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 = 7print(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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