Count pairs from two sorted arrays whose sum is equal to a given value x

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.
?>

Javascript

<script>

// Javascript 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)
        {
            let mid = Math.floor((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
    // value1
    function countPairs(arr1,arr2,m,n,x)
    {
        let count = 0;
        for (let i = 0; i < m; i++)
        {

            // for each arr1[i]
            let 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
    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 avanitrachhadiya2155

</script>

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

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;

    let us = new Set();

    // insert all the elements
    // of 1st array in the hash
    // table(unordered_set 'us')
    for (let i = 0; i < m; i++)
        us.add(arr1[i]);

    // for each element of 'arr2[]
    for (let j = 0; j < n; j++)

        // find (x - arr2[j]) in 'us'
        if (us.has(x - arr2[j]))
            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 rag2127

</script>

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

?>

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;
        let 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;
    }

    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));

</script>

Output :

Count = 2

Time Complexity : O(m + n) 
Auxiliary space : O(1)
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