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Count pairs from two sorted arrays whose sum is equal to a given value x
  • Difficulty Level : Easy

Given two sorted arrays of size m and n of distinct elements. Given a value x. The problem is to count all pairs from both arrays whose sum is equal to x
Note: The pair has an element from each array.
Examples : 
 

Input : arr1[] = {1, 3, 5, 7}
        arr2[] = {2, 3, 5, 8}
        x = 10

Output : 2
The pairs are:
(5, 5) and (7, 3)

Input : arr1[] = {1, 2, 3, 4, 5, 7, 11} 
        arr2[] = {2, 3, 4, 5, 6, 8, 12} 
        x = 9

Output : 5

 

Method 1 (Naive Approach): Using two loops pick elements from both the arrays and check whether the sum of the pair is equal to x or not.
 

C++




// C++ implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
#include <bits/stdc++.h>
using namespace std;
 
// function to count all pairs
// from both the sorted arrays
// whose sum is equal to a given
// value
int countPairs(int arr1[], int arr2[],
               int m, int n, int x)
{
    int count = 0;
     
    // generating pairs from
    // both the arrays
    for (int i = 0; i < m; i++)
        for (int j = 0; j < n; j++)
 
            // if sum of pair is equal
            // to 'x' increment count
            if ((arr1[i] + arr2[j]) == x)
                count++;
     
    // required count of pairs    
    return count;
}
 
// Driver Code
int main()
{
    int arr1[] = {1, 3, 5, 7};
    int arr2[] = {2, 3, 5, 8};
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    int x = 10;
    cout << "Count = "
         << countPairs(arr1, arr2, m, n, x);
    return 0;    
}

Java




// Java implementation to count pairs from
// both sorted arrays whose sum is equal
// to a given value
import java.io.*;
 
class GFG {
         
    // function to count all pairs
    // from both the sorted arrays
    // whose sum is equal to a given
    // value
    static int countPairs(int []arr1,
             int []arr2, int m, int n, int x)
    {
        int count = 0;
         
        // generating pairs from
        // both the arrays
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
     
                // if sum of pair is equal
                // to 'x' increment count
                if ((arr1[i] + arr2[j]) == x)
                    count++;
         
        // required count of pairs
        return count;
    }
     
    // Driver Code
 
    public static void main (String[] args)
    {
        int arr1[] = {1, 3, 5, 7};
        int arr2[] = {2, 3, 5, 8};
        int m = arr1.length;
        int n = arr2.length;
        int x = 10;
         
        System.out.println( "Count = "
        + countPairs(arr1, arr2, m, n, x));
    }
}
 
// This code is contributed by anuj_67.

Python3




# python implementation to count
# pairs from both sorted arrays
# whose sum is equal to a given
# value
 
# function to count all pairs from
# both the sorted arrays whose sum
# is equal to a given value
def countPairs(arr1, arr2, m, n, x):
    count = 0
 
    # generating pairs from both
    # the arrays
    for i in range(m):
        for j in range(n):
 
            # if sum of pair is equal
            # to 'x' increment count
            if arr1[i] + arr2[j] == x:
                count = count + 1
 
    # required count of pairs
    return count
 
# Driver Program
arr1 = [1, 3, 5, 7]
arr2 = [2, 3, 5, 8]
m = len(arr1)
n = len(arr2)
x = 10
print("Count = ",
        countPairs(arr1, arr2, m, n, x))
 
# This code is contributed by Shrikant13.

C#




// C# implementation to count pairs from
// both sorted arrays whose sum is equal
// to a given value
using System;
 
class GFG {
         
    // function to count all pairs
    // from both the sorted arrays
    // whose sum is equal to a given
    // value
    static int countPairs(int []arr1,
            int []arr2, int m, int n, int x)
    {
        int count = 0;
         
        // generating pairs from
        // both the arrays
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
     
                // if sum of pair is equal
                // to 'x' increment count
                if ((arr1[i] + arr2[j]) == x)
                    count++;
         
        // required count of pairs
        return count;
    }
     
    // Driver Code
 
    public static void Main ()
    {
        int []arr1 = {1, 3, 5, 7};
        int []arr2 = {2, 3, 5, 8};
        int m = arr1.Length;
        int n = arr2.Length;
        int x = 10;
         
        Console.WriteLine( "Count = "
        + countPairs(arr1, arr2, m, n, x));
    }
}
 
// This code is contributed by anuj_67.

