# Count pairs with given sum

Given an array of N integers, and an integer K, the task is to find the number of pairs of integers in the array whose sum is equal to K.

Examples:

Input: arr[] = {1, 5, 7, -1}, K = 6
Output:  2
Explanation: Pairs with sum 6 are (1, 5) and (7, -1).

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

Input: arr[] = {1, 1, 1, 1}, K = 2
Output:  6
Explanation: Pairs with sum 2 are (1, 1), (1, 1), (1, 1), (1, 1), (1, 1).

Input: arr[] = {10, 12, 10, 15, -1, 7, 6, 5, 4, 2, 1, 1, 1}, K = 11
Output:  9
Explanation: Pairs with sum 11 are (10, 1), (10, 1), (10, 1), (12, -1), (10, 1), (10, 1), (10, 1), (7, 4), (6, 5).

NaÃ¯ve Approach:

A simple solution is to user nested loops, one for traversal of each element and other for checking if there’s another number in the array which can be added to it to give K.

Illustration:

Consider an array arr [ ] = {1 ,5 ,7 ,1} and K = 6

• initialize count = 0
• First iteration
• i = 0
• j = 1
• arr[i]+arr[j] == K
• count = 1
• i = 0
• j = 2
• arr[i] + arr[j] != K
• count = 1
• i = 0
• j = 3
• arr[i] + arr[j] != K
• count = 1
• Second iteration
• i = 1
• j = 2
• arr[i] + arr[j] != K
• count = 1
• i = 1
• j = 3
• arr[i] + arr[j] == K
• count = 2
• Third iteration
• i = 2
• j = 3
• arr[i] + arr[j] != K
• count = 2
• Hence Output is 2 .

Follow the steps below to solve the given problem:

Follow the steps below for implementation:

• Initialize the count variable with 0 which stores the result.
• Use two nested loop and check if arr[i] + arr[j] == K, then increment the count variable.
• Return the count.

Below is the implementation of the above approach.

## C++

 `// C++ implementation of simple method to find count of``// pairs with given sum K.``#include ``using` `namespace` `std;` `// Returns number of pairs in arr[0..n-1] with sum equal``// to 'K'``int` `getPairsCount(``int` `arr[], ``int` `n, ``int` `K)``{``    ``// Initialize result``    ``int` `count = 0;` `    ``// 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] == K)``                ``count++;` `    ``return` `count;``}` `// Driver function to test the above function``int` `main()``{``    ``int` `arr[] = { 1, 5, 7, -1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `K = 6;``    ``cout << ``"Count of pairs is "``         ``// Function Call``         ``<< getPairsCount(arr, n, K);``    ``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 ` `// Returns number of pairs in arr[0..n-1] with sum equal``// to 'sum'``int` `getPairsCount(``int` `arr[], ``int` `n, ``int` `K)``{``   ``// Initialize result``    ``int` `count = 0;` `    ``// 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] == K)``                ``count++;` `    ``return` `count;``}` `// Driver function to test the above function``int` `main()``{``    ``int` `arr[] = { 1, 5, 7, -1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `K = 6;``    ``printf``(``"Count of pairs is %d"``,``           ``// Function Call``           ``getPairsCount(arr, n, K));``    ``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` `K = ``6``;``      ``// Function Call``        ``getPairsCount(arr, K);``    ``}` `    ``// Prints number of pairs in arr[0..n-1] with sum equal``    ``// to 'sum'``    ``public` `static` `void` `getPairsCount(``int``[] arr, ``int` `K)``    ``{``        ``// Initialize result``        ``int` `count = ``0``; ` `        ``// 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]) == K)``                    ``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, K):` `    ``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] ``=``=` `K:``                ``count ``+``=` `1` `    ``return` `count`  `# Driver function``arr ``=` `[``1``, ``5``, ``7``, ``-``1``]``n ``=` `len``(arr)``K ``=` `6``print``(``"Count of pairs is"``,``      ``getPairsCount(arr, n, K))` `# 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`

## Javascript

 ``

## PHP

 ``

Output
```Count of pairs is 2
```

Time Complexity: O(n2), traversing the array for each element
Auxiliary Space: O(1)

## Count pairs with given sum using Binary Search.

If the array is sorted then for each array element , we can find the number of pairs by searching all the values (K – arr[i]) which are situated after ith index using Binary Search.

