array(n);
        for(let i=0;i<temp;i++)
        {
            temp[i]=0;
        }
        return _mergeSort(arr, temp, 0, n - 1);
    }
     
    // Driver code
    let arr=[5, 0, 10, 2, 4, 1, 6];
    let n = arr.length;
    document.write( "Count of Pairs : "
                          + countPairs(arr, n));
    // This code is contributed by rag2127
</script>


Output

Count of Pairs : 5

Time Complexity: O(n log n)
Auxiliary Space: O(n)

Another efficient approach (Using policy-based data structures in C++ STL):

In this approach, we first store the value i*arr[i] for every index i and let name this array obtained as a modified array. So now instead of i*arr[i]>j*arr[j] in the original array, we have to find x>y where x and y are now elements of the modified array where the index of x is smaller than the index of y. In other words for every index i in the modified array, the total number of such pairs is the count of all the smaller elements to the right side of that index.

Hence the problem is transformed into another problem where we need to find the count of the smaller elements to the right side for each index i in the modified array.

So we can solve this problem efficiently by maintaining a policy-based data structure of pairs.

The steps involved are:

  1.  Traverse through the array and for each index i store i*arr[i].
  2.  Then traverse through the modified array from the end and every time we find the order_of_key of the current element to find the number of smaller elements to the right of it.
  3. Then we will add the obtained value from order_of_key of the current element into the ans variable.
  4. After that, we will insert the current element into our policy-based data structure.  

Below is the code for the above approach:

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pbds                                             \
    tree<pair<int, int>, null_type, less<pair<int, int> >, \
        rb_tree_tag, tree_order_statistics_node_update>
using namespace __gnu_pbds;
using namespace std;
 
// Print the count of pair in given array
// such that  i*arr[i] > j*arr[j]
// for 0<=i<j<n
void countPrint(int arr[], int n)
{
    // storing i*arr[i]
    // for every index
    for (int i=0;i<n;i++)
    {
        arr[i]=i*arr[i];
    }
    pbds s;
     
    // variable to store the count
    int ans=0;
     
    for (int i = n - 1; i >= 0; i--) {
        // for the last element count is
        // zero so excluding it.
        if(i!=n-1) {
          // add the count of the smaller elements
          // to the right of current element into
          // the ans variable
            ans+=s.order_of_key({ arr[i], i });
        }
        // insert current element
        s.insert({ arr[i], i });
    }
     
    cout << ans << endl;
}
 
// Driver Code
int main()
{
    int n = 7;
    int arr[] = { 5 , 0, 10, 2, 4, 1, 6 };
    countPrint(arr, n);
    return 0;
}
 
// This code is contributed by Pushpesh raj


Java




import java.util.*;
import java.util.AbstractMap.SimpleEntry;
 
public class Main {
 
  // Print the count of pair in given array
  // such that i*arr[i] > j*arr[j]
  // for 0 <= i < j < n
  static void countPrint(int[] arr, int n)
  {
 
    // storing i*arr[i]
    // for every index
    for (int i = 0; i < n; i++) {
      arr[i] = i * arr[i];
    }
    TreeMap<Integer, Integer> s = new TreeMap<>();
 
    // variable to store the count
    int ans = 0;
 
    for (int i = n - 1; i >= 0; i--)
    {
 
      // for the last element count is
      // zero so excluding it.
      if (i != n - 1)
      {
 
        // add the count of the smaller elements
        // to the right of current element into
        // the ans variable
        ans += s.headMap(arr[i]).size();
      }
      // insert current element
      s.put(arr[i], i);
    }
 
    System.out.println(ans);
  }
 
  // Driver Code
  public static void main(String[] args) {
    int n = 7;
    int[] arr = { 5, 0, 10, 2, 4, 1, 6 };
    countPrint(arr, n);
  }
}


Python3




from bisect import bisect_left
 
# Print the count of pairs in given array
# such that i*arr[i] > j*arr[j]
# for 0 <= i < j < n
def countPrint(arr, n):
   
    # storing i*arr[i] for every index
    for i in range(n):
        arr[i] = i * arr[i]
 
    s = []
    ans = 0
 
    for i in range(n - 1, -1, -1):
       
        # for the last element, the count is zero, so excluding it
        if i != n - 1:
            # add the count of smaller elements to the right of
            # the current element into the ans variable
            ans += bisect_left(s, arr[i])
 
        # insert current element
        s.insert(bisect_left(s, arr[i]), arr[i])
 
    print(ans)
 
# Driver Code
if __name__ == '__main__':
    n = 7
    arr = [5, 0, 10, 2, 4, 1, 6]
    countPrint(arr, n)
 
# This code is contributed by Aditya Sharma


C#




using System;
using System.Collections.Generic;
 
class MainClass {
 
    // Print the count of pairs in given array
    // such that i*arr[i] > j*arr[j]
    // for 0 <= i < j < n
    static void countPrint(int[] arr, int n)
    {
        // storing i*arr[i] for every index
        for (int i = 0; i < n; i++) {
            arr[i] = i * arr[i];
        }
 
        List<int> s = new List<int>();
        int ans = 0;
 
        for (int i = n - 1; i >= 0; i--) {
            // for the last element, the count is zero, so
            // excluding it
            if (i != n - 1) {
                // add the count of smaller elements to the
                // right of the current element into the ans
                // variable
                ans += s.BinarySearch(arr[i]) >= 0
                           ? s.BinarySearch(arr[i])
                           : ~s.BinarySearch(arr[i]);
            }
 
            // insert current element
            s.Insert(s.BinarySearch(arr[i]) >= 0
                         ? s.BinarySearch(arr[i])
                         : ~s.BinarySearch(arr[i]),
                     arr[i]);
        }
 
        Console.WriteLine(ans);
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int n = 7;
        int[] arr = { 5, 0, 10, 2, 4, 1, 6 };
        countPrint(arr, n);
    }
}


Javascript




function countPrint(arr, n) {
  // storing i*arr[i]
  // for every index
  for (let i = 0; i < n; i++) {
    arr[i] = i * arr[i];
  }
 
  // using a map to store values of arr[i]
  // and their corresponding counts
  const s = new Map();
 
  // variable to store the count
  let ans = 0;
 
  // iterating over the array in reverse order
  for (let i = n - 1; i >= 0; i--) {
    // for the last element count is
    // zero so excluding it.
    if (i != n - 1) {
      // add the count of the smaller elements
      // to the right of current element into
      // the ans variable
      for (const [key, value] of s.entries()) {
        if (key < arr[i]) {
          ans += value;
        }
      }
    }
    // insert current element
    if (s.has(arr[i])) {
      s.set(arr[i], s.get(arr[i]) + 1);
    } else {
      s.set(arr[i], 1);
    }
  }
 
  console.log(ans);
}
 
// Driver Code
const n = 7;
const arr = [5, 0, 10, 2, 4, 1, 6];
countPrint(arr, n);


Output

5

Time Complexity: O(n*log(n))
Auxiliary Space: O(n)

 



Last Updated : 13 Mar, 2023
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