PHP




<?php
// PHP implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
 
 
// function to count all pairs
// from both the sorted arrays
// whose sum is equal to a given
// value
function countPairs( $arr1, $arr2,
                     $m, $n, $x)
{
    $count = 0;
     
    // generating pairs from
    // both the arrays
    for ( $i = 0; $i < $m; $i++)
        for ( $j = 0; $j < $n; $j++)
 
            // if sum of pair is equal
            // to 'x' increment count
            if (($arr1[$i] + $arr2[$j]) == $x)
                $count++;
     
    // required count of pairs
    return $count;
}
 
// Driver Code
$arr1 = array(1, 3, 5, 7);
$arr2 = array(2, 3, 5, 8);
$m = count($arr1);
$n = count($arr2);
$x = 10;
echo "Count = ",
      countPairs($arr1, $arr2,
                   $m,$n, $x);
 
// This code is contributed by anuj_67.
?>

Javascript




<script>
 
// JavaScript implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
 
// function to count all pairs
// from both the sorted arrays
// whose sum is equal to a given
// value
function countPairs(arr1, arr2, m, n, x)
{
    let count = 0;
     
    // generating pairs from
    // both the arrays
    for (let i = 0; i < m; i++)
        for (let j = 0; j < n; j++)
 
            // if sum of pair is equal
            // to 'x' increment count
            if ((arr1[i] + arr2[j]) == x)
                count++;
     
    // required count of pairs    
    return count;
}
 
// Driver Code
    let arr1 = [1, 3, 5, 7];
    let arr2 = [2, 3, 5, 8];
    let m = arr1.length;
    let n = arr2.length;
    let x = 10;
    document.write("Count = "
        + countPairs(arr1, arr2, m, n, x));
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>

Output : 

Count = 2

Time Complexity : O(mn) 
Auxiliary space : O(1)
Method 2 (Binary Search): For each element arr1[i], where 1 <= i <= m, search the value (x – arr1[i]) in arr2[]. If search is successful, increment the count.
 



C++




// C++ implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
#include <bits/stdc++.h>
using namespace std;
 
// function to search 'value'
// in the given array 'arr[]'
// it uses binary search technique
// as  'arr[]' is sorted
bool isPresent(int arr[], int low,
               int high, int value)
{
    while (low <= high)
    {
        int mid = (low + high) / 2;
         
        // value found
        if (arr[mid] == value)
            return true;    
             
        else if (arr[mid] > value)
            high = mid - 1;
        else
            low = mid + 1;
    }
     
    // value not found
    return false;
}
 
// function to count all pairs
// from both the sorted arrays
// whose sum is equal to a given
// value
int countPairs(int arr1[], int arr2[],
               int m, int n, int x)
{
    int count = 0;    
    for (int i = 0; i < m; i++)
    {
        // for each arr1[i]
        int value = x - arr1[i];
         
        // check if the 'value'
        // is present in 'arr2[]'
        if (isPresent(arr2, 0, n - 1, value))
            count++;
    }
     
    // required count of pairs    
    return count;
}
 
// Driver Code
int main()
{
    int arr1[] = {1, 3, 5, 7};
    int arr2[] = {2, 3, 5, 8};
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    int x = 10;
    cout << "Count = "
         << countPairs(arr1, arr2, m, n, x);
    return 0;    
}

Java




// Java implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
import java.io.*;
class GFG {
 
// function to search 'value'
// in the given array 'arr[]'
// it uses binary search technique
// as 'arr[]' is sorted
static boolean isPresent(int arr[], int low,
                         int high, int value)
{
    while (low <= high)
    {
        int mid = (low + high) / 2;
         
        // value found
        if (arr[mid] == value)
            return true;    
             
        else if (arr[mid] > value)
            high = mid - 1;
        else
            low = mid + 1;
    }
     