Illustration:

Given arr[ ] = {1, 5, 7, -1} and K = 6

• sort the array = {-1,1,5,7}
• initialize count = 0
• At index = 0:
• val = K – arr[0] = 6 – (-1) = 7
• count = count + upperBound(1, 3, 7) – lowerBound(1, 3, 7)
• count = 1
• At index = 1:
• val = K – arr[1] = 6 – 1 = 5
• count = count + upperBound(2, 3, 5) – lowerBound(2, 3, 5)
• count = 2
• At index = 2:
• val = K – arr[2] = 6 – 5 = 1
• count = count + upperBound(3, 3, 1) – lowerBound(3, 3, 1)
• count = 2
• Number of pairs = 2

Follow the steps below to solve the given problem:

• 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 (K – arr[i]) using lower bound.
• Find the index of the first element having value just greater than (K – arr[i]) using upper bound.
• The gap between these two indices is the number of elements with value same as (K – 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 ``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 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `k = 6;` `    ``// Function call``    ``cout << ``"Count of pairs is "``         ``<< getPairsCount(arr, n, k);``    ``return` `0;``}`

## Java

 `// Java code for the above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `  ``// lowerBound implementation``  ``public` `static` `int` `lowerBound(``int``[] arr, ``int` `start,``                               ``int` `end, ``int` `key)``  ``{``    ``while` `(start < end) {``      ``int` `mid = start + (end - start) / ``2``;``      ``if` `(arr[mid] < key) {``        ``start = mid + ``1``;``      ``}``      ``else` `{``        ``end = mid;``      ``}``    ``}``    ``return` `start;``  ``}` `  ``// upperBound implementation``  ``public` `static` `int` `upperBound(``int``[] arr, ``int` `start,``                               ``int` `end, ``int` `key)``  ``{``    ``while` `(start < end) {``      ``int` `mid = start + (end - start) / ``2``;``      ``if` `(arr[mid] <= key) {``        ``start = mid + ``1``;``      ``}``      ``else` `{``        ``end = mid;``      ``}``    ``}``    ``return` `start;``  ``}` `  ``// Function to find the count of pairs``  ``public` `static` `int` `getPairsCount(``int``[] arr, ``int` `n, ``int` `k)``  ``{``    ``Arrays.sort(arr);``    ``int` `c = ``0``;``    ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {``      ``int` `x = k - arr[i];``      ``int` `y = lowerBound(arr, i + ``1``, n, x);``      ``int` `z = upperBound(arr, i + ``1``, n, x);``      ``c = c + z - y;``    ``}``    ``return` `c;``  ``}` `  ``public` `static` `void` `main(String[] args)``  ``{``    ``int``[] arr = { ``1``, ``5``, ``7``, -``1``};``    ``int` `n = arr.length;``    ``int` `k = ``6``;` `    ``// Function call``    ``System.out.println(``"Count of pairs is "``                       ``+ getPairsCount(arr, n, k));``  ``}``}` `// This code is contributed by lokeshmvs21.`

## 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``]``n ``=` `len``(arr)``k ``=` `6` `# Function call``print``(``"Count of pairs is"``, getPairsCount(arr, n, k))` `# This code is contributed by Pushpesh Raj`

## C#

 `// C# code for the above approach` `using` `System;``using` `System.Collections;` `public` `class` `GFG {` `    ``// lowerBound implementation``    ``public` `static` `int` `lowerBound(``int``[] arr, ``int` `start,``                                 ``int` `end, ``int` `key)``    ``{``        ``while` `(start < end) {``            ``int` `mid = start + (end - start) / 2;``            ``if` `(arr[mid] < key) {``                ``start = mid + 1;``            ``}``            ``else` `{``                ``end = mid;``            ``}``        ``}``        ``return` `start;``    ``}` `    ``// upperBound implementation``    ``public` `static` `int` `upperBound(``int``[] arr, ``int` `start,``                                 ``int` `end, ``int` `key)``    ``{``        ``while` `(start < end) {``            ``int` `mid = start + (end - start) / 2;``            ``if` `(arr[mid] <= key) {``                ``start = mid + 1;``            ``}``            ``else` `{``                ``end = mid;``            ``}``        ``}``        ``return` `start;``    ``}` `    ``// Function to find the count of pairs``    ``public` `static` `int` `getPairsCount(``int``[] arr, ``int` `n, ``int` `k)``    ``{``        ``Array.Sort(arr);``        ``int` `c = 0;``        ``for` `(``int` `i = 0; i < n - 1; i++) {``            ``int` `x = k - arr[i];``            ``int` `y = lowerBound(arr, i + 1, n, x);``            ``int` `z = upperBound(arr, i + 1, n, x);``            ``c = c + z - y;``        ``}``        ``return` `c;``    ``}` `    ``static` `public` `void` `Main()``    ``{` `        ``// Code``        ``int``[] arr = { 1, 5, 7, -1};``        ``int` `n = arr.Length;``        ``int` `k = 6;` `        ``// Function call``        ``Console.WriteLine(``"Count of pairs is "``                          ``+ getPairsCount(arr, n, k));``    ``}``}` `// This code is contributed by lokesh.`

