Zähle Paare aus zwei sortierten Arrays, deren Summe gleich einem gegebenen Wert x ist
Gegeben sind zwei sortierte Arrays der Größe m und n unterschiedlicher Elemente. Gegeben sei ein Wert x . Das Problem besteht darin, alle Paare aus beiden Arrays zu zählen, deren Summe gleich x ist .
Hinweis: Das Paar hat ein Element aus jedem Array.
Beispiele:
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
Methode 1 (naiver Ansatz): Verwenden Sie zwei Schleifen, um Elemente aus beiden Arrays auszuwählen und zu prüfen, ob die Summe des Paars gleich x ist oder nicht.
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>
Ausgabe :
Count = 2
Zeitkomplexität : O(mn) Hilfsraum
: O(1)
Methode 2 (binäre Suche): Suchen Sie für jedes Element arr1[i] , wobei 1 <= i <= m , den Wert (x – arr1[i]) in arr2[] . Wenn die Suche erfolgreich ist, erhöhen Sie den Zähler .
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>
Ausgabe :
Count = 2
Zeitkomplexität: O(mlogn), die Suche sollte auf das Array angewendet werden, das größer ist, um die Zeitkomplexität zu reduzieren.
Hilfsraum: O(1)
Methode 3 (Hashing): Die Hash-Tabelle wird mit unordered_set in C++ implementiert . Wir speichern alle ersten Array-Elemente in der Hash-Tabelle. Für Elemente des zweiten Arrays subtrahieren wir jedes Element von x und überprüfen das Ergebnis in der Hash-Tabelle. Wenn result vorhanden ist, erhöhen wir den 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>
Ausgabe :
Count = 2
Zeitkomplexität: O(m+n)
Hilfsraum: O(m), Hash-Tabelle sollte aus dem Array mit kleinerer Größe erstellt werden, um die Raumkomplexität zu reduzieren.
Methode 4 (effizienter Ansatz): Dieser Ansatz verwendet das Konzept von zwei Zeigern, einen zum Durchlaufen des ersten Arrays von links nach rechts und einen anderen zum Durchlaufen des 2. Arrays von rechts nach links.
Algorithmus:
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>
Ausgabe :
Count = 2
Zeitkomplexität: O(m + n)
Hilfsraum: O(1)
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