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
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 codeint 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 Codelet 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’
- First traversal array element, i goes from 0 to n-1
- Multiple each element with its index arr[i] = arr[i] * i
- 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 codeint 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 codeif __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:
- Traverse through the array and for each index i store i*arr[i].
- 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.
- Then we will add the obtained value from order_of_key of the current element into the ans variable.
- 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<nvoid 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 Codeint 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 < ndef 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 Codeif __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 Codeconst 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)
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