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

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: 0

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `?> `

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 ` `?> `

Output:

```3
```

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