Count pairs with given sum
- Difficulty Level : Medium
- Last Updated : 21 Jul, 2022
Given an array of integers, and a number ‘sum’, find the number of pairs of integers in the array whose sum is equal to ‘sum’.
Examples:

Input: arr[] = {1, 5, 7, -1}, sum = 6
Output: 2
Explanation: Pairs with sum 6 are (1, 5) and (7, -1)Input: arr[] = {1, 5, 7, -1, 5}, sum = 6
Output: 3
Explanation: Pairs with sum 6 are (1, 5), (7, -1) & (1, 5)Input: arr[] = {1, 1, 1, 1}, sum = 2
Output: 6
Explanation: There are 3! pairs with sum 2.Input: arr[] = {10, 12, 10, 15, -1, 7, 6, 5, 4, 2, 1, 1, 1}, sum = 11
Output: 9
Expected time complexity O(n)
Naive Solution – A simple solution is to traverse each element and check if there’s another number in the array which can be added to it to give sum.
C++
// C++ implementation of simple method to find count of // pairs with given sum. #include <bits/stdc++.h> using namespace std; // Returns number of pairs in arr[0..n-1] with sum equal // to 'sum' int getPairsCount( int arr[], int n, int sum) { int count = 0; // Initialize result // Consider all possible pairs and check their sums for ( int i = 0; i < n; i++) for ( int j = i + 1; j < n; j++) if (arr[i] + arr[j] == sum) count++; return count; } // Driver function to test the above function int main() { int arr[] = { 1, 5, 7, -1, 5 }; int n = sizeof (arr) / sizeof (arr[0]); int sum = 6; cout << "Count of pairs is " << getPairsCount(arr, n, sum); return 0; } // This code is contributed by Aditya Kumar (adityakumar129) |
C
// C implementation of simple method to find count of // pairs with given sum. #include <stdio.h> // Returns number of pairs in arr[0..n-1] with sum equal // to 'sum' int getPairsCount( int arr[], int n, int sum) { int count = 0; // Initialize result // Consider all possible pairs and check their sums for ( int i = 0; i < n; i++) for ( int j = i + 1; j < n; j++) if (arr[i] + arr[j] == sum) count++; return count; } // Driver function to test the above function int main() { int arr[] = { 1, 5, 7, -1, 5 }; int n = sizeof (arr) / sizeof (arr[0]); int sum = 6; printf ( "Count of pairs is %d" ,getPairsCount(arr, n, sum)); return 0; } // This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java implementation of simple method to find count of // pairs with given sum. public class find { public static void main(String args[]) { int [] arr = { 1 , 5 , 7 , - 1 , 5 }; int sum = 6 ; getPairsCount(arr, sum); } // Prints number of pairs in arr[0..n-1] with sum equal // to 'sum' public static void getPairsCount( int [] arr, int sum) { int count = 0 ; // Initialize result // Consider all possible pairs and check their sums for ( int i = 0 ; i < arr.length; i++) for ( int j = i + 1 ; j < arr.length; j++) if ((arr[i] + arr[j]) == sum) count++; System.out.printf( "Count of pairs is %d" , count); } } // This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# Python3 implementation of simple method # to find count of pairs with given sum. # Returns number of pairs in arr[0..n-1] # with sum equal to 'sum' def getPairsCount(arr, n, sum ): count = 0 # Initialize result # Consider all possible pairs # and check their sums for i in range ( 0 , n): for j in range (i + 1 , n): if arr[i] + arr[j] = = sum : count + = 1 return count # Driver function arr = [ 1 , 5 , 7 , - 1 , 5 ] n = len (arr) sum = 6 print ( "Count of pairs is" , getPairsCount(arr, n, sum )) # This code is contributed by Smitha Dinesh Semwal |
C#
// C# implementation of simple // method to find count of // pairs with given sum. using System; class GFG { public static void getPairsCount( int [] arr, int sum) { int count = 0; // Initialize result // Consider all possible pairs // and check their sums for ( int i = 0; i < arr.Length; i++) for ( int j = i + 1; j < arr.Length; j++) if ((arr[i] + arr[j]) == sum) count++; Console.WriteLine( "Count of pairs is " + count); } // Driver Code static public void Main() { int [] arr = { 1, 5, 7, -1, 5 }; int sum = 6; getPairsCount(arr, sum); } } // This code is contributed // by Sach_Code |
PHP
<?php // PHP implementation of simple // method to find count of // pairs with given sum. // Returns number of pairs in // arr[0..n-1] with sum equal // to 'sum' function getPairsCount( $arr , $n , $sum ) { // Initialize result $count = 0; // Consider all possible pairs // and check their sums for ( $i = 0; $i < $n ; $i ++) for ( $j = $i + 1; $j < $n ; $j ++) if ( $arr [ $i ] + $arr [ $j ] == $sum ) $count ++; return $count ; } // Driver Code $arr = array (1, 5, 7, -1, 5) ; $n = sizeof( $arr ); $sum = 6; echo "Count of pairs is " , getPairsCount( $arr , $n , $sum ); // This code is contributed by nitin mittal. ?> |
Javascript
<script> // Javascript implementation of simple method to find count of // pairs with given sum. // Returns number of pairs in arr[0..n-1] with sum equal // to 'sum' function getPairsCount(arr, n, sum) { let count = 0; // Initialize result // Consider all possible pairs and check their sums for (let i = 0; i < n; i++) for (let j = i + 1; j < n; j++) if (arr[i] + arr[j] == sum) count++; return count; } // Driver function to test the above function let arr = [ 1, 5, 7, -1, 5 ]; let n = arr.length; let sum = 6; document.write( "Count of pairs is " + getPairsCount(arr, n, sum)); // This code is contributed by Mayank Tyagi </script> |
Count of pairs is 3
Time Complexity: O(n2)
Auxiliary Space: O(1)
Efficient Approach: This approach is based on the following idea:
If the array is sorted then for each array element arr[i] we can find the number of pairs by finding all the values (sum – arr[i]) which are situated after ith index.