    // value not found
    return false;
}
 
// function to count all pairs
// from both the sorted arrays
// whose sum is equal to a given
// value
static int countPairs(int arr1[], int arr2[],
                      int m, int n, int x)
{
    int count = 0;
    for (int i = 0; i < m; i++)
    {
         
        // for each arr1[i]
        int value = x - arr1[i];
         
        // check if the 'value'
        // is present in 'arr2[]'
        if (isPresent(arr2, 0, n - 1, value))
            count++;
    }
     
    // required count of pairs
    return count;
}
 
    // Driver Code
    public static void main (String[] args)
    {
        int arr1[] = {1, 3, 5, 7};
        int arr2[] = {2, 3, 5, 8};
        int m = arr1.length;
        int n = arr2.length;
        int x = 10;
        System.out.println("Count = "
              + countPairs(arr1, arr2, m, n, x));
    }
}
 
// This code is contributed by anuj_67.

Python 3




# Python 3 implementation to count
# pairs from both sorted arrays
# whose sum is equal to a given
# value
 
# function to search 'value'
# in the given array 'arr[]'
# it uses binary search technique
# as 'arr[]' is sorted
def isPresent(arr, low, high, value):
 
    while (low <= high):
     
        mid = (low + high) // 2
         
        # value found
        if (arr[mid] == value):
            return True
             
        elif (arr[mid] > value) :
            high = mid - 1
        else:
            low = mid + 1
     
    # value not found
    return False
 
# function to count all pairs
# from both the sorted arrays
# whose sum is equal to a given
# value
def countPairs(arr1, arr2, m, n, x):
    count = 0
    for i in range(m):
        # for each arr1[i]
        value = x - arr1[i]
         
        # check if the 'value'
        # is present in 'arr2[]'
        if (isPresent(arr2, 0, n - 1, value)):
            count += 1
     
    # required count of pairs    
    return count
 
# Driver Code
if __name__ == "__main__":
    arr1 = [1, 3, 5, 7]
    arr2 = [2, 3, 5, 8]
    m = len(arr1)
    n = len(arr2)
    x = 10
    print("Count = ",
           countPairs(arr1, arr2, m, n, x))
 
# This code is contributed
# by ChitraNayal

C#




// C# implementation to count pairs from both
// sorted arrays whose sum is equal to a given
// value
using System;
 
class GFG {
 
    // function to search 'value' in the given
    // array 'arr[]' it uses binary search
    // technique as 'arr[]' is sorted
    static bool isPresent(int []arr, int low,
                         int high, int value)
    {
        while (low <= high)
        {
            int mid = (low + high) / 2;
             
            // value found
            if (arr[mid] == value)
                return true;    
                 
            else if (arr[mid] > value)
                high = mid - 1;
            else
                low = mid + 1;
        }
         
        // value not found
        return false;
    }
     
    // function to count all pairs
    // from both the sorted arrays
    // whose sum is equal to a given
    // value
    static int countPairs(int []arr1, int []arr2,
                             int m, int n, int x)
    {
        int count = 0;
         
        for (int i = 0; i < m; i++)
        {
             
            // for each arr1[i]
            int value = x - arr1[i];
             
            // check if the 'value'
            // is present in 'arr2[]'
            if (isPresent(arr2, 0, n - 1, value))
                count++;
        }
         
        // required count of pairs
        return count;
    }
 
    // Driver Code
    public static void Main ()
    {
        int []arr1 = {1, 3, 5, 7};
        int []arr2 = {2, 3, 5, 8};
        int m = arr1.Length;
        int n = arr2.Length;
        int x = 10;
        Console.WriteLine("Count = "
            + countPairs(arr1, arr2, m, n, x));
    }
}
 
// This code is contributed by anuj_67.

