Count number of pairs with the given Comparator
Last Updated :
19 Sep, 2023
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)
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
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 -
- When comparison operator between pairs is of greater than or equal to.
- When comparison operator between pairs is of less than or equal to.
- When comparison 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++
// 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;
}
// 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));
}
}
# Python3 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 c
# 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 c
# 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
# Driver Code
if __name__ == "__main__":
arr = [4, 3, 2, 1]
g = 1
out = findElementsOnRight(arr, 0, \
len(arr)-1, g)
print(out)
// C# implementation to find the
// elements on the right with the
// given custom comparator
using System;
using System.Collections.Generic;
class GFG{
// comparator to check
// if two elements are equal
public static 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
public static int findIntraArrayCount(int[] arr, int s1,
int e1, int s2,
int e2, int g)
{
// Base Case
if (s1 == e1 && s2 == e2)
{
int cc = 0;
if (compare(arr, s1, s2))
{
cc += 1;
}
if (arr[s1] > arr[s2])
{
int temp = arr[s1];
arr[s1] = arr[s2];
arr[s2] = temp;
}
return cc;
}
// 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
List<int> aux = new List<int>();
// 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[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
static public void Main()
{
int[] arr = { 4, 3, 2, 1 };
int g = 1;
Console.WriteLine(findElementsOnRight(
arr, 0, 3, g));
}
}
// This code is contributed by offbeat
<script>
// Javascript implementation to find the
// elements on the right with the given
// custom comparator
// comparator to check
// if two elements are equal
function 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
function findIntraArrayCount(arr, s1, e1, s2, e2, g) {
// Base Case
if (s1 == e1 && s2 == e2) {
let c = 0;
if (compare(arr, s1, s2)) {
c += 1;
}
if (arr[s1] > arr[s2]) {
let temp = arr[s1];
arr[s1] = arr[s2];
arr[s2] = temp;
}
return c;
}
// Variable for keeping
// the count of the pair
let c = 0;
let s = s1, e = e2, s3 = s1;
let 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)) {
let c1 = 0;
while (s1 <= e1 &&
compare(arr, s1, s2)) {
c1 += 1;
s1 += 1;
}
s1 -= 1;
let 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
let aux = new Array();
// Merge the two subarrays
while (s1 <= e1 && s2 <= e2) {
if (arr[s1] <= arr[s2]) {
aux.push(arr[s1]);
s1 += 1;
}
else {
aux.push(arr[s2]);
s2 += 1;
}
}
// Copy subarray 1 elements
while (s1 <= e1) {
aux.push(arr[s1]);
s1 += 1;
}
// Copy subarray 2 elements
while (s2 <= e2) {
aux.push(arr[s2]);
s2 += 1;
}
// Update the original array
for (let i = s; i <= e; i++) {
arr[i] = aux[i - s];
}
return c;
}
// Function to find such pairs with
// any custom comparator function
function findElementsOnRight(arr, s, e, g) {
if (s >= e) {
return 0;
}
let mid = Math.floor((s + e) / 2);
// Recursive call for inter-array
// count of pairs in left subarrays
let c1 = findElementsOnRight(
arr, s, mid, g);
// Recursive call for inter-array
// count of pairs in right sub-arrays
let c2 = findElementsOnRight(
arr, mid + 1, e, g);
// Call for intra-array pairs
let c3 = findIntraArrayCount(
arr, s, mid, mid + 1, e, g);
return c1 + c2 + c3;
}
// Driver code
let arr = [4, 3, 2, 1];
let g = 1;
document.write(findElementsOnRight(arr, 0, 3, g));
// This code is contributed by gfgking
</script>
Time Complexity: The above method takes O(N*logN) time.
Auxiliary Space: O(N)
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