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
// 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
# 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#
// 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 |
Javascript
<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> |
Output:
6
Time Complexity: The above method takes O(N*logN) time.
Auxiliary Space: O(N)
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