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# Count number of pairs (i, j) such that arr[i] * arr[j] > arr[i] + arr[j]

• Difficulty Level : Basic
• Last Updated : 11 Aug, 2021

Given an array arr[] of non-negative integers, the task is to count pairs of indices (i, j such that arr[i] * arr[j] > arr[i] + arr[j] where i < j.
Examples:

Input: arr[] = { 5, 0, 3, 1, 2 }
Output: 3
Input: arr[] = { 1, 1, 1 }
Output:

Naive Approach: Run two nested loops and check for every pair whether the condition is satisfied. If the condition is satisfied for any pair then update count = count + 1 and print the count in the end.
Below is the implementation of the above approach:

## C++

 `// C++ program to count pairs (i, j)``// such that arr[i] * arr[j] > arr[i] + arr[j]``#include ``using` `namespace` `std;` `// Function to return the count of pairs``// such that arr[i] * arr[j] > arr[i] + arr[j]``long` `countPairs(``int` `arr[], ``int` `n)``{``    ``long` `count = 0;``    ``for` `(``int` `i = 0; i < n - 1; i++) {``        ``for` `(``int` `j = i + 1; j < n; j++) {` `            ``// If condition is satisfied``            ``if` `(arr[i] * arr[j] > arr[i] + arr[j])``                ``count++;``        ``}``    ``}``    ``return` `count;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 5, 0, 3, 1, 2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << countPairs(arr, n);``    ``return` `0;``}`

## Java

 `// Java program to count pairs (i, j)``// such that arr[i] * arr[j] > arr[i] + arr[j]``import` `java.util.*;` `class` `solution``{` `// Function to return the count of pairs``// such that arr[i] * arr[j] > arr[i] + arr[j]``static` `long` `countPairs(``int` `arr[], ``int` `n)``{``    ``long` `count = ``0``;``    ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {``        ``for` `(``int` `j = i + ``1``; j < n; j++) {` `            ``// If condition is satisfied``            ``if` `(arr[i] * arr[j] > arr[i] + arr[j])``                ``count++;``        ``}``    ``}``    ``return` `count;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `arr[] = { ``5``, ``0``, ``3``, ``1``, ``2` `};``    ``int` `n = arr.length;``    ``System.out.println(countPairs(arr, n));``    ` `}``}` `// This code is contributed by``// Surendra_Gangwar`

## Python3

 `# Python3 program to count pairs(i,j)``# such that arr[i]*arr[j]>arr[i]+arr[j]``import` `math as mt` `# function to return the count of pairs``# such that arr[i]*arr[j]>arr[i]+arr[j]``def` `countPairs(arr, n):``    ``count ``=` `0``    ` `    ``for` `i ``in` `range``(n):``        ``for` `j ``in` `range``(i ``+` `1``, n):``            ` `            ``# if condition is satisified``            ``if` `arr[i] ``*` `arr[j] > arr[i] ``+` `arr[j]:``                ``count ``+``=` `1``            ` `    ``return` `count` `# Driver code``arr ``=` `[``5``, ``0``, ``3``, ``1``, ``2``]``n ``=` `len``(arr)` `print``(countPairs(arr, n))``        ` `# This code is contributed``# by Mohit Kumar 29`

## C#

 `// C# program to count pairs (i, j)``// such that arr[i] * arr[j] > arr[i] + arr[j]` `using` `System;` `public` `class` `GFG{``    ` `// Function to return the count of pairs``// such that arr[i] * arr[j] > arr[i] + arr[j]``static` `long` `countPairs(``int` `[]arr, ``int` `n)``{``    ``long` `count = 0;``    ``for` `(``int` `i = 0; i < n - 1; i++) {``        ``for` `(``int` `j = i + 1; j < n; j++) {` `            ``// If condition is satisfied``            ``if` `(arr[i] * arr[j] > arr[i] + arr[j])``                ``count++;``        ``}``    ``}``    ``return` `count;``}` `// Driver code``    ``static` `public` `void` `Main (){``    ``int` `[]arr = { 5, 0, 3, 1, 2 };``    ``int` `n = arr.Length;``    ``Console.WriteLine (countPairs(arr, n));``    ``}``}`

## PHP

 ` arr[i] + arr[j]` `// Function to return the count of pairs``// such that arr[i] * arr[j] > arr[i] + arr[j]``function` `countPairs(``\$arr``, ``\$n``)``{``    ``\$count` `= 0;``    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n` `- 1; ``\$i``++)``    ``{``        ``for` `(``\$j` `= ``\$i` `+ 1; ``\$j` `< ``\$n``; ``\$j``++)``        ``{` `            ``// If condition is satisfied``            ``if` `(``\$arr``[``\$i``] *``                ``\$arr``[``\$j``] > ``\$arr``[``\$i``] +``                           ``\$arr``[``\$j``])``                ``\$count``++;``        ``}``    ``}``    ``return` `\$count``;``}` `// Driver code``\$arr` `= ``array``( 5, 0, 3, 1, 2 );``\$n` `= sizeof(``\$arr``) ;` `echo` `countPairs(``\$arr``, ``\$n``);` `// This code is contributed by Ryuga``?>`

## Javascript

 ``
Output:

`3`

Efficient Approach: Consider the following cases:

