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# Count pairs with given sum

• Difficulty Level : Medium
• Last Updated : 10 Jan, 2023

Given an array of N integers, and a number sum, the task is to 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: 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}, sum = 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).

Recommended Practice

Naive Approach:

• 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.
• This can be achieved by nested loops.

Illustration:

Given arr[] = {1, 5, 7, -1}, sum = 6
count = 0

• First Iteration : For index = 0
{1, 5, 7, -1}, pair = (1, 5), count = 1
• Second Iteration : For index = 1
{1, 5, 7, -1}, count = 1
• Third Iteration : For index = 2
{1, 5, 7, -1}, count = 2

Hence output is 2

Follow the steps below to solve the given problem:

• Initialize the count variable with 0 which stores the result.
• Iterate arr and if the sum of ith and jth [i + 1…..n – 1] element is equal to sum i.e. arr[i] + arr[j] == sum, 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. ` `#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 < 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 ` ` `  `// 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

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## Javascript

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Output

`Count of pairs is 3`

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

## Count pairs with given sum using Binary Search

This approach is based on the following idea:

• If the array is sorted then for each array element arr[i], find the number of pairs by finding all the values (sum – arr[i]) which are situated after ith index.
• This can be achieved using Binary Search.

Illustration:

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

Array after sorting: arr[] = {-1, 1, 5, 7}
count = 0

At index = 0: val = sum – arr[0] = 6 – (-1) = 7
count = count + upperBound(1, 3, 7) – lowerBound(1, 3, 7)
count = 1

At index = 1: val = sum – arr[1] = 6 – 1 = 5
count = count + upperBound(2, 3, 5) – lowerBound(2, 3, 5)
count = 2

At index = 2: val = sum – 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 (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 ` `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; ` `}`

## 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``, ``5` `}; ` `    ``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``, ``5``] ` `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, 5 }; ` `        ``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.`

Output

`Count of pairs is 3`

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

## Count pairs with given sum using Hashing

This approach is based on the following idea:

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

Illustration:

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

• Store the frequency of every element:
freq[arr[i]] = freq[arr[i]] + 1
freq[1] : 1
freq[5] : 1
freq[7] : 1
freq[-1] : 1

• Initialise a variable count with 0 to find the required count of pairs
• At index = 0: freq[sum – arr[0]] = freq[6 – 1] = freq[5] = 1
count = 1
• At index = 1: freq[sum – arr[1]] = freq[6 – 5] = freq[1] = 1
count = 2
• At index = 2: freq[sum – arr[2]] = freq[6 – 7] = freq[-1] = 1
count = 3
• At index = 3: freq[sum – arr[3]] = freq[6 – (-1)] = freq[7] = 1
count = 4
• The above also contains repeated pairs from front and last, i.e. pair (a, b) and (b, a) are considered as different pairs till now.
Therefore, we will reduce the count by half to determine the count of unique pairs.
count = count / 2 = 2
• Therefore, required Number of pairs with given sum = 2

Follow the steps below to solve the given problem:

• Create a map to store the 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 the second traversal, we’d have twice the required value stored in counter because every pair is counted two times. Hence divide the count by 2 and return.

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` `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 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

 ``

Output

`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

## Count pairs with given sum using Hashing in Single loop

This approach is based on the following idea:

• The idea is to solve in single loop.
• Check the frequency of sum – arr[i] in the arr
• This can be achieved using Hashing.

Illustration:

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

count = 0

At index = 0: freq[sum – arr[0]] = freq[6 – 1] = freq[5] = 0
count = 0
freq[arr[0]] = freq[1] = 1

At index = 1: freq[sum – arr[1]] = freq[6 – 5] = freq[1] = 1
count = 1
freq[arr[1]] = freq[5] = 1

At index = 2: freq[sum – arr[2]] = freq[6 – 7] = freq[-1] = 0
count = 1
freq[arr[2]] = freq[7] = 1

At index = 3: freq[sum – arr[3]] = freq[6 – (-1)] = freq[7] = 1
count = 2
freq[arr[3]] = freq[-1] = 1

count = 2

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 (sum – 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, 5 }; ` `    ``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``, ``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

 ``

Output

`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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