# Count of pairs (x, y) in an array such that x < y

Given an array of N distinct integers, the task is to find the number of pairs (x, y) such that x < y.

Example:

Input: arr[] = {2, 4, 3, 1}
Output: 6
Possible pairs are (1, 2), (1, 3), (1, 4), (2, 3), (2, 4) and (3, 4)

Input: arr[] = {5, 10}
Output: 1
Only possible pair is (5, 10)

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

Naive approach: Find every possible pair and check whether it satisfies the given condition or not.

## Java

 `// Java implementation of the approach ` `class` `GFG { ` ` `  `    ``// Function to return the number of ` `    ``// pairs (x, y) such that x < y ` `    ``static` `int` `getPairs(``int` `a[]) ` `    ``{ ` `        ``// To store the number of valid pairs ` `        ``int` `count = ``0``; ` `        ``for` `(``int` `i = ``0``; i < a.length; i++) { ` `            ``for` `(``int` `j = ``0``; j < a.length; j++) { ` ` `  `                ``// If a valid pair is found ` `                ``if` `(a[i] < a[j]) ` `                    ``count++; ` `            ``} ` `        ``} ` ` `  `        ``// Return the count of valid pairs ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `a[] = { ``2``, ``4``, ``3``, ``1` `}; ` `        ``System.out.println(getPairs(a)); ` `    ``} ` `} `

## Python 3

 `# Python3 implementation of the approach ` ` `  `# Function to return the number of ` `# pairs (x, y) such that x < y ` `def` `getPairs(a): ` `     `  `    ``# To store the number of valid pairs ` `    ``count ``=` `0` `    ``for` `i ``in` `range``(``len``(a)): ` `        ``for` `j ``in` `range``(``len``(a)): ` ` `  `            ``# If a valid pair is found ` `            ``if` `(a[i] < a[j]): ` `                ``count ``+``=` `1` ` `  `    ``# Return the count of valid pairs ` `    ``return` `count ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``a ``=` `[ ``2``, ``4``, ``3``, ``1` `] ` `    ``print``(getPairs(a)) ` ` `  `# This code is contributed by ita_c `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` ` `  `    ``// Function to return the number of  ` `    ``// pairs (x, y) such that x < y  ` `    ``static` `int` `getPairs(``int` `[]a)  ` `    ``{  ` `        ``// To store the number of valid pairs  ` `        ``int` `count = 0;  ` `        ``for` `(``int` `i = 0; i < a.Length; i++)  ` `        ``{  ` `            ``for` `(``int` `j = 0; j < a.Length; j++)  ` `            ``{  ` ` `  `                ``// If a valid pair is found  ` `                ``if` `(a[i] < a[j])  ` `                    ``count++;  ` `            ``}  ` `        ``}  ` ` `  `        ``// Return the count of valid pairs  ` `        ``return` `count;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `[]a = { 2, 4, 3, 1 };  ` `        ``Console.WriteLine(getPairs(a));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Ryuga `

## PHP

 ` `

Output:

```6
```

Time Complexity: O(n2)

Efficient approach: For an element x. In order to find the count of valid pairs of the form (x, y1), (x, y2), …, (x, yn), we need to count the elements which are greater than x. For the smallest element, there will be n – 1 elements greater than it. Similarly, the second smallest element can form n – 2 pairs and so on. Therefore, the desired count of valid pairs will be (n – 1) + (n – 2) + …. + 1 = n * (n – 1) / 2 where n is the length of the array.

Below is the implementation of the above approach:

## Java

 `// Java implementation of the approach ` `class` `GFG { ` ` `  `    ``// Function to return the number of ` `    ``// pairs (x, y) such that x < y ` `    ``static` `int` `getPairs(``int` `a[]) ` `    ``{ ` `        ``// Length of the array ` `        ``int` `n = a.length; ` ` `  `        ``// Calculate the number of valid pairs ` `        ``int` `count = (n * (n - ``1``)) / ``2``; ` ` `  `        ``// Return the count of valid pairs ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `a[] = { ``2``, ``4``, ``3``, ``1` `}; ` `        ``System.out.println(getPairs(a)); ` `    ``} ` `} `

## C#

 `// C# implementation of the approach ` `using` `System; ` `class` `GFG  ` `{ ` ` `  `    ``// Function to return the number of ` `    ``// pairs (x, y) such that x < y ` `    ``static` `int` `getPairs(``int` `[]a) ` `    ``{ ` `        ``// Length of the array ` `        ``int` `n = a.Length; ` ` `  `        ``// Calculate the number of valid pairs ` `        ``int` `count = (n * (n - 1)) / 2; ` ` `  `        ``// Return the count of valid pairs ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]a = { 2, 4, 3, 1 }; ` `        ``Console.Write(getPairs(a)); ` `    ``} ` `} ` ` `  `// This code is contributed  ` `// by Akanksha Rai `

Output:

```6
```

Time Complexity: O(1)

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