1) arr[i] = 0 or arr[i] = 1 and arr[j] = any element
In this case, arr[j] * arr[i] will always be less than arr[i] + arr[j].
Hence we can discard all pairs which have one element either 0 or 1.
2) arr[i] = 2 and arr[j] <= 2
In this case, arr[j] * arr[i] will always be less than or equal to arr[i] + arr[j].
Hence again we can discard all such pairs.
3) arr[i] = 2 and arr[j] > 2
This case will produce valid pairs. If count_2 is the count of ‘2’s and count_others
is the count of elements greater than 2,
then number of pairs will be count_2 * count_others.
4) arr[i] > 2 and arr[j] > 2
This case will also produce valid pairs. Let count_others be the number of elements
greater than 2, then every two elements among these count_others elements
will form a valid pair. Hence the number of pairs will be Therefore, total count = (count_2 * count_others) + (count_others * (count_others – 1)) / 2

Below is the implementation of the above approach:

## C++

 `// C++ program to count pairs (i, j)``// such that arr[i] * arr[j] > arr[i] + arr[j]``#include ``using` `namespace` `std;` `// Function to return the count of pairs``// such that arr[i] * arr[j] > arr[i] + arr[j]``long` `countPairs(``const` `int``* arr, ``int` `n)``{``    ``int` `count_2 = 0, count_others = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] == 2)``            ``count_2++;``        ``else` `if` `(arr[i] > 2)``            ``count_others++;``    ``}``    ``long` `ans``        ``= 1L * count_2 * count_others``          ``+ (1L * count_others * (count_others - 1)) / 2;``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 5, 0, 3, 1, 2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << countPairs(arr, n);``    ``return` `0;``}`

## Java

 `// Java program to count pairs (i, j)``// such that arr[i] * arr[j] > arr[i] + arr[j]` `class` `GFG``{``    ``// Function to return the count of pairs``    ``// such that arr[i] * arr[j] > arr[i] + arr[j]``    ``static` `long` `countPairs(``int``[] arr, ``int` `n)``    ``{``        ``int` `count_2 = ``0``, count_others = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``if` `(arr[i] == ``2``)``            ``{``                ``count_2++;``            ``}``            ``else` `if` `(arr[i] > ``2``)``            ``{``                ``count_others++;``            ``}``        ``}``        ` `        ``long` `ans = 1L * count_2 * count_others +``                ``(1L * count_others * (count_others - ``1``)) / ``2``;``        ``return` `ans;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = {``5``, ``0``, ``3``, ``1``, ``2``};``        ``int` `n = arr.length;``        ``System.out.println(countPairs(arr, n));``    ``}``}` `// This code is contributed by``// 29AjayKumar`

## Python3

 `# Python3 program to count pairs(i,j)``# such that arr[i]*arr[j]>arr[i]+arr[j]``import` `math as mt` `# function to return the count of pairs``# such that arr[i]*arr[j]>arr[i]+arr[j]``def` `countPairs(arr, n):``    ``count_2, count_others ``=` `0``, ``0``    ` `    ``for` `i ``in` `range``(n):``        ``if` `arr[i] ``=``=` `2``:``            ``count_2 ``+``=` `1``        ``elif` `arr[i] > ``2``:``            ``count_others ``+``=` `1``    ``ans ``=` `(count_2 ``*` `count_others ``+``          ``(count_others ``*``          ``(count_others ``-` `1``)) ``/``/` `2``)``    ``return` `ans` `# Driver code``arr ``=` `[``5``, ``0``, ``3``, ``1``, ``2``]``n ``=` `len``(arr)` `print``(countPairs(arr, n))``        ` `# This code is contributed``# by Mohit Kumar`

## C#

 `// C# program to count pairs (i, j) such``// that arr[i] * arr[j] > arr[i] + arr[j]``using` `System;` `class` `GFG``{``    ``// Function to return the count of pairs``    ``// such that arr[i] * arr[j] > arr[i] + arr[j]``    ``static` `long` `countPairs(``int``[] arr, ``int` `n)``    ``{``        ``int` `count_2 = 0, count_others = 0;``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``if` `(arr[i] == 2)``            ``{``                ``count_2++;``            ``}``            ``else` `if` `(arr[i] > 2)``            ``{``                ``count_others++;``            ``}``        ``}``        ` `        ``long` `ans = 1L * count_2 * count_others +``                  ``(1L * count_others *``                       ``(count_others - 1)) / 2;``        ``return` `ans;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = {5, 0, 3, 1, 2};``        ``int` `n = arr.Length;``        ``Console.WriteLine(countPairs(arr, n));``    ``}``}` `// This code is contributed by``// Mukul Singh`

## PHP

 ` arr[i] + arr[j]` `// Function to return the count of pairs``// such that arr[i] * arr[j] > arr[i] + arr[j]``function` `countPairs(``\$arr``, ``\$n``)``{``    ``\$count_2` `= 0; ``\$count_others` `= 0;``    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)``    ``{``        ``if` `(``\$arr``[``\$i``] == 2)``            ``\$count_2``++;``        ``else` `if` `(``\$arr``[``\$i``] > 2)``            ``\$count_others``++;``    ``}``    ``\$ans` `= ``\$count_2` `* ``\$count_others` `+``                     ``(``\$count_others` `*``                     ``(``\$count_others` `- 1)) / 2;``    ``return` `\$ans``;``}` `// Driver code``\$arr` `= ``array``( 5, 0, 3, 1, 2 );``\$n` `= sizeof(``\$arr``);``echo` `countPairs(``\$arr``, ``\$n``);` `// This code is contributed``// by Akanksha Rai``?>`

## Javascript

 ``
Output:
`3`

Time Complexity: O(N)
Auxiliary Space: O(1)

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