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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