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)

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<bits/stdc++.h> ` `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

`<?php ` `// PHP 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. ` `function` `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` `++; ` ` ` `} ` ` ` ` ` `// Move to smaller value ` ` ` `else` ` ` `$r` `--; ` ` ` `} ` ` ` ` ` `return` `$result` `; ` `} ` ` ` ` ` `// Driver Code ` ` ` `$arr` `= ` `array` `(1, 2, 3, 4, 5, 6, 7, 8); ` ` ` `$n` `= sizeof(` `$arr` `) / sizeof(` `$arr` `[0]); ` ` ` `$x` `= 7; ` ` ` `echo` `findPairs(` `$arr` `, ` `$n` `, ` `$x` `); ` ` ` `return` `0; ` ` ` `// This code is contributed by nitin mittal. ` `?> ` |

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.

**Related Articles:**

Count all distinct pairs with difference equal to k

Count pairs with given sum

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