Count number of pairs with the given Comparator

Given an array arr[], the task is to count the number of pairs (arr[i], arr[j]) on the right of every element with any custom comparator.

Comparator can be of any type, some of them are given below –

arr[i] > arr[j],    where i < j
arr[i] < arr[j],    where i  2 * arr[j], where i < j

Examples:



Input: arr[] = {5, 4, 3, 2, 1}, comp = arr[i] > arr[j]
Output: 10
Explanation:
There are 10 such pairs, in which right element is smaller than the left element –
{(5, 4), (5, 3), (5, 2), (5, 1), (4, 3), (4, 2), (4, 1), (3, 2), (3, 1), (2, 1)}

Input: arr[] = {3, 4, 3}, comp = arr[i] == arr[j]
Output: 1
Explanation:
There is only one such pair such that elements are equal. That is (3, 3)


Naive Solution: Iterate over every pairs of the elements, such that i < j and check for each pair that the custom comparator is true or not. If yes, then increment the count.

Time Complexity: O(N2)

Efficient Approach: The idea is to customize merge sort, to compute such pairs at the time of merging two sub-arrays. There will be two types of count for every array that is –

  • Inter-Array Pairs: Pairs those are present in the left subarray itself.
  • Intra-Array Pairs: Pairs those are present in the right subarray.

For Left subarrays, the count can be calculated recursively from bottom to top whereas the main task will be to count the intra-array pairs.

Therefore, Total such pairs can be defined as –

Total Pairs = Inter-Array pairs in Left Sub-array +
      Inter-Array pairs in Right Sub-array +
      Intra-Array pairs from left to right sub-array 

Below is the illustration of the intra-array pairs of the array from left sub-array to right sub-array –

  • Base Case: The base case for this problem will be when there is only one element in the two sub-arrays and we wanted to check the intra-array pairs. Then, check that those two elements form one such pair then increment the count and also place the smaller element at its position.
    if start1 == end1 and start2 == end2:
        if compare(arr, start1, start2):
            c += 1
    
  • Recursive Case: This problem can be divided into three types on the basis of the comparator function –
    1. When comparision operator between pairs is of greater than or equal to.
    2. When comparision operator between pairs is of less than or equal to.
    3. When comparision operator between pairs is equal to.

    Therefore, These all the three cases can be calculated individually for such pairs.

    • Case 1: In case greater than or equal to, if we find any such pair then all the elements to right of that subarray will also form pair with the current element. Due to which the count of such pairs is incremented by the number of elements left in the left sub-array.
      if compare(arr, start1, start2):
          count += end1 - start1 + 1
      
    • Case 2: In case less than or equal to, if we find any such pair then all the elements to right of that subarray will also form pair with the current element. Due to which the count of such pairs is incremented by the number of elements left in the right sub-array.
      if compare(arr, start1, start2):
          count += end2 - start2 + 1
      
    • Case 3: In case equal to, If we find any such pair, then we try to find all such pairs possible in the left subarray with the help of a while loop. In each such possible pair increment the count by 1.
      if compare(arr, start1, start2):
          while compare(arr, start1, start2):
              count += 1
              start1 += 1
      
  • Finally, Merge the two subarrays as it is done in the merge sort.

Below is the implementation of the above approach:

C++

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// C++ implementation to find the 
// elements on the right with the given 
// custom comparator 
  
#include <bits/stdc++.h>
  
using namespace std;
  
// comparator to check 
// if two elements are equal 
bool compare(int arr[], int s1, int s2){
    if (arr[s1] > arr[s2]){
        return true;
    }
    else{
        return false;
    }
}
  
// Function to find the Intra-Array 
// Count in the two subarrays 
int findIntraArrayCount(int arr[], int s1,
       int e1, int s2, int e2, int g){
             
    // Base Case
    if (s1 == e1 && s2 == e2){
        int c = 0;
        if (compare(arr, s1, s2)){
            c += 1;
        }
        if (arr[s1] > arr[s2]){
            int temp = arr[s1];
            arr[s1] = arr[s2];
            arr[s2] = temp;
        }
        return c;
    }
      
    // Variable for keeping 
    // the count of the pair
    int c = 0;
    int s = s1, e = e2, s3 = s1;
    int e3 = e1, s4 = s2, e4 = e2;
      
    while (s1 <= e1 && s2 <= e2){
          
        // Condition when we have to use the 
        // Greater than comparator 
        if (g == 1){
            if (compare(arr, s1, s2)){
                c += e1 - s1 + 1;
                s2 += 1;
            }
            else{
                s1 += 1;
            }
        }
          
        // Condition when we have to use the 
        // Less than comparator
        else if (g == 0){
            if (compare(arr, s1, s2)){
                c += e2 - s2 + 1;
                s1 += 1;
            }
            else {
                s2 += 1;
            }
        }
          
        // Condition when we have to use the 
        // Equal to Comparator 
        else if (g == -1){
            if (compare(arr, s1, s2)){
                int c1 = 0;
                while (s1 <= e1 && 
                       compare(arr, s1, s2)){
                    c1 += 1;
                    s1 += 1;
                }
                s1 -= 1;
                int c2 = 0;
                while (s2 <= e2 && 
                       compare(arr, s1, s2)){
                    c2 += 1;
                    s2 += 1;
                }
                c += c1 * c2;
            }
            else {
                if (arr[s1] > arr[s2]){
                    s2 += 1;
                }
                else{
                    s1 += 1;
                }
            }
        }
    }
    s1 = s3; e1 = e3;
    s2 = s4; e2 = e4;
      
    // Array to store 
    // the sorted subarray
    vector<int> aux;
      
    // Merge the two subarrays
    while (s1 <= e1 && s2 <= e2){
        if (arr[s1] <= arr[s2]){
            aux.push_back(arr[s1]);
            s1 += 1;
        }
        else{
            aux.push_back(arr[s2]);
            s2 += 1;
        }
    }
      
    // Copy subarray 1 elements 
    while (s1 <= e1){
        aux.push_back(arr[s1]);
        s1 += 1;
    }
      
    // Copy subarray 2 elements 
    while (s2 <= e2){
        aux.push_back(arr[s2]);
        s2 += 1;
    }
      
    // Update the original array
    for (int i = s; i <= e; i++){
        arr[i] = aux[i-s];
    }
    return c;
}
  
// Function to find such pairs with 
// any custom comparator function 
int findElementsOnRight(int arr[], int s,
                           int e, int g){
    if (s >= e){
        return 0;
    }
    int mid = (s+e)/2;
      
    // Recursive call for inter-array 
    // count of pairs in left subarrays
    int c1 = findElementsOnRight(
                      arr, s, mid, g);
      
    // Recursive call for inter-array 
    // count of pairs in right sub-arrays 
    int c2 = findElementsOnRight(
                   arr, mid + 1, e, g);
      
    // Call for intra-array pairs 
    int c3 = findIntraArrayCount(
              arr, s, mid, mid+1, e, g);
      
    return c1 + c2 + c3;
}
  
  
// Driver code
int main()
{
    int arr[] = {4, 3, 2, 1};
    int g = 1;
    cout << findElementsOnRight(arr, 0, 3, g);
    return 0;
}

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Java

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// Java implementation to find the 
// elements on the right with the given 
// custom comparator 
  
import java.io.*;
import java.lang.*;
import java.util.*;
  
class GFG {
      
    // comparator to check 
    // if two elements are equal 
    public static boolean compare(
          int[] arr, int s1, int s2){
        if (arr[s1] > arr[s2]){
            return true;
        }
        else{
            return false;
        }
    }
      
    // Function to find the Intra-Array 
    // Count in the two subarrays 
    public static int findIntraArrayCount(
        int[] arr, int s1, int e1, int s2, 
                           int e2, int g){
                                 
        // Base Case
        if (s1 == e1 && s2 == e2){
            int c = 0;
            if (compare(arr, s1, s2)){
                c += 1;
            }
            if (arr[s1] > arr[s2]){
                int temp = arr[s1];
                arr[s1] = arr[s2];
                arr[s2] = temp;
            }
            return c;
        }
          
        // Variable for keeping 
        // the count of the pair
        int c = 0;
        int s = s1, e = e2, s3 = s1;
        int e3 = e1, s4 = s2, e4 = e2;
          
        while (s1 <= e1 && s2 <= e2){
              
            // Condition when we have to use the 
            // Greater than comparator
            if (g == 1){
                if (compare(arr, s1, s2)){
                    c += e1 - s1 + 1;
                    s2 += 1;
                }
                else{
                    s1 += 1;
                }
            }
              
            // Condition when we have to use the 
            // Equal to Comparator 
            else if (g == 0){
                if (compare(arr, s1, s2)){
                    c += e2 - s2 + 1;
                    s1 += 1;
                }
                else {
                    s2 += 1;
                }
            }
              
            // Condition when we have to use the 
            // Equal to Comparator 
            else if (g == -1){
                if (compare(arr, s1, s2)){
                    int c1 = 0;
                    while (s1 <= e1 && 
                         compare(arr, s1, s2)){
                        c1 += 1;
                        s1 += 1;
                    }
                    s1 -= 1;
                    int c2 = 0;
                    while (s2 <= e2 && 
                          compare(arr, s1, s2)){
                        c2 += 1;
                        s2 += 1;
                    }
                    c += c1 * c2;
                }
                else {
                    if (arr[s1] > arr[s2]){
                        s2 += 1;
                    }
                    else{
                        s1 += 1;
                    }
                }
            }
        }
        s1 = s3; e1 = e3;
        s2 = s4; e2 = e4;
          
        // Array to store 
        // the sorted subarray
        ArrayList<Integer> aux = 
                     new ArrayList<>();
                       
        // Merge the two subarrays
        while (s1 <= e1 && s2 <= e2){
            if (arr[s1] <= arr[s2]){
                aux.add(arr[s1]);
                s1 += 1;
            }
            else{
                aux.add(arr[s2]);
                s2 += 1;
            }
        }
          
        // Copy subarray 1 elements 
        while (s1 <= e1){
            aux.add(arr[s1]);
            s1 += 1;
        }
          
        // Copy subarray 2 elements
        while (s2 <= e2){
            aux.add(arr[s2]);
            s2 += 1;
        }
          
        // Update the original array
        for (int i = s; i <= e; i++){
            arr[i] = aux.get(i-s);
        }
        return c;
    }
      
    // Function to find such pairs with 
    // any custom comparator function 
    public static int findElementsOnRight(
        int[] arr, int s, int e, int g){
        if (s >= e){
            return 0;
        }
        int mid = (s+e)/2;
          
        // Recursive call for inter-array 
        // count of pairs in left subarrays
        int c1 = findElementsOnRight(arr, s, 
                                     mid, g);
          
        // Recursive call for inter-array 
        // count of pairs in right sub-arrays 
        int c2 = findElementsOnRight(arr, mid + 1
                                         e, g);
          
        // Call for intra-array pairs 
        int c3 = findIntraArrayCount(arr, s, 
                            mid, mid+1, e, g);
          
        return c1 + c2 + c3;                                   
    }
      
    // Driver code
    public static void main (String[] args) {
        int[] arr = {4, 3, 2, 1};
        int g = 1;
        System.out.println(
            findElementsOnRight(arr, 0, 3, g));
    }
}

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Python

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# Python implementation to find the 
# elements on the right with the given 
# custom comparator 
  
import random, math 
from copy import deepcopy as dc 
  
# comparator to check 
# if two elements are equal 
def compare(arr, s1, s2): 
    if arr[s1] > arr[s2]: 
        return True
    else
        return False
  
  
# Function to find the Intra-Array 
# Count in the two subarrays 
def findIntraArrayCount(arr, s1, \
                e1, s2, e2, g): 
      
    # Base Case 
    if s1 == e1 and s2 == e2: 
        c = 0
        if compare(arr, s1, s2): 
            c += 1
        if arr[s1] > arr[s2]: 
            arr[s1], arr[s2] = arr[s2], arr[s1] 
        return
  
  
    # Variable for keeping 
    # the count of the pair 
    c = 0
      
    # Total subarray length 
    s = dc(s1); e = dc(e2) 
      
    # length of subarray 1 
    s3 = dc(s1); e3 = dc(e1) 
      
    # length of subarray 2 
    s4 = dc(s2); e4 = dc(e2) 
      
    while s1 <= e1 and s2 <= e2: 
  
  
        # Condition when we have to use the 
        # Greater than comparator 
        if g == 1
            if compare(arr, s1, s2): 
                c += e1 - s1 + 1
                s2 += 1
            else
                s1 += 1
  
        # Condition when we have to use the 
        # Less than comparator 
        elif g == 0
            if compare(arr, s1, s2): 
                c += e2 - s2 + 1
                s1 += 1
            else
                s2 += 1
                  
        # Condition when we have to use the 
        # Equal to Comparator 
        elif g == -1
            if compare(arr, s1, s2):
                c1 = 0
                while s1 <= e1 and\
                   compare(arr, s1, s2): 
                    c1 += 1
                    s1 += 1
                s1 -= 1
                c2 = 0
                while s2 <= e2 and\
                    compare(arr, s1, s2):
                    c2 += 1
                    s2 += 1
                c += c1 * c2
            else
                if arr[s1] > arr[s2]:
                    s2 += 1
                else:
                    s1 += 1
  
    s1 = dc(s3); e1 = dc(e3) 
      
    s2 = dc(s4); e2 = dc(e4) 
      
    # Array to store the sorted subarray 
    aux = [] 
      
      
    # Merge the two subarrays 
    while s1 <= e1 and s2 <= e2: 
        if arr[s1] <= arr[s2]: 
            aux.append(arr[s1]) 
            s1 += 1
        else
            aux.append(arr[s2]) 
            s2 += 1
      
    # Copy subarray 1 elements 
    while s1 <= e1: 
        aux.append(arr[s1]) 
        s1 += 1
          
      
    # Copy subarray 2 elements 
    while s2 <= e2: 
        aux.append(arr[s2]) 
        s2 += 1
          
      
    # Update the original array 
    for i in range(s, e + 1): 
        arr[i] = aux[i-s] 
    return
  
  
  
# Function to find such pairs with 
# any custom comparator function 
def findElementsOnRight(arr, s, e, g): 
    if s >= e: 
        return 0
    mid = (s + e)//2
      
    # Recursive call for inter-array 
    # count of pairs in left subarrays 
    c1 = findElementsOnRight(arr, s, \
                            mid, g) 
      
    # Recursive call for inter-array 
    # count of pairs in right sub-arrays 
    c2 = findElementsOnRight(arr, mid + 1, \
                                e, g) 
      
    # Call for intra-array pairs 
    c3 = findIntraArrayCount(arr, s, mid, \
                            mid + 1, e, g) 
    return c1 + c2 + c3 
  
  
  
  
# Drive Code 
if __name__ == "__main__"
    arr = [4, 3, 2, 1]
    g = 1
    out = findElementsOnRight(arr, 0, \
                        len(arr)-1, g) 
    print(out) 

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Output:

6

Time Complexity: The above method takes O(N*logN) time.

Auxillary Space: O(N)

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