## Javascript

 `// JavaScript code for the above approach` `// lowerBound implementation``function` `lowerBound(arr, start, end, key) {``  ``while` `(start < end) {``    ``var` `mid = start + Math.floor((end - start) / 2);``    ``if` `(arr[mid] < key) {``      ``start = mid + 1;``    ``} ``else` `{``      ``end = mid;``    ``}``  ``}``  ``return` `start;``}` `// upperBound implementation``function` `upperBound(arr, start, end, key) {``  ``while` `(start < end) {``    ``var` `mid = start + Math.floor((end - start) / 2);``    ``if` `(arr[mid] <= key) {``      ``start = mid + 1;``    ``} ``else` `{``      ``end = mid;``    ``}``  ``}``  ``return` `start;``}` `// Function to find the count of pairs``function` `getPairsCount(arr, n, k) {``  ``arr.sort((a, b) => a - b);``  ``var` `c = 0;``  ``for` `(``var` `i = 0; i < n - 1; i++) {``    ``var` `x = k - arr[i];``    ``var` `y = lowerBound(arr, i + 1, n, x);``    ``var` `z = upperBound(arr, i + 1, n, x);``    ``c = c + z - y;``  ``}``  ``return` `c;``}` `var` `arr = [1, 5, 7, -1];``var` `n = arr.length;``var` `k = 6;` `// Function call``console.log(``"Count of pairs is "` `+ getPairsCount(arr, n, k));`

Output
```Count of pairs is 2
```

Time Complexity: O(n * log(n) ), applying binary search on each element
Auxiliary Space: O(1)

## Count pairs with given sum using Hashing

• Check the frequency of K – arr[i] in the arr
• This can be achieved using Hashing.

Illustration:

Given arr[] = {1, 5, 7, -1} and K = 6

• initialize count = 0
• At index = 0:
• freq[K – arr[0]] = freq[6 – 1] = freq[5] = 0
• count = 0
• freq[arr[0]] = freq[1] = 1
• At index = 1:
• freq[K – arr[1]] = freq[6 – 5] = freq[1] = 1
• count = 1
• freq[arr[1]] = freq[5] = 1
• At index = 2:
• freq[K – arr[2]] = freq[6 – 7] = freq[-1] = 0
• count = 1
• freq[arr[2]] = freq[7] = 1
• At index = 3:
• freq[K – arr[3]] = freq[6 – (-1)] = freq[7] = 1
• count = 2
• freq[arr[3]] = freq[-1] = 1
• Number of pairs  = 2

Follow the steps below to solve the given problem:

• Create a map to store the frequency of each number in the array.
• Check if (K – arr[i]) is present in the map, if present then increment the count variable by its frequency.
• After traversal is over, return the count.

Below is the implementation of the 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` `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 code``int` `main()``{``    ``int` `arr[] = { 1, 5, 7, -1};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `sum = 6;` `    ``// Function Call``    ``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 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``};``        ``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``]``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 };``        ``int` `n = arr.Length;``        ``int` `sum = 6;``        ``Console.Write(``"Count of pairs is "``                      ``+ getPairsCount(arr, n, sum));``    ``}``}` `// This code is contributed by umadevi9616`

## Javascript

 ``

Output
```Count of pairs is 2
```

Time Complexity: O(n), to iterate over the array
Auxiliary Space: O(n), to make a map of size n

Previous
Next