# Count pairs with given sum

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
Pairs with sum 6 are (1, 5) and (7, -1)

Input  :  arr[] = {1, 5, 7, -1, 5},
sum = 6
Output :  3
Pairs with sum 6 are (1, 5), (7, -1) &
(1, 5)

Input  :  arr[] = {1, 1, 1, 1},
sum = 2
Output :  6
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)

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A simple solution is be 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 ` `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

## 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 program is contributed by Jyotsna `

## 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

 ` `

Output :

`Count of pairs is 3`

Time Complexity : O(n2)
Auxiliary Space : O(1)

A better solution is possible in O(n) time.

Below is the Algorithm.

1. Create a map to store frequency of each number in the array. (Single traversal is required)
2. 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.
3. 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 ` `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

## 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 hm = ``new` `HashMap<>(); ` ` `  `        ``// Store counts of all elements in map hm ` `        ``for` `(``int` `i=``0``; i

## 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]] ` `        ``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]] != ``null``) ` `        ``{ ` `            ``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 `

Output :

`Count of pairs is 3`