Follow the steps below to implement this approach:
- Sort the array arr[] in increasing order.
- Loop from i = 0 to N-1.
- Find the index of the first element having value same or just greater than (sum – arr[i]) using lower bound.
- Find the index of the first element having value just greater than (sum – arr[i]) using upper bound.
- The gap between these two indices is the number of elements with value same as (sum – arr[i]).
- Add this with the final count of pairs.
- Return the final count after the iteration is over.
Below is the implementation of the above approach.
C++
// C++ code to implement the approach #include <bits/stdc++.h> using namespace std; // Function to find the count of pairs int getPairsCount( int arr[], int n, int k) { sort(arr, arr + n); int x = 0, c = 0, y, z; for ( int i = 0; i < n - 1; i++) { x = k - arr[i]; // Lower bound from i+1 int y = lower_bound(arr + i + 1, arr + n, x) - arr; // Upper bound from i+1 int z = upper_bound(arr + i + 1, arr + n, x) - arr; c = c + z - y; } return c; } // Driver code int main() { int arr[] = { 1, 5, 7, -1, 5 }; int n = sizeof (arr) / sizeof (arr[0]); int k = 6; // Function call cout << "Count of pairs is " << getPairsCount(arr, n, k); return 0; } |
Python3
# Python code to implement the approach import bisect # Function to find the count of pairs def getPairsCount(arr,n,k): arr.sort() x,c = 0 , 0 for i in range (n - 1 ): x = k - arr[i] # Lower bound from i+1 y = bisect.bisect_left(arr,x,i + 1 ,n) # Upper bound from i+1 z = bisect.bisect(arr,x,i + 1 ,n) c = c + z - y return c # Driver function arr = [ 1 , 5 , 7 , - 1 , 5 ] n = len (arr) k = 6 # Function call print ( "Count of pairs is" ,getPairsCount(arr, n, k)) # This code is contributed by Pushpesh Raj |
Count of pairs is 3
Time Complexity: O(n * log(n) )
Auxiliary Space: O(1)
This approach is contributed by tene2902
Most Efficient Approach:
A better solution is possible in O(n) time. Below is the Algorithm –
- Create a map to store frequency of each number in the array. (Single traversal is required)
- In the next traversal, for every element check if it can be combined with any other element (other than itself!) to give the desired sum. Increment the counter accordingly.
- After completion of second traversal, we’d have twice the required value stored in counter because every pair is counted two times. Hence divide count by 2 and return.
Below is the implementation of above idea :
C++
// C++ implementation of simple method to find count of // pairs with given sum. #include <bits/stdc++.h> using namespace std; // Returns number of pairs in arr[0..n-1] with sum equal // to 'sum' int getPairsCount( int arr[], int n, int sum) { unordered_map< int , int > m; // Store counts of all elements in map m for ( int i = 0; i < n; i++) m[arr[i]]++; int twice_count = 0; // iterate through each element and increment the // count (Notice that every pair is counted twice) for ( int i = 0; i < n; i++) { twice_count += m[sum - arr[i]]; // if (arr[i], arr[i]) pair satisfies the condition, // then we need to ensure that the count is // decreased by one such that the (arr[i], arr[i]) // pair is not considered if (sum - arr[i] == arr[i]) twice_count--; } // return the half of twice_count return twice_count / 2; } // Driver function to test the above function int main() { int arr[] = { 1, 5, 7, -1, 5 }; int n = sizeof (arr) / sizeof (arr[0]); int sum = 6; cout << "Count of pairs is " << getPairsCount(arr, n, sum); return 0; } |
Java
/* Java implementation of simple method to find count of pairs with given sum*/ import java.util.HashMap; class Test { static int arr[] = new int [] { 1 , 5 , 7 , - 1 , 5 }; // Returns number of pairs in arr[0..n-1] with sum equal // to 'sum' static int getPairsCount( int n, int sum) { HashMap<Integer, Integer> hm = new HashMap<>(); // Store counts of all elements in map hm for ( int i = 0 ; i < n; i++) { // initializing value to 0, if key not found if (!hm.containsKey(arr[i])) hm.put(arr[i], 0 ); hm.put(arr[i], hm.get(arr[i]) + 1 ); } int twice_count = 0 ; // iterate through each element and increment the // count (Notice that every pair is counted twice) for ( int i = 0 ; i < n; i++) { if (hm.get(sum - arr[i]) != null ) twice_count += hm.get(sum - arr[i]); // if (arr[i], arr[i]) pair satisfies the // condition, then we need to ensure that the // count is decreased by one such that the // (arr[i], arr[i]) pair is not considered if (sum - arr[i] == arr[i]) twice_count--; } // return the half of twice_count return twice_count / 2 ; } // Driver method to test the above function public static void main(String[] args) { int sum = 6 ; System.out.println( "Count of pairs is " + getPairsCount(arr.length, sum)); } } // This code is contributed by Gaurav Miglani |
Python3
# Python 3 implementation of simple method # to find count of pairs with given sum. import sys # Returns number of pairs in arr[0..n-1] # with sum equal to 'sum' def getPairsCount(arr, n, sum ): m = [ 0 ] * 1000 # Store counts of all elements in map m for i in range ( 0 , n): m[arr[i]] + = 1 twice_count = 0 # Iterate through each element and increment # the count (Notice that every pair is # counted twice) for i in range ( 0 , n): twice_count + = m[ sum - arr[i]] # if (arr[i], arr[i]) pair satisfies the # condition, then we need to ensure that # the count is decreased by one such # that the (arr[i], arr[i]) pair is not # considered if ( sum - arr[i] = = arr[i]): twice_count - = 1 # return the half of twice_count return int (twice_count / 2 ) # Driver function arr = [ 1 , 5 , 7 , - 1 , 5 ] n = len (arr) sum = 6 print ( "Count of pairs is" , getPairsCount(arr, n, sum )) # This code is contributed by # Smitha Dinesh Semwal |
C#
// C# implementation of simple method to // find count of pairs with given sum using System; using System.Collections.Generic; class GFG { public static int [] arr = new int [] { 1, 5, 7, -1, 5 }; // Returns number of pairs in arr[0..n-1] // with sum equal to 'sum' public static int getPairsCount( int n, int sum) { Dictionary< int , int > hm = new Dictionary< int , int >(); // Store counts of all elements // in map hm for ( int i = 0; i < n; i++) { // initializing value to 0, // if key not found if (!hm.ContainsKey(arr[i])) { hm[arr[i]] = 0; } hm[arr[i]] = hm[arr[i]] + 1; } int twice_count = 0; // iterate through each element and // increment the count (Notice that // every pair is counted twice) for ( int i = 0; i < n; i++) { if (hm[sum - arr[i]] != 0) { twice_count += hm[sum - arr[i]]; } // if (arr[i], arr[i]) pair satisfies // the condition, then we need to ensure // that the count is decreased by one // such that the (arr[i], arr[i]) // pair is not considered if (sum - arr[i] == arr[i]) { twice_count--; } } // return the half of twice_count return twice_count / 2; } // Driver Code public static void Main( string [] args) { int sum = 6; Console.WriteLine( "Count of pairs is " + getPairsCount(arr.Length, sum)); } } // This code is contributed by Shrikant13 |
Javascript
<script> /* javascript implementation of simple method to find count of pairs with given sum*/ var arr = [ 1, 5, 7, -1, 5 ]; // Returns number of pairs in arr[0..n-1] with sum equal // to 'sum' function getPairsCount(n , sum) { var hm = new Map(); // Store counts of all elements in map hm for ( var i = 0; i < n; i++) { // initializing value to 0, if key not found if (!hm.has(arr[i])) hm.set(arr[i], 0); hm.set(arr[i], hm.get(arr[i]) + 1); } var twice_count = 0; // iterate through each element and increment the // count (Notice that every pair is counted twice) for (i = 0; i < n; i++) { if (hm.get(sum - arr[i]) != null ) twice_count += hm.get(sum - arr[i]); // if (arr[i], arr[i]) pair satisfies the // condition, then we need to ensure that the // count is decreased by one such that the // (arr[i], arr[i]) pair is not considered if (sum - arr[i] == arr[i]) twice_count--; } // return the half of twice_count return twice_count / 2; } // Driver method to test the above function var sum = 6; document.write( "Count of pairs is " + getPairsCount(arr.length, sum)); // This code is contributed by umadevi9616 </script> |
Count of pairs is 3
Time Complexity: O(n), to iterate over the array
Auxiliary Space: O(n), to make a map of size n
This article is contributed by Ashutosh Kumar.