PHP




<?php
// PHP implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
 
// function to search 'value'
// in the given array 'arr[]'
// it uses binary search technique
// as 'arr[]' is sorted
function isPresent($arr, $low,
                   $high, $value)
{
    while ($low <= $high)
    {
        $mid = ($low + $high) / 2;
         
        // value found
        if ($arr[$mid] == $value)
            return true;    
             
        else if ($arr[$mid] > $value)
            $high = $mid - 1;
        else
            $low = $mid + 1;
    }
     
    // value not found
    return false;
}
 
// function to count all pairs
// from both the sorted arrays
// whose sum is equal to a given
// value
function countPairs($arr1, $arr2,
                    $m, $n, $x)
{
    $count = 0;
    for ($i = 0; $i < $m; $i++)
    {
         
        // for each arr1[i]
        $value = $x - $arr1[$i];
         
        // check if the 'value'
        // is present in 'arr2[]'
        if (isPresent($arr2, 0,
                      $n - 1, $value))
            $count++;
    }
     
    // required count of pairs
    return $count;
}
 
    // Driver Code
    $arr1 = array(1, 3, 5, 7);
    $arr2 = array(2, 3, 5, 8);
    $m = count($arr1);
    $n = count($arr2);
    $x = 10;
    echo "Count = "
        , countPairs($arr1, $arr2, $m, $n, $x);
 
// This code is contributed by anuj_67.
?>

Output : 
 

Count = 2

Time Complexity : O(mlogn), searching should be applied on the array which is of greater size so as to reduce the time complexity. 
Auxiliary space : O(1)
Method 3 (Hashing): Hash table is implemented using unordered_set in C++. We store all first array elements in hash table. For elements of second array, we subtract every element from x and check the result in hash table. If result is present, we increment the count.
 

C++




// C++ implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
#include <bits/stdc++.h>
using namespace std;
 
// function to count all pairs
// from both the sorted arrays
// whose sum is equal to a given
// value
int countPairs(int arr1[], int arr2[],
               int m, int n, int x)
{
    int count = 0;
     
    unordered_set<int> us;
     
    // insert all the elements
    // of 1st array in the hash
    // table(unordered_set 'us')
    for (int i = 0; i < m; i++)
        us.insert(arr1[i]);
     
    // for each element of 'arr2[]
    for (int j = 0; j < n; j++)
 
        // find (x - arr2[j]) in 'us'
        if (us.find(x - arr2[j]) != us.end())
            count++;
     
    // required count of pairs    
    return count;
}
 
// Driver Code
int main()
{
    int arr1[] = {1, 3, 5, 7};
    int arr2[] = {2, 3, 5, 8};
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    int x = 10;
    cout << "Count = "
         << countPairs(arr1, arr2, m, n, x);
    return 0;    
}

Java




import java.util.*;
// Java implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
 
class GFG
{
 
// function to count all pairs
// from both the sorted arrays
// whose sum is equal to a given
// value
static int countPairs(int arr1[], int arr2[],
            int m, int n, int x)
{
    int count = 0;
     
    HashSet<Integer> us = new HashSet<Integer>();
     
    // insert all the elements
    // of 1st array in the hash
    // table(unordered_set 'us')
    for (int i = 0; i < m; i++)
        us.add(arr1[i]);
     
    // for each element of 'arr2[]
    for (int j = 0; j < n; j++)
 
        // find (x - arr2[j]) in 'us'
        if (us.contains(x - arr2[j]))
            count++;
     
    // required count of pairs
    return count;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr1[] = {1, 3, 5, 7};
    int arr2[] = {2, 3, 5, 8};
    int m = arr1.length;
    int n = arr2.length;
    int x = 10;
    System.out.print("Count = "
        + countPairs(arr1, arr2, m, n, x));
}
}
 
// This code has been contributed by 29AjayKumar

Python3




# Python3 implementation to count
# pairs from both sorted arrays
# whose sum is equal to a given value
 
# function to count all pairs from 
# both the sorted arrays whose sum
# is equal to a given value
def countPairs(arr1, arr2, m, n, x):
    count = 0
    us = set()
 
    # insert all the elements
    # of 1st array in the hash
    # table(unordered_set 'us')
    for i in range(m):
        us.add(arr1[i])
 
    # or each element of 'arr2[]
    for j in range(n):
 
        # find (x - arr2[j]) in 'us'
        if x - arr2[j] in us:
            count += 1
 
    # required count of pairs
    return count
 
# Driver code
arr1 = [1, 3, 5, 7]
arr2 = [2, 3, 5, 8]
m = len(arr1)
n = len(arr2)
x = 10
print("Count =",
       countPairs(arr1, arr2, m, n, x))
 
