# Given an array A[] and a number x, check for pair in A[] with sum as x

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

```Input: arr[] = {0, -1, 2, -3, 1}
sum = -2
Output: -3, 1
If we calculate the sum of the output,
1 + (-3) = -2

Input: arr[] = {1, -2, 1, 0, 5}
sum = 0
Output: -1
No valid pair exists.
```

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

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 left pointer to increase the value of required sum and if the sum is lesser than the required value, shift the right pointer to decrease the value. 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 whole array – return 0

## 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); ` ` `  `    ``// 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 satisified ` ` `  `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 ` `?> `

Output:

`Array has two elements with the given sum`

Complexity Analysis:

• Time Complexity: Depends on what sorting algorithm we use.
• If Merge Sort or Heap Sort is used then (-)(nlogn) in worst case.
• If Quick Sort is used then O(n^2) in 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]
```

## 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 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); ` ` `  `    ``// 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 ` ` `  `// Works only if range elements is limited ` `#include ` `#define MAX 100000 ` ` `  `void` `printPairs(``int` `arr[], ``int` `arr_size, ``int` `sum) ` `{ ` `    ``int` `i, temp; ` ` `  `    ``/*initialize hash set as 0*/` `    ``bool` `s[MAX] = { 0 }; ` ` `  `    ``for` `(i = 0; i < arr_size; i++) { ` `        ``temp = sum - arr[i]; ` `        ``if` `(s[temp] == 1) ` `            ``printf``( ` `                ``"Pair with given sum %d is (%d, %d) n"``, ` `                ``sum, arr[i], temp); ` `        ``s[arr[i]] = 1; ` `    ``} ` `} ` ` `  `/* 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); ` ` `  `    ``printPairs(A, arr_size, n); ` ` `  `    ``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]); ` `        ``} ` `    ``} ` ` `  `    ``// Main to test the above function ` `    ``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 `

## Python

 `# Python program to find if there are ` `# two elements wtih given sum ` ` `  `# function to check for the given sum ` `# in the array ` `def` `printPairs(arr, arr_size, ``sum``): ` `     `  `    ``# Create an empty hash set ` `    ``s ``=` `set``() ` `     `  `    ``for` `i ``in` `range``(``0``, arr_size): ` `        ``temp ``=` `sum``-``arr[i] ` `        ``if` `(temp ``in` `s): ` `            ``print` `"Pair with given sum "``+` `str``(``sum``) ``+` `" is ("` `+` `str``(arr[i]) ``+` `", "` `+` `str``(temp) ``+` `")"` `        ``s.add(arr[i]) ` ` `  `# driver program to check the above function ` `A ``=` `[``1``, ``4``, ``45``, ``6``, ``10``, ``8``] ` `n ``=` `16` `printPