# Count pairs in a sorted array whose sum is less than x

Given a sorted integer array and number x, the task is to count pairs in array whose sum is less than x.

Examples:

```Input  : arr[] = {1, 3, 7, 9, 10, 11}
x = 7
Output : 1
There is only one pair (1, 3)

Input  : arr[] = {1, 2, 3, 4, 5, 6, 7, 8}
x = 7
Output : 6
Pairs are (1, 2), (1, 3), (1, 4), (1, 5)
(2, 3) and (2, 4)
```

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

A simple solution of this problem run two loops to generate all pairs and one by one and check if current pair’s sum is less than x or not.

An Efficient solution of this problem is take initial and last value of index in l and r variable.

```1) Initialize two variables l and r to find the candidate
elements in the sorted array.
(a) l = 0
(b) r = n - 1
2) Initialize : result = 0
2) Loop while l < r.

// If current left and current
// right have sum smaller than x,
// the all elements from l+1 to r
// form a pair with current
(a) If (arr[l] + arr[r] < x)
result = result + (r - l)

(b) Else
r--;

3) Return result
```

Below is the implementation of above steps.

## C++

 `// C++ program to count pairs in an array ` `// whose sum is less than given number x ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count pairs in array ` `// with sum less than x. ` `int` `findPairs(``int` `arr[],``int` `n,``int` `x) ` `{ ` `    ``int` `l = 0, r = n-1; ` `    ``int` `result = 0; ` ` `  `    ``while` `(l < r) ` `    ``{ ` `        ``// If current left and current ` `        ``// right have sum smaller than x, ` `        ``// the all elements from l+1 to r ` `        ``// form a pair with current l. ` `        ``if` `(arr[l] + arr[r] < x) ` `        ``{ ` `            ``result += (r - l); ` `            ``l++; ` `        ``} ` ` `  `        ``// Move to smaller value ` `        ``else` `            ``r--; ` `    ``} ` ` `  `    ``return` `result; ` `} ` ` `  `// Driven code ` `int` `main() ` `{ ` `    ``int` `arr[] = {1, 2, 3, 4, 5, 6, 7, 8}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(``int``); ` `    ``int` `x = 7; ` `    ``cout << findPairs(arr, n, x); ` `    ``return` `0; ` `} `

## Java

 `// Java program to count pairs in an array ` `// whose sum is less than given number x ` `class` `GFG { ` `     `  `    ``// Function to count pairs in array ` `    ``// with sum less than x. ` `    ``static` `int` `findPairs(``int` `arr[], ``int` `n, ``int` `x) ` `    ``{ ` `         `  `        ``int` `l = ``0``, r = n - ``1``; ` `        ``int` `result = ``0``; ` `     `  `        ``while` `(l < r) ` `        ``{ ` `             `  `            ``// If current left and current ` `            ``// right have sum smaller than x, ` `            ``// the all elements from l+1 to r ` `            ``// form a pair with current l. ` `            ``if` `(arr[l] + arr[r] < x) ` `            ``{ ` `                ``result += (r - l); ` `                ``l++; ` `            ``} ` `     `  `            ``// Move to smaller value ` `            ``else` `                ``r--; ` `        ``} ` `     `  `        ``return` `result; ` `    ``} ` `     `  `    ``// Driver method ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `         `  `        ``int` `arr[] = {``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``8``}; ` `        ``int` `n = arr.length; ` `        ``int` `x = ``7``; ` `         `  `        ``System.out.print(findPairs(arr, n, x)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python3 program to count pairs ` `# in an array whose sum is less ` `# than given number x ` ` `  `# Function to count pairs in array ` `# with sum less than x. ` `def` `findPairs(arr, n, x): ` ` `  `    ``l ``=` `0``; r ``=` `n``-``1` `    ``result ``=` `0` ` `  `    ``while` `(l < r): ` `     `  `        ``# If current left and current ` `        ``# right have sum smaller than x, ` `        ``# the all elements from l+1 to r ` `        ``# form a pair with current l. ` `        ``if` `(arr[l] ``+` `arr[r] < x): ` `         `  `            ``result ``+``=` `(r ``-` `l) ` `            ``l ``+``=` `1` `         `  ` `  `        ``# Move to smaller value ` `        ``else``: ` `            ``r ``-``=` `1` ` `  `    ``return` `result ` `     `  `# Driver Code ` `arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``8``] ` `n ``=` `len``(arr) ` `x ``=` `7` `print``(findPairs(arr, n, x)) ` ` `  `# This code is contributed by Anant Agarwal. `

## C#

 `// C# program to count pairs in  ` `// an array whose sum is less ` `// than given number x ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Function to count pairs in array ` `    ``// with sum less than x. ` `    ``static` `int` `findPairs(``int` `[]arr, ``int` `n, ` `                         ``int` `x) ` `    ``{ ` `         `  `        ``int` `l = 0, r = n - 1; ` `        ``int` `result = 0; ` `     `  `        ``while` `(l < r) ` `        ``{ ` `             `  `            ``// If current left and current ` `            ``// right have sum smaller than x, ` `            ``// the all elements from l+1 to r ` `            ``// form a pair with current l. ` `            ``if` `(arr[l] + arr[r] < x) ` `            ``{ ` `                ``result += (r - l); ` `                ``l++; ` `            ``} ` `     `  `            ``// Move to smaller value ` `            ``else` `                ``r--; ` `        ``} ` `     `  `        ``return` `result; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `         `  `        ``int` `[]arr = {1, 2, 3, 4, 5, 6, 7, 8}; ` `        ``int` `n = arr.Length; ` `        ``int` `x = 7; ` `         `  `        ``Console.Write(findPairs(arr, n, x)); ` `    ``} ` `} ` ` `  `// This code is contributed by parashar... `

## PHP

 ` `

Output:

```6
```

Time complexity : O(n)
Space complexity : O(1)

Extension:
If array is unsorted, then we can sort the array first and then apply above method to solve in O(n Log n) time.

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Improved By : parashar, nitin mittal

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