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<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:

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.

Reference: https://www.quora.com/Given-a-sorted-integer-array-and-number-z-count-all-pairs-x-y-in-the-array-so-that-x-+-y-z-Can-it-be-done-in-O-n

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My Personal Notes

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

C++

Java

Python3

C#

PHP

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