# This code is contributed by Shrikant13

C#




// C# implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
using System;
using System.Collections.Generic;
 
class GFG
{
 
// function to count all pairs
// from both the sorted arrays
// whose sum is equal to a given
// value
static int countPairs(int []arr1, int []arr2,
            int m, int n, int x)
{
    int count = 0;
     
    HashSet<int> us = new HashSet<int>();
     
    // insert all the elements
    // of 1st array in the hash
    // table(unordered_set 'us')
    for (int i = 0; i < m; i++)
        us.Add(arr1[i]);
     
    // for each element of 'arr2[]
    for (int j = 0; j < n; j++)
 
        // find (x - arr2[j]) in 'us'
        if(us.Contains(x - arr2[j]))
            count++;
     
    // required count of pairs
    return count;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr1 = {1, 3, 5, 7};
    int []arr2 = {2, 3, 5, 8};
    int m = arr1.Length;
    int n = arr2.Length;
    int x = 10;
    Console.Write("Count = "
        + countPairs(arr1, arr2, m, n, x));
}
}
 
// This code contributed by Rajput-Ji

Output : 
 

Count = 2

Time Complexity : O(m+n) 
Auxiliary space : O(m), hash table should be created of the array having smaller size so as to reduce the space complexity.
Method 4 (Efficient Approach): This approach uses the concept of two pointers, one to traverse 1st array from left to right and another to traverse the 2nd array from right to left.
Algorithm : 
 

countPairs(arr1, arr2, m, n, x)

     Initialize l = 0, r = n - 1
     Initialize count = 0

     loop while l = 0
        if (arr1[l] + arr2[r]) == x
           l++, r--
           count++
        else if (arr1[l] + arr2[r]) < x
           l++
        else
           r--

     return count 

 

C++




// C++ implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
#include <bits/stdc++.h>
using namespace std;
 
// function to count all pairs
// from both the sorted arrays
// whose sum is equal to a given
// value
int countPairs(int arr1[], int arr2[],
               int m, int n, int x)
{
    int count = 0;
    int l = 0, r = n - 1;
     
    // traverse 'arr1[]' from
    // left to right
    // traverse 'arr2[]' from
    // right to left
    while (l < m && r >= 0)
    {
        // if this sum is equal
        // to 'x', then increment 'l',
        // decrement 'r' and
        // increment 'count'
        if ((arr1[l] + arr2[r]) == x)
        {
            l++; r--;
            count++;        
        }
         
        // if this sum is less
        // than x, then increment l
        else if ((arr1[l] + arr2[r]) < x)
            l++;
             
        // else decrement 'r'
        else
            r--;
    }
     
    // required count of pairs    
    return count;
}
 
// Driver Code
int main()
{
    int arr1[] = {1, 3, 5, 7};
    int arr2[] = {2, 3, 5, 8};
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    int x = 10;
    cout << "Count = "
          << countPairs(arr1, arr2, m, n, x);
    return 0;    
}

Java




// Java implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
import java.io.*;
 
class GFG {
 
    // function to count all pairs
    // from both the sorted arrays
    // whose sum is equal to a given
    // value
    static int countPairs(int arr1[],
         int arr2[], int m, int n, int x)
    {
        int count = 0;
        int l = 0, r = n - 1;
         
        // traverse 'arr1[]' from
        // left to right
        // traverse 'arr2[]' from
        // right to left
        while (l < m && r >= 0)
        {
             
            // if this sum is equal
            // to 'x', then increment 'l',
            // decrement 'r' and
            // increment 'count'
            if ((arr1[l] + arr2[r]) == x)
            {
                l++; r--;
                count++;        
            }
             
            // if this sum is less
            // than x, then increment l
            else if ((arr1[l] + arr2[r]) < x)
                l++;
                 
            // else decrement 'r'
            else
                r--;
        }
         
        // required count of pairs
        return count;
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        int arr1[] = {1, 3, 5, 7};
        int arr2[] = {2, 3, 5, 8};
        int m = arr1.length;
        int n = arr2.length;
        int x = 10;
        System.out.println( "Count = "
         + countPairs(arr1, arr2, m, n, x));
    }
}
 
// This code is contributed by anuj_67.

