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

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: Please solve it on “PRACTICE ” first, before moving on to the solution.

A Simple solution is to run two loops. Pick each element of array one-by-one and for each element find element on right side of array that hold condition, then increment counter and last return counter value.
Below is the implementation of above idea:

## C++

 `// C++ program to count all pair that ` `// hold condition i*arr[i] > j*arr[j] ` `#include ` `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 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); ` `    ``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

 `# C# 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

 ` 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 ` `?> `

Output:

```Count of Pairs : 5
```

Time Complexity: O(n2)

An efficient solution of this problem takes O(n log n) time. The idea is based on an interesting fact about this problem that after modifying the array such that every element is multiplied with its index, this problem convert 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
a) 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 the implementation of above idea

## C++

 `// C++ program to count all pair that ` `// hold condition i*arr[i] > j*arr[j] ` `#include ` `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; ``/* ndex 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

## 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;  ` `        ``/* ndex 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 ` `        ``// https://www.geeksforgeeks.org/counting-inversions/ ` `        ``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 `

## Python 3

 `# 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 ``# ndex 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 ` `    ``# https://www.geeksforgeeks.org/counting-inversions/ ` `    ``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;  ` `        ``/* ndex 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 ` `        ``// https://www.geeksforgeeks.org/counting-inversions/ ` `        ``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. `

Output:

```Count of Pairs : 5
```

Time Complexity: O(n log n)

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