# Given an array A[] and a number x, check for pair in A[] with sum as x (aka Two Sum)

• Difficulty Level : Easy
• Last Updated : 15 Jun, 2022

Write a program that, given an array A[] of n numbers and another number x, determines whether or not there exist two elements in A[] whose sum is exactly x.

Examples:

Input: arr[] = {0, -1, 2, -3, 1}
x= -2
Output: Pair with a given sum -2 is (-3, 1)
Valid pair exists
Explanation:  If we calculate the sum of the output,1 + (-3) = -2

Input: arr[] = {1, -2, 1, 0, 5}
x = 0
Output: No valid pair exists for 0

Method: Using simple logic by calculating the array’s elements themselves.

## C++

 `// C++ program for the above approach``#include ` `using` `namespace` `std;` `// Function to find and print pair``bool` `chkPair(``int` `A[], ``int` `size, ``int` `x) {``  ``for` `(``int` `i = 0; i < (size - 1); i++) {``    ``for` `(``int` `j = (i + 1); j < size; j++) {``      ``if` `(A[i] + A[j] == x) {``        ``cout << ``"Pair with a given sum "` `<< x << +``        ``" is ("` `<< A[i] << ``", "` `<< A[j] << ``")"` `<<``          ``endl;` `        ``return` `1;``      ``}``    ``}``  ``}` `  ``return` `0;``}` `int` `main() {``  ``int` `A[] = {``    ``0,``    ``-1,``    ``2,``    ``-3,``    ``1``  ``};``  ``int` `x = -2;``  ``int` `size = ``sizeof``(A) / ``sizeof``(A[0]);` `  ``if` `(chkPair(A, size, x)) {``    ``cout << ``"Valid pair exists"` `<< endl;``  ``} ``else` `{``    ``cout << ``"No valid pair exists for "` `<< x << endl;``  ``}` `  ``return` `0;``}` `// This code is contributed by Samim Hossain Mondal.`

## C

 `/*``  ``* This C program tells if there exists a pair in array whose sum results in x.``  ``*/` `#include ` `// Function to find and print pair``int` `chkPair(``int` `A[], ``int` `size, ``int` `x) {``    ``for` `(``int` `i = 0; i < (size - 1); i++) {``        ``for` `(``int` `j = (i + 1); j < size; j++) {``            ``if` `(A[i] + A[j] == x) {``                ``printf``(``"Pair with a given sum %d is (%d, %d)\n"``, x, A[i], A[j]);` `                ``return` `1;``            ``}``        ``}``    ``}` `    ``return` `0;``}` `int` `main(``void``) {``    ``int` `A[] = {0, -1, 2, -3, 1};``    ``int` `x = -2;``    ``int` `size = ``sizeof``(A) / ``sizeof``(A[0]);` `    ``if` `(chkPair(A, size, x)) {``        ``printf``(``"Valid pair exists\n"``);``    ``}``    ``else` `{``        ``printf``(``"No valid pair exists for %d\n"``, x);``    ``}` `    ``return` `0;``}``// This code is contributed by Manish Kumar (mkumar2789)`

## Java

 `// Java program to check if there exists a pair``// in array whose sum results in x.``class` `GFG{` `  ``// Function to find and print pair``  ``static` `boolean` `chkPair(``int` `A[], ``int` `size, ``int` `x) {``      ``for` `(``int` `i = ``0``; i < (size - ``1``); i++) {``          ``for` `(``int` `j = (i + ``1``); j < size; j++) {``              ``if` `(A[i] + A[j] == x) {``                  ``System.out.println(``"Pair with a given sum "` `+ x +``                                     ``" is ("` `+ A[i] + ``", "` `+ A[j] + ``")"``);``  ` `                  ``return` `true``;``              ``}``          ``}``      ``}` `      ``return` `false``;``  ``}` `  ``public` `static` `void` `main(String [] args) {``    ` `      ``int` `A[] = {``0``, -``1``, ``2``, -``3``, ``1``};``      ``int` `x = -``2``;``      ``int` `size = A.length;` `      ``if` `(chkPair(A, size, x)) {``          ``System.out.println(``"Valid pair exists"``);``      ``}``      ``else` `{``          ``System.out.println(``"No valid pair exists for "` `+ x );``      ``}``  ``}``}` `// This code is contributed by umadevi9616`

## Python3

 `# This python program tells if there exists a pair in array whose sum results in x.` `# Function to find and print pair``def` `chkPair(A, size, x):``    ``for` `i ``in` `range``(``0``, size ``-` `1``):``        ``for` `j ``in` `range``(i ``+` `1``, size):``            ``if` `(A[i] ``+` `A[j] ``=``=` `x):``                ``print``(f``"Pair with a given sum {x} is ({A[i]},{A[j]})"``)` `                ``return` `1``    ``return` `0` `if` `__name__ ``=``=` `"__main__"``:``    ``A ``=` `[``0``, ``-``1``, ``2``, ``-``3``, ``1``]``    ``x ``=` `-``2``    ``size ``=` `len``(A)` `    ``if` `(chkPair(A, size, x)):``        ``print``(``"Valid pair exists"``)` `    ``else``:``        ``print``(f``"No valid pair exists for {x}"``)` `    ``# This code is contributed by rakeshsahni`