Python3




# Python 3 implementation to count
# pairs from both sorted arrays
# whose sum is equal to a given
# value
 
# function to count all pairs
# from both the sorted arrays
# whose sum is equal to a given
# value
def countPairs(arr1, arr2, m, n, x):
    count, l, r = 0, 0, n - 1
     
    # traverse 'arr1[]' from
    # left to right
    # traverse 'arr2[]' from
    # right to left
    while (l < m and r >= 0):
         
        # if this sum is equal
        # to 'x', then increment 'l',
        # decrement 'r' and
        # increment 'count'
        if ((arr1[l] + arr2[r]) == x):
            l += 1
            r -= 1
            count += 1
             
        # if this sum is less
        # than x, then increment l
        elif ((arr1[l] + arr2[r]) < x):
            l += 1
             
        # else decrement 'r'
        else:
            r -= 1
             
    # required count of pairs
    return count
 
# Driver Code
if __name__ == '__main__':
    arr1 = [1, 3, 5, 7]
    arr2 = [2, 3, 5, 8]
    m = len(arr1)
    n = len(arr2)
    x = 10
    print("Count =",
            countPairs(arr1, arr2,
                          m, n, x))
 
# This code is contributed
# by PrinciRaj19992

C#




// C# implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
using System;
 
class GFG {
 
    // function to count all pairs
    // from both the sorted arrays
    // whose sum is equal to a given
    // value
    static int countPairs(int []arr1,
        int []arr2, int m, int n, int x)
    {
        int count = 0;
        int l = 0, r = n - 1;
         
        // traverse 'arr1[]' from
        // left to right
        // traverse 'arr2[]' from
        // right to left
        while (l < m && r >= 0)
        {
             
            // if this sum is equal
            // to 'x', then increment 'l',
            // decrement 'r' and
            // increment 'count'
            if ((arr1[l] + arr2[r]) == x)
            {
                l++; r--;
                count++;        
            }
             
            // if this sum is less
            // than x, then increment l
            else if ((arr1[l] + arr2[r]) < x)
                l++;
                 
            // else decrement 'r'
            else
                r--;
        }
         
        // required count of pairs
        return count;
    }
     
    // Driver Code
    public static void Main ()
    {
        int []arr1 = {1, 3, 5, 7};
        int []arr2 = {2, 3, 5, 8};
        int m = arr1.Length;
        int n = arr2.Length;
        int x = 10;
        Console.WriteLine( "Count = "
        + countPairs(arr1, arr2, m, n, x));
    }
}
 
// This code is contributed by anuj_67.

PHP




<?php
// PHP implementation to count
// pairs from both sorted arrays
// whose sum is equal to a given
// value
 
 
// function to count all pairs
// from both the sorted arrays
// whose sum is equal to a given
// value
function  countPairs( $arr1$arr2,
          $m$n$x)
{
     $count = 0;
     $l = 0; $r = $n - 1;
     
    // traverse 'arr1[]' from
    // left to right
    // traverse 'arr2[]' from
    // right to left
    while ($l < $m and $r >= 0)
    {
        // if this sum is equal
        // to 'x', then increment 'l',
        // decrement 'r' and
        // increment 'count'
        if (($arr1[$l] + $arr2[$r]) == $x)
        {
            $l++; $r--;
            $count++;        
        }
         
        // if this sum is less
        // than x, then increment l
        else if (($arr1[$l] + $arr2[$r]) < $x)
            $l++;
             
        // else decrement 'r'
        else
            $r--;
    }
     
    // required count of pairs    
    return $count;
}
 
// Driver Code
     $arr1 = array(1, 3, 5, 7);
     $arr2 = array(2, 3, 5, 8);
     $m = count($arr1);
     $n = count($arr2);
     $x = 10;
     echo "Count = "
    , countPairs($arr1, $arr2, $m, $n, $x);
// This code is contributed by anuj_67
 
?>

Output : 
 

Count = 2

Time Complexity : O(m + n) 
Auxiliary space : O(1)
This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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