More efficient solution in one loop:-
C++
// C++ implementation of simple method to find count of // pairs with given sum. #include <bits/stdc++.h> using namespace std; // Returns number of pairs in arr[0..n-1] with sum equal // to 'sum' int getPairsCount( int arr[], int n, int k) { unordered_map< int , int > m; int count = 0; for ( int i = 0; i < n; i++) { if (m.find(k - arr[i]) != m.end()) { count += m[k - arr[i]]; } m[arr[i]]++; } return count; } // Driver function to test the above function int main() { int arr[] = { 1, 5, 7, -1, 5}; int n = sizeof (arr) / sizeof (arr[0]); int sum = 6; cout << "Count of pairs is " << getPairsCount(arr, n, sum); return 0; } |
Java
// Java implementation of simple method to find count of // pairs with given sum. import java.util.*; class GFG{ // Returns number of pairs in arr[0..n-1] with sum equal // to 'sum' static int getPairsCount( int arr[], int n, int k) { HashMap<Integer,Integer> m = new HashMap<>(); int count = 0 ; for ( int i = 0 ; i < n; i++) { if (m.containsKey(k - arr[i])) { count += m.get(k - arr[i]); } if (m.containsKey(arr[i])){ m.put(arr[i], m.get(arr[i])+ 1 ); } else { m.put(arr[i], 1 ); } } return count; } // Driver function to test the above function public static void main(String[] args) { int arr[] = { 1 , 5 , 7 , - 1 , 5 }; int n = arr.length; int sum = 6 ; System.out.print( "Count of pairs is " + getPairsCount(arr, n, sum)); } } // This code is contributed by umadevi9616 |
Python3
# Python implementation of simple method to find count of # pairs with given sum. # Returns number of pairs in arr[0..n-1] with sum equal to 'sum' def getPairsCount(arr, n, sum ): unordered_map = {} count = 0 for i in range (n): if sum - arr[i] in unordered_map: count + = unordered_map[ sum - arr[i]] if arr[i] in unordered_map: unordered_map[arr[i]] + = 1 else : unordered_map[arr[i]] = 1 return count # Driver code arr = [ 1 , 5 , 7 , - 1 , 5 ] n = len (arr) sum = 6 print ( 'Count of pairs is' , getPairsCount(arr, n, sum )) # This code is contributed by Manish Thapa |
C#
// C# implementation of simple method to find count of // pairs with given sum. using System; using System.Collections.Generic; public class GFG { // Returns number of pairs in arr[0..n-1] with sum equal // to 'sum' static int getPairsCount( int []arr, int n, int k) { Dictionary< int , int > m = new Dictionary< int , int >(); int count = 0; for ( int i = 0; i < n; i++) { if (m.ContainsKey(k - arr[i])) { count += m[k - arr[i]]; } if (m.ContainsKey(arr[i])) { m[arr[i]] = m[arr[i]] + 1; } else { m.Add(arr[i], 1); } } return count; } // Driver function to test the above function public static void Main(String[] args) { int []arr = { 1, 5, 7, -1, 5 }; int n = arr.Length; int sum = 6; Console.Write( "Count of pairs is " + getPairsCount(arr, n, sum)); } } // This code is contributed by umadevi9616 |
Javascript
<script> // javascript implementation of simple method to find count of // pairs with given sum. // Returns number of pairs in arr[0..n-1] with sum equal // to 'sum' function getPairsCount(arr , n , k) { var m = new Map(); var count = 0; for ( var i = 0; i < n; i++) { if (m.has(k - arr[i])) { count += m.get(k - arr[i]); } if (m.has(arr[i])) { m.set(arr[i], m.get(arr[i]) + 1); } else { m.set(arr[i], 1); } } return count; } // Driver function to test the above function var arr = [ 1, 5, 7, -1, 5 ]; var n = arr.length; var sum = 6; document.write( "Count of pairs is " + getPairsCount(arr, n, sum)); // This code is contributed by umadevi9616 </script> |
Count of pairs is 3
Time Complexity: O(n), to iterate over the array
Auxiliary Space: O(n), to make a map of size n
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