## C#

 `// C# program to check if there exists a pair``// in array whose sum results in x.``using` `System;``class` `GFG{` `  ``// Function to find and print pair``  ``static` `bool` `chkPair(``int` `[] A, ``int` `size, ``int` `x) {``      ``for` `(``int` `i = 0; i < (size - 1); i++) {``          ``for` `(``int` `j = (i + 1); j < size; j++) {``              ``if` `(A[i] + A[j] == x) {``                  ``Console.WriteLine(``"Pair with a given sum "` `+ x +``                                     ``" is ("` `+ A[i] + ``", "` `+ A[j] + ``")"``);``  `  `                  ``return` `true``;``              ``}``          ``}``      ``}` `      ``return` `false``;``  ``}` `  ``public` `static` `void` `Main()``  ``{``      ``int` `[] A = {0, -1, 2, -3, 1};``      ``int` `x = -2;``      ``int` `size = A.Length;` `      ``if` `(chkPair(A, size, x)) {``          ``Console.WriteLine(``"Valid pair exists"``);``      ``}``      ``else` `{``          ``Console.WriteLine(``"No valid pair exists for "` `+ x );``      ``} ``  ``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Javascript

 ``

## Go

 `package` `main``import` `(``"fmt"``)`  `func` `twoSum(nums [``5``]int, target int) {``    ``var` `flag bool = false``    ``result:=make([]int,``2``)``    ``for` `i:=``0``;i

Output

```Pair with a given sum -2 is (-3, 1)
Valid pair exists```

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

Method 1: Sorting and Two-Pointers technique.

Approach: A tricky approach to solve this problem can be to use the two-pointer technique. But for using two pointer technique, the array must be sorted. Once the array is sorted the two pointers can be taken which mark the beginning and end of the array respectively. If the sum is greater than the sum of those two elements, shift the right pointer to decrease the value of the required sum and if the sum is lesser than the required value, shift the left pointer to increase the value of the required sum. Let’s understand this using an example.

Let an array be {1, 4, 45, 6, 10, -8} and sum to find be 16
After sorting the array
A = {-8, 1, 4, 6, 10, 45}
Now, increment ‘l’ when the sum of the pair is less than the required sum and decrement ‘r’ when the sum of the pair is more than the required sum.
This is because when the sum is less than the required sum then to get the number which could increase the sum of pair, start moving from left to right(also sort the array) thus “l++” and vice versa.
Initialize l = 0, r = 5
A[l] + A[r] ( -8 + 45) > 16 => decrement r. Now r = 4
A[l] + A[r] ( -8 + 10) increment l. Now l = 1
A[l] + A[r] ( 1 + 10) increment l. Now l = 2
A[l] + A[r] ( 4 + 10) increment l. Now l = 3
A[l] + A[r] ( 6 + 10) == 16 => Found candidates (return 1)

Note: If there is more than one pair having the given sum then this algorithm reports only one. Can be easily extended for this though.

Algorithm:

1. hasArrayTwoCandidates (A[], ar_size, sum)
2. Sort the array in non-decreasing order.
3. Initialize two index variables to find the candidate
elements in the sorted array.
1. Initialize first to the leftmost index: l = 0
2. Initialize second the rightmost index: r = ar_size-1
4. Loop while l < r.
1. If (A[l] + A[r] == sum) then return 1
2. Else if( A[l] + A[r] < sum ) then l++
3. Else r–
5. No candidates in the whole array – return 0

Below is the implementation of the above approach:

## C++

 `// C++ program to check if given array``// has 2 elements whose sum is equal``// to the given value` `#include ``using` `namespace` `std;` `// Function to check if array has 2 elements``// whose sum is equal to the given value``bool` `hasArrayTwoCandidates(``int` `A[], ``int` `arr_size,``                           ``int` `sum)``{``    ``int` `l, r;` `    ``/* Sort the elements */``    ``sort(A, A + arr_size);` `    ``/* Now look for the two candidates in``       ``the sorted array*/``    ``l = 0;``    ``r = arr_size - 1;``    ``while` `(l < r) {``        ``if` `(A[l] + A[r] == sum)``            ``return` `1;``        ``else` `if` `(A[l] + A[r] < sum)``            ``l++;``        ``else` `// A[i] + A[j] > sum``            ``r--;``    ``}``    ``return` `0;``}` `/* Driver program to test above function */``int` `main()``{``    ``int` `A[] = { 1, 4, 45, 6, 10, -8 };``    ``int` `n = 16;``    ``int` `arr_size = ``sizeof``(A) / ``sizeof``(A[0]);` `    ``// Function calling``    ``if` `(hasArrayTwoCandidates(A, arr_size, n))``        ``cout << ``"Array has two elements"``                ``" with given sum"``;``    ``else``        ``cout << ``"Array doesn't have two"``                ``" elements with given sum"``;` `    ``return` `0;``}`

## C

 `// C program to check if given array``// has 2 elements whose sum is equal``// to the given value` `#include ``#define bool int` `void` `quickSort(``int``*, ``int``, ``int``);` `bool` `hasArrayTwoCandidates(``    ``int` `A[], ``int` `arr_size, ``int` `sum)``{``    ``int` `l, r;` `    ``/* Sort the elements */``    ``quickSort(A, 0, arr_size - 1);` `    ``/* Now look for the two candidates in the sorted``       ``array*/``    ``l = 0;``    ``r = arr_size - 1;``    ``while` `(l < r) {``        ``if` `(A[l] + A[r] == sum)``            ``return` `1;``        ``else` `if` `(A[l] + A[r] < sum)``            ``l++;``        ``else` `// A[i] + A[j] > sum``            ``r--;``    ``}``    ``return` `0;``}` `/* FOLLOWING FUNCTIONS ARE ONLY FOR SORTING``    ``PURPOSE */``void` `exchange(``int``* a, ``int``* b)``{``    ``int` `temp;``    ``temp = *a;``    ``*a = *b;``    ``*b = temp;``}` `int` `partition(``int` `A[], ``int` `si, ``int` `ei)``{``    ``int` `x = A[ei];``    ``int` `i = (si - 1);``    ``int` `j;` `    ``for` `(j = si; j <= ei - 1; j++) {``        ``if` `(A[j] <= x) {``            ``i++;``            ``exchange(&A[i], &A[j]);``        ``}``    ``}``    ``exchange(&A[i + 1], &A[ei]);``    ``return` `(i + 1);``}` `/* Implementation of Quick Sort``A[] --> Array to be sorted``si  --> Starting index``ei  --> Ending index``*/``void` `quickSort(``int` `A[], ``int` `si, ``int` `ei)``{``    ``int` `pi; ``/* Partitioning index */``    ``if` `(si < ei) {``        ``pi = partition(A, si, ei);``        ``quickSort(A, si, pi - 1);``        ``quickSort(A, pi + 1, ei);``    ``}``}` `/* Driver program to test above function */``int` `main()``{``    ``int` `A[] = { 1, 4, 45, 6, 10, -8 };``    ``int` `n = 16;``    ``int` `arr_size = 6;` `    ``if` `(hasArrayTwoCandidates(A, arr_size, n))``        ``printf``(``"Array has two elements with given sum"``);``    ``else``        ``printf``(``"Array doesn't have two elements with given sum"``);` `    ``getchar``();``    ``return` `0;``}`

## Java

 `// Java program to check if given array``// has 2 elements whose sum is equal``// to the given value``import` `java.util.*;` `class` `GFG {``    ``// Function to check if array has 2 elements``    ``// whose sum is equal to the given value``    ``static` `boolean` `hasArrayTwoCandidates(``        ``int` `A[],``        ``int` `arr_size, ``int` `sum)``    ``{``        ``int` `l, r;` `        ``/* Sort the elements */``        ``Arrays.sort(A);` `        ``/* Now look for the two candidates``        ``in the sorted array*/``        ``l = ``0``;``        ``r = arr_size - ``1``;``        ``while` `(l < r) {``            ``if` `(A[l] + A[r] == sum)``                ``return` `true``;``            ``else` `if` `(A[l] + A[r] < sum)``                ``l++;``            ``else` `// A[i] + A[j] > sum``                ``r--;``        ``}``        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `A[] = { ``1``, ``4``, ``45``, ``6``, ``10``, -``8` `};``        ``int` `n = ``16``;``        ``int` `arr_size = A.length;` `        ``// Function calling``        ``if` `(hasArrayTwoCandidates(A, arr_size, n))``            ``System.out.println(``"Array has two "``                               ``+ ``"elements with given sum"``);``        ``else``            ``System.out.println(``"Array doesn't have "``                               ``+ ``"two elements with given sum"``);``    ``}``}`

## Python

 `# Python program to check for the sum``# condition to be satisfied` `def` `hasArrayTwoCandidates(A, arr_size, ``sum``):``    ` `    ``# sort the array``    ``quickSort(A, ``0``, arr_size``-``1``)``    ``l ``=` `0``    ``r ``=` `arr_size``-``1``    ` `    ``# traverse the array for the two elements``    ``while` `l Array to be sorted``# si  --> Starting index``# ei  --> Ending index``def` `quickSort(A, si, ei):``    ``if` `si < ei:``        ``pi ``=` `partition(A, si, ei)``        ``quickSort(A, si, pi``-``1``)``        ``quickSort(A, pi ``+` `1``, ei)` `# Utility function for partitioning``# the array(used in quick sort)``def` `partition(A, si, ei):``    ``x ``=` `A[ei]``    ``i ``=` `(si``-``1``)``    ``for` `j ``in` `range``(si, ei):``        ``if` `A[j] <``=` `x:``            ``i ``+``=` `1``            ` `            ``# This operation is used to swap``            ``# two variables is python``            ``A[i], A[j] ``=` `A[j], A[i]` `        ``A[i ``+` `1``], A[ei] ``=` `A[ei], A[i ``+` `1``]``        ` `    ``return` `i ``+` `1``    `  `# Driver program to test the functions``A ``=` `[``1``, ``4``, ``45``, ``6``, ``10``, ``-``8``]``n ``=` `16``if` `(hasArrayTwoCandidates(A, ``len``(A), n)):``    ``print``(``"Array has two elements with the given sum"``)``else``:``    ``print``("Array doesn't have two elements``                                  ``with the given ``sum``")` `## This code is contributed by __Devesh Agrawal__`

## C#

 `// C# program to check for pair``// in A[] with sum as x` `using` `System;` `class` `GFG {``    ``static` `bool` `hasArrayTwoCandidates(``int``[] A,``                       ``int` `arr_size, ``int` `sum)``    ``{``        ``int` `l, r;` `        ``/* Sort the elements */``        ``sort(A, 0, arr_size - 1);` `        ``/* Now look for the two candidates``        ``in the sorted array*/``        ``l = 0;``        ``r = arr_size - 1;``        ``while` `(l < r) {``            ``if` `(A[l] + A[r] == sum)``                ``return` `true``;``            ``else` `if` `(A[l] + A[r] < sum)``                ``l++;``            ``else` `// A[i] + A[j] > sum``                ``r--;``        ``}``        ``return` `false``;``    ``}` `    ``/* Below functions are only to sort the``    ``array using QuickSort */` `    ``/* This function takes last element as pivot,``    ``places the pivot element at its correct``    ``position in sorted array, and places all``    ``smaller (smaller than pivot) to left of``    ``pivot and all greater elements to right``    ``of pivot */``    ``static` `int` `partition(``int``[] arr, ``int` `low, ``int` `high)``    ``{``        ``int` `pivot = arr[high];` `        ``// index of smaller element``        ``int` `i = (low - 1);``        ``for` `(``int` `j = low; j <= high - 1; j++) {``            ``// If current element is smaller``            ``// than or equal to pivot``            ``if` `(arr[j] <= pivot) {``                ``i++;` `                ``// swap arr[i] and arr[j]``                ``int` `temp = arr[i];``                ``arr[i] = arr[j];``                ``arr[j] = temp;``            ``}``        ``}` `        ``// swap arr[i+1] and arr[high] (or pivot)``        ``int` `temp1 = arr[i + 1];``        ``arr[i + 1] = arr[high];``        ``arr[high] = temp1;` `        ``return` `i + 1;``    ``}` `    ``/* The main function that``    ``implements QuickSort()``    ``arr[] --> Array to be sorted,``    ``low --> Starting index,``    ``high --> Ending index */``    ``static` `void` `sort(``int``[] arr, ``int` `low, ``int` `high)``    ``{``        ``if` `(low < high) {``            ``/* pi is partitioning index, arr[pi]``            ``is now at right place */``            ``int` `pi = partition(arr, low, high);` `            ``// Recursively sort elements before``            ``// partition and after partition``            ``sort(arr, low, pi - 1);``            ``sort(arr, pi + 1, high);``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] A = { 1, 4, 45, 6, 10, -8 };``        ``int` `n = 16;``        ``int` `arr_size = 6;` `        ``if` `(hasArrayTwoCandidates(A, arr_size, n))``            ``Console.Write(``"Array has two elements"``                          ``+ ``" with given sum"``);``        ``else``            ``Console.Write(``"Array doesn't have "``                          ``+ ``"two elements with given sum"``);``    ``}``}` `// This code is contributed by Sam007`

## PHP

 ` sum``            ``\$r``--;``    ``}``    ``return` `0;``}` `// Driver Code``\$A` `= ``array` `(1, 4, 45, 6, 10, -8);``\$n` `= 16;``\$arr_size` `= sizeof(``\$A``);` `// Function calling``if``(hasArrayTwoCandidates(``\$A``, ``\$arr_size``, ``\$n``))``    ``echo` `"Array has two elements "` `.``                   ``"with given sum"``;``else``    ``echo` `"Array doesn't have two "` `.``          ``"elements with given sum"``;``    ` `// This code is contributed by m_kit``?>`

## Javascript

 ``

Output

`Array has two elements with given sum`

Complexity Analysis:

• Time Complexity: Depends on what sorting algorithm we use.
• If Merge Sort or Heap Sort is used then (-)(nlogn) in the worst case.
• If Quick Sort is used then O(n^2) in the worst case.
• Auxiliary Space: This too depends on sorting algorithm. The auxiliary space is O(n) for merge sort and O(1) for Heap Sort.

Method 2: Hashing.

Approach: This problem can be solved efficiently by using the technique of hashing. Use a hash_map to check for the current array value x(let), if there exists a value target_sum-x which on adding to the former gives target_sum. This can be done in constant time. Let’s look at the following example.

arr[] = {0, -1, 2, -3, 1}
sum = -2
Now start traversing:
Step 1: For ‘0’ there is no valid number ‘-2’ so store ‘0’ in hash_map.
Step 2: For ‘-1’ there is no valid number ‘-1’ so store ‘-1’ in hash_map.
Step 3: For ‘2’ there is no valid number ‘-4’ so store ‘2’ in hash_map.
Step 4: For ‘-3’ there is no valid number ‘1’ so store ‘-3’ in hash_map.
Step 5: For ‘1’ there is a valid number ‘-3’ so answer is 1, -3

Algorithm:

1. Initialize an empty hash table s.
2. Do following for each element A[i] in A[]
1. If s[x – A[i]] is set then print the pair (A[i], x – A[i])
2. Insert A[i] into s.

Pseudo Code :

```unordered_set s
for(i=0 to end)
if(s.find(target_sum - arr[i]) == s.end)
insert(arr[i] into s)
else
print arr[i], target-arr[i]```

Implementation:

## C++

 `// C++ program to check if given array``// has 2 elements whose sum is equal``// to the given value``#include ` `using` `namespace` `std;` `void` `printPairs(``int` `arr[], ``int` `arr_size, ``int` `sum)``{``    ``unordered_set<``int``> s;``    ``for` `(``int` `i = 0; i < arr_size; i++)``    ``{``        ``int` `temp = sum - arr[i];` `        ``if` `(s.find(temp) != s.end())``            ``cout << ``"Pair with given sum "``                 ``<< sum << ``" is ("` `<< arr[i] << ``","``                    ``<< temp << ``")"` `<< endl;` `        ``s.insert(arr[i]);``    ``}``}` `/* Driver Code */``int` `main()``{``    ``int` `A[] = { 1, 4, 45, 6, 10, 8 };``    ``int` `n = 16;``    ``int` `arr_size = ``sizeof``(A) / ``sizeof``(A[0]);` `    ``// Function calling``    ``printPairs(A, arr_size, n);` `    ``return` `0;``}`

## C

 `// C program to check if given array``// has 2 elements whose sum is equal``// to the given value` `#include ``#define MAX 100000` `// NOTE: Works only if range elements is limited``// target - arr[i] >= 0 && target - arr[i] < MAX`   `void` `printPairs(``int` `arr[], ``int` `arr_size, ``int` `target)``{``    ``int` `i, temp;` `    ``/*initialize hash set as 0*/``    ``int` `s[MAX] = { 0 };` `    ``for` `(i = 0; i < arr_size; i++)``    ``{``        ``temp = target - arr[i];``        ``if` `(s[temp] == 1)``            ``printf``(``                ``"Pair with given sum %d is (%d, %d) \n"``,``                ``target, arr[i], temp);``        ``s[arr[i]] = 1;``    ``}``}` `/* Driver Code */``int` `main()``{``    ``int` `A[] = { 1, 4, 45, 6, 10, 8 };``    ``int` `target = 16;``    ``int` `arr_size = ``sizeof``(A) / ``sizeof``(A[0]);` `    ``printPairs(A, arr_size, target);` `    ``getchar``();``    ``return` `0;``}`

## Java

 `// Java implementation using Hashing``import` `java.io.*;``import` `java.util.HashSet;` `class` `PairSum {``    ``static` `void` `printpairs(``int` `arr[], ``int` `sum)``    ``{``        ``HashSet s = ``new` `HashSet();``        ``for` `(``int` `i = ``0``; i < arr.length; ++i)``        ``{``            ``int` `temp = sum - arr[i];` `            ``// checking for condition``            ``if` `(s.contains(temp)) {``                ``System.out.println(``                    ``"Pair with given sum "``                    ``+ sum + ``" is ("` `+ arr[i]``                    ``+ ``", "` `+ temp + ``")"``);``            ``}``            ``s.add(arr[i]);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `A[] = { ``1``, ``4``, ``45``, ``6``, ``10``, ``8` `};``        ``int` `n = ``16``;``        ``printpairs(A, n);``    ``}``}` `// This article is contributed by Aakash Hasija`

## Python3

 `# Python program to find if there are``# two elements with given sum` `# function to check for the given sum``# in the array``def` `printPairs(arr, arr_size, ``sum``):``    ` `    ``# Create an empty hash map``    ``# using an hashmap allows us to store the indices``    ``hashmap ``=` `{}``    ` `    ``for` `i ``in` `range``(``0``, arr_size):``        ``temp ``=` `sum``-``arr[i]``        ``if` `(temp ``in` `hashmap):``            ``print` `(f``'Pair with given sum {sum} is ({temp},{arr[i]}) at indices ({hashmap[temp]},{i})'``)``        ``hashmap[arr[i]] ``=` `i` `# driver code``A ``=` `[``1``, ``4``, ``45``, ``6``, ``10``, ``8``]``n ``=` `16``printPairs(A, ``len``(A), n)` `# This code will also work in case the array has the same number twice``# and target is the sum of those numbers``# Eg: Array = [4,6,4] Target = 8` `# This code is contributed by __Achyut Upadhyay__`

## C#

 `// C# implementation using Hashing``using` `System;``using` `System.Collections.Generic;` `class` `GFG {``    ``static` `void` `printpairs(``int``[] arr,``                           ``int` `sum)``    ``{``        ``HashSet<``int``> s = ``new` `HashSet<``int``>();``        ``for` `(``int` `i = 0; i < arr.Length; ++i)``        ``{``            ``int` `temp = sum - arr[i];` `            ``// checking for condition``            ``if` `(s.Contains(temp)) {``                ``Console.Write(``"Pair with given sum "` `+``           ``sum + ``" is ("` `+ arr[i] + ``", "` `+ temp + ``")"``);``            ``}``            ``s.Add(arr[i]);``        ``}``    ``}` `    ``// Driver Code``    ``static` `void` `Main()``    ``{``        ``int``[] A = ``new` `int``[] { 1, 4, 45,``                              ``6, 10, 8 };``        ``int` `n = 16;``        ``printpairs(A, n);``    ``}``}` `// This code is contributed by``// Manish Shaw(manishshaw1)`

## Javascript

 ``

Output

`Pair with given sum 16 is (10,6)`

Complexity Analysis:

• Time Complexity: O(n).
As the whole array is needed to be traversed only once.
• Auxiliary Space: O(n).
A hash map has been used to store array elements.

Note: The solution will work even if the range of numbers includes negative numbers + if the pair is formed by numbers recurring twice in array eg: array = [3,4,3]; pair = (3,3); target sum = 6.

Method 3: Using remainders of the elements less than x.

Approach:
The idea is to count the elements with remainders when divided by x, i.e 0 to x-1, each remainder separately. Suppose we have x as 6, then the numbers which are less than 6 and have remainders which add up to 6 gives sum as 6 when added. For example, we have elements, 2,4 in the array and 2%6 = 2 and 4%6 =4, and these remainders add up to give 6. Like that we have to check for pairs with remainders (1,5),(2,4),(3,3). if we have one or more elements with remainder 1 and one or more elements with remainder 5, then surely we get a sum as 6. Here we do not consider (0,6) as the elements for the resultant pair should be less than 6. when it comes to (3,3) we have to check if we have two elements with remainder 3, then we can say that “There exists a pair whose sum is x”.

Algorithm:

1. Create an array with size x.

2. Initialize all rem elements to zero.

3. Traverse the given array

• Do the following if arr[i] is less than x:
• r=arr[i]%x which is done to get the remainder.
• rem[r]=rem[r]+1 i.e. increasing the count of elements that have remainder r when divided with x.

4. Now, traverse the rem array from 1 to x/2.

• If(rem[i]> 0 and rem[x-i]>0) then print “YES” and come out of the loop. This means that we have a pair that results in x upon doing.

5. Now when we reach at x/2 in the above loop

• If x is even, for getting a pair we should have two elements with remainder x/2.
• If rem[x/2]>1 then print “YES” else print “NO”
• If it is not satisfied that is x is odd, it will have a separate pair with x-x/2.
• If rem[x/2]>1 and rem[x-x/2]>1 , then print “Yes” else, print”No”;

Implementation:

## C++

 `// Code in cpp to tell if there exists a pair in array whose``// sum results in x.``#include ``using` `namespace` `std;` `// Function to print pairs``void` `printPairs(``int` `a[], ``int` `n, ``int` `x)``{``    ``int` `i;``    ``int` `rem[x];``    ``// initializing the rem values with 0's.``    ``for` `(i = 0; i < x; i++)``        ``rem[i] = 0;``    ``// Perform the remainder operation only if the element``    ``// is x, as numbers greater than x can't be used to get``    ``// a sum x. Updating the count of remainders.``    ``for` `(i = 0; i < n; i++)``        ``if` `(a[i] < x)``            ``rem[a[i] % x]++;` `    ``// Traversing the remainder list from start to middle to``    ``// find pairs``    ``for` `(i = 1; i < x / 2; i++) {``        ``if` `(rem[i] > 0 && rem[x - i] > 0) {``            ``// The elements with remainders i and x-i will``            ``// result to a sum of x. Once we get two``            ``// elements which add up to x , we print x and``            ``// break.``            ``cout << ``"Yes\n"``;``            ``break``;``        ``}``    ``}` `    ``// Once we reach middle of remainder array, we have to``    ``// do operations based on x.``    ``if` `(i >= x / 2) {``        ``if` `(x % 2 == 0) {``            ``// if x is even and we have more than 1 elements``            ``// with remainder x/2, then we will have two``            ``// distinct elements which add up to x. if we``            ``// dont have more than 1 element, print "No".``            ``if` `(rem[x / 2] > 1)``                ``cout << ``"Yes\n"``;``            ``else``                ``cout << ``"No\n"``;``        ``}``        ``else` `{``            ``// When x is odd we continue the same process``            ``// which we did in previous loop.``            ``if` `(rem[x / 2] > 0 && rem[x - x / 2] > 0)``                ``cout << ``"Yes\n"``;``            ``else``                ``cout << ``"No\n"``;``        ``}``    ``}``}` `/* Driver Code */``int` `main()``{``    ``int` `A[] = { 1, 4, 45, 6, 10, 8 };``    ``int` `n = 16;``    ``int` `arr_size = ``sizeof``(A) / ``sizeof``(A[0]);` `    ``// Function calling``    ``printPairs(A, arr_size, n);` `    ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `// Code in c to tell if there exists a pair in array whose``// sum results in x.``#include ` `// Function to print pairs``void` `printPairs(``int` `a[], ``int` `n, ``int` `x)``{``    ``int` `i;``    ``int` `rem[x];``    ``// initializing the rem values with 0's.``    ``for` `(i = 0; i < x; i++)``        ``rem[i] = 0;``    ``// Perform the remainder operation only if the element``    ``// is x, as numbers greater than x can't be used to get``    ``// a sum x. Updating the count of remainders.``    ``for` `(i = 0; i < n; i++)``        ``if` `(a[i] < x)``            ``rem[a[i] % x]++;` `    ``// Traversing the remainder list from start to middle to``    ``// find pairs``    ``for` `(i = 1; i < x / 2; i++) {``        ``if` `(rem[i] > 0 && rem[x - i] > 0) {``            ``// The elements with remainders i and x-i will``            ``// result to a sum of x. Once we get two``            ``// elements which add up to x , we print x and``            ``// break.``            ``printf``(``"Yes\n"``);``            ``break``;``        ``}``    ``}` `    ``// Once we reach middle of remainder array, we have to``    ``// do operations based on x.``    ``if` `(i >= x / 2) {``        ``if` `(x % 2 == 0) {``            ``// if x is even and we have more than 1 elements``            ``// with remainder x/2, then we will have two``            ``// distinct elements which add up to x. if we``            ``// dont have more than 1 element, print "No".``            ``if` `(rem[x / 2] > 1)``                ``printf``(``"Yes\n"``);``            ``else``                ``printf``(``"No\n"``);``        ``}``        ``else` `{``            ``// When x is odd we continue the same process``            ``// which we did in previous loop.``            ``if` `(rem[x / 2] > 0 && rem[x - x / 2] > 0)``                ``printf``(``"Yes\n"``);``            ``else``                ``printf``(``"No\n"``);``        ``}``    ``}``}` `/* Driver Code */``int` `main()``{``    ``int` `A[] = { 1, 4, 45, 6, 10, 8 };``    ``int` `n = 16;``    ``int` `arr_size = ``sizeof``(A) / ``sizeof``(A[0]);` `    ``// Function calling``    ``printPairs(A, arr_size, n);` `    ``return` `0;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// Code in Java to tell if there exists a pair in array``// whose sum results in x.``import` `java.util.*;` `class` `GFG {` `    ``// Function to print pairs``    ``static` `void` `printPairs(``int` `a[], ``int` `n, ``int` `x)``    ``{``        ``int` `i;``        ``int``[] rem = ``new` `int``[x];``        ``// initializing the rem values with 0's.``        ``for` `(i = ``0``; i < x; i++)``            ``rem[i] = ``0``;``        ``// Perform the remainder operation only if``        ``// the element is x, as numbers greater than``        ``// x can't be used to get a sum x. Updating``        ``// the count of remainders.``        ``for` `(i = ``0``; i < n; i++)``            ``if` `(a[i] < x)``                ``rem[a[i] % x]++;` `        ``// Traversing the remainder list from start to``        ``// middle to find pairs``        ``for` `(i = ``1``; i < x / ``2``; i++) {``            ``if` `(rem[i] > ``0` `&& rem[x - i] > ``0``) {``                ``// The elements with remainders i and x-i``                ``// will result to a sum of x. Once we get``                ``// two elements which add up to x , we print``                ``// x and break.``                ``System.out.println(``"Yes"``);``                ``break``;``            ``}``        ``}` `        ``// Once we reach middle of remainder array, we have``        ``// to do operations based on x.``        ``if` `(i >= x / ``2``) {``            ``if` `(x % ``2` `== ``0``) {``                ``// if x is even and we have more than 1``                ``// elements with remainder x/2, then we``                ``// will have two distinct elements which``                ``// add up to x. if we dont have more``                ``// than 1 element, print "No".``                ``if` `(rem[x / ``2``] > ``1``)``                    ``System.out.println(``"Yes"``);``                ``else``                    ``System.out.println(``"No"``);``            ``}``            ``else` `{` `                ``// When x is odd we continue the same``                ``// process which we did in previous loop.``                ``if` `(rem[x / ``2``] > ``0` `&& rem[x - x / ``2``] > ``0``)``                    ``System.out.println(``"Yes"``);``                ``else``                    ``System.out.println(``"No"``);``            ``}``        ``}``    ``}` `    ``/* Driver Code */``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `A[] = { ``1``, ``4``, ``45``, ``6``, ``10``, ``8` `};``        ``int` `n = ``16``;``        ``int` `arr_size = A.length;` `        ``// Function calling``        ``printPairs(A, arr_size, n);``    ``}``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## Python3

 `# Code in Python3 to tell if there``# exists a pair in array whose``# sum results in x.` `# Function to print pairs``def` `printPairs(a, n, x):``    ` `    ``rem ``=` `[]``    ` `    ``for` `i ``in` `range``(x):` `        ``# Initializing the rem``        ``# values with 0's.``        ``rem.append(``0``)` `    ``for` `i ``in` `range``(n):``        ``if` `(a[i] < x):` `            ``# Perform the remainder operation``            ``# only if the element is x, as``            ``# numbers greater than x can't``            ``# be used to get a sum x.Updating``            ``# the count of remainders.``            ``rem[a[i] ``%` `x] ``+``=` `1` `    ``# Traversing the remainder list from``    ``# start to middle to find pairs``    ``for` `i ``in` `range``(``1``, x ``/``/` `2``):``        ``if` `(rem[i] > ``0` `and` `rem[x ``-` `i] > ``0``):` `            ``# The elements with remainders``            ``# i and x-i will result to a``            ``# sum of x. Once we get two``            ``# elements which add up to x,``            ``# we print x and break.``            ``print``(``"Yes"``)``            ``break` `    ``# Once we reach middle of``    ``# remainder array, we have to``    ``# do operations based on x.``    ``if` `(i >``=` `x ``/``/` `2``):``        ``if` `(x ``%` `2` `=``=` `0``):``            ``if` `(rem[x ``/``/` `2``] > ``1``):` `                ``# If x is even and we have more``                ``# than 1 elements with remainder``                ``# x/2, then we will have two``                ``# distinct elements which add up``                ``# to x. if we dont have than 1``                ``# element, print "No".``                ``print``(``"Yes"``)``            ``else``:``                ``print``(``"No"``)``        ``else``:` `            ``# When x is odd we continue``            ``# the same process which we``            ``# did in previous loop.``            ``if` `(rem[x ``/``/` `2``] > ``0` `and``                ``rem[x ``-` `x ``/``/` `2``] > ``0``):``                ``print``(``"Yes"``)``            ``else``:``                ``print``(``"No"``)` `# Driver Code``A ``=` `[ ``1``, ``4``, ``45``, ``6``, ``10``, ``8` `]``n ``=` `16``arr_size ``=` `len``(A)` `# Function calling``printPairs(A, arr_size, n)` `# This code is contributed by subhammahato348`

## C#

 `// C# Code in C# to tell if there``// exists a pair in array whose``// sum results in x.``using` `System;``class` `GFG``{``  ` `// Function to print pairs``static` `void` `printPairs(``int` `[]a, ``int` `n, ``int` `x)``{``  ``int` `i;``  ``int` `[]rem = ``new` `int``[x];``  ``for` `(i = 0; i < x; i++)``  ``{``    ` `    ``// initializing the rem``    ``// values with 0's.``    ``rem[i] = 0;``  ``}``  ``for` `(i = 0; i < n; i++)``  ``{``    ``if` `(a[i] < x)``    ``{` `      ``// Perform the remainder``      ``// operation only if the``      ``// element is x, as numbers``      ``// greater than x can't``      ``// be used to get a sum x.``      ``// Updating the count of remainders.``      ``rem[a[i] % x]++;``    ``}``  ``}` `  ``// Traversing the remainder list``  ``// from start to middle to``  ``// find pairs``  ``for` `(i = 1; i < x / 2; i++)``  ``{``    ``if` `(rem[i] > 0 && rem[x - i] > 0)``    ``{` `      ``// The elements with remainders``      ``// i and x-i will``      ``// result to a sum of x.``      ``// Once we get two``      ``// elements which add up to x ,``      ``// we print x and``      ``// break.``      ``Console.Write(``"Yes"` `+ ``"\n"``);``      ``break``;``    ``}``  ``}` `  ``// Once we reach middle of``  ``// remainder array, we have to``  ``// do operations based on x.``  ``if` `(i >= x / 2)``  ``{``    ``if` `(x % 2 == 0)``    ``{``      ``if` `(rem[x / 2] > 1)``      ``{` `        ``// if x is even and``        ``// we have more than 1``        ``// elements with remainder``        ``// x/2, then we will``        ``// have two distinct elements``        ``// which add up``        ``// to x. if we dont have``        ``//more than 1``        ``// element, print "No".``        ``Console.Write(``"Yes"` `+ ``"\n"``);``      ``}``      ``else``      ``{``        ``Console.Write(``"No"``                         ``+ ``"\n"``);``      ``}``    ``}``    ``else``    ``{` `      ``// When x is odd we continue``      ``// the same process``      ``// which we did in previous loop.``      ``if` `(rem[x / 2] > 0 &&``          ``rem[x - x / 2] > 0)``      ``{``        ``Console.Write(``"Yes"``                         ``+ ``"\n"``);``      ``}``      ``else``      ``{``        ``Console.WriteLine(``"No"``                         ``+ ``"\n"``);``      ``}``    ``}``  ``}``}` `/* Driver Code */``public` `static` `void` `Main(``string``[] args)``{``    ``int``[] A = { 1, 4, 45, 6, 10, 8 };``    ``int` `n = 16;``    ``int` `arr_size = A.Length;` `    ``// Function calling``    ``printPairs(A, arr_size, n);``}``}` `// This code is contributed by SoumikMondal`

## Javascript

 ``

Output

`Yes`

Time Complexity: O(n+x)
Auxiliary Space: O(x)

Similarly, the indices of a pair that add up to a given sum can also be calculated by an unordered map. The only change here is that we also have to store indices of elements as values for each element as key.
Implementation:

## C++14

 `#include ``using` `namespace` `std;` `pair<``int``,``int``> findSum(``int` `*arr,``int``& n,``int``& target)``{``    ``int` `i,findElement;``    ``unordered_map<``int``,``int``>mp;``    ``pair<``int``,``int``>result;``    ``for``(i=0;ians=findSum(arr,n,search);``    ``cout<

## Java

 `// Java Code for above Approach to tell if there exists a pair in array``// whose sum results in x.``import` `java.io.*;``import` `java.util.*; ` `class` `GFG``{` `  ``public` `static` `int``[] findSum(``int` `arr[], ``int` `n, ``int` `target)``  ``{``    ``int` `i, findElement;``    ``Map mp=``new` `HashMap();   ``    ``int` `result[] = {``0``,``0``};``    ``for``(i = ``0``; i < n; i++)``    ``{``      ``findElement = target-arr[i];``      ``if``(mp.containsKey(findElement))``      ``{``        ``result[``0``] = i - ``1``;``        ``result[``1``] = mp.get(findElement) - ``1``;``        ``break``;``      ``}``      ``else` `mp.put(arr[i],i);``    ``}``    ``return` `result;``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `A[] = {``1``,``5``,``4``,``3``,``7``,``9``,``2``};``    ``int` `n = A.length;``    ``int` `search = ``7``;``    ``int` `ans[] = findSum(A,n,search);``    ``System.out.println(Math.min(ans[``0``], ans[``1``]) + ``" "` `+Math.max(ans[``0``], ans[``1``]));``  ``}``}` `// This code is contributed by kothavvsaakash`

## Python3

 `def` `findSum(arr, n, target):` `    ``mp ``=` `{}``    ``result ``=` `[``0``]``*``2``    ``for` `i ``in` `range``(n):` `        ``findElement ``=` `target``-``arr[i]``        ``if``(findElement ``in` `mp):``            ``result[``0``] ``=` `i ``-` `1``            ``result[``1``] ``=` `mp[findElement] ``-` `1``            ``break` `        ``else``:``            ``mp[arr[i]] ``=` `i``    ``return` `result` `# driver code``arr ``=` `[``1``,``5``,``4``,``3``,``7``,``9``,``2``]``n ``=` `len``(arr)``search ``=` `7``ans ``=` `findSum(arr,n,search)``print``(f``"{min(ans[0], ans[1])} {max(ans[0], ans[1])}"``)` `# This code is contributed by shinjanpatra`

## C#

 `// C# Code for above Approach to tell if there exists a pair``// in array whose sum results in x.` `using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG {``  ``// function to find the pair whose sum is target``  ``public` `static` `int``[] findSum(``int``[] arr, ``int` `n,``                              ``int` `target)``  ``{``    ``int` `i, findElement;` `    ``// creating a dictionary``    ``IDictionary<``int``, ``int``> mp``      ``= ``new` `Dictionary<``int``, ``int``>();` `    ``// the default result pair``    ``int``[] result = { 0, 0 };` `    ``// building the dictionary``    ``for` `(i = 0; i < n; i++) {``      ``findElement = target - arr[i];``      ``if` `(mp.ContainsKey(findElement)) {``        ``result[0] = i - 1;``        ``result[1] = mp[findElement] - 1;``        ``break``;``      ``}``      ``else``        ``mp[arr[i]] = i;``    ``}``    ``return` `result;``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(``string``[] args)``  ``{``    ``{``      ``int``[] A = { 1, 5, 4, 3, 7, 9, 2 };``      ``int` `n = A.Length;``      ``int` `search = 7;` `      ``// Function call``      ``int``[] ans = findSum(A, n, search);``      ``Console.WriteLine(Math.Min(ans[0], ans[1]) + ``" "``                        ``+ Math.Max(ans[0], ans[1]));``    ``}``  ``}``}`  `//this code is contributed by phasing17`

## Javascript

 ``

Output

`1 2`

Time Complexity: O(n)
Auxiliary Space: O(n)

Related Problems:

Please write comments if you find any of the above codes/algorithms incorrect, or find other ways to solve the same problem.

My Personal Notes arrow_drop_up