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; } |
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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. |
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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 = 10print("Count = ", countPairs(arr1, arr2, m, n, x)) # This code is contributed by Shrikant13. |
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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. |
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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. ?> |
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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; } |
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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. |
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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 |
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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. |
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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. ?> |
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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; } |
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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 |
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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 = 10print("Count =", countPairs(arr1, arr2, m, n, x)) # This code is contributed by Shrikant13 |
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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 |
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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; } |
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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. |
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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 |
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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. |
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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 ?> |
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Output :
Count = 2
Time Complexity : O(m + n)
Auxiliary space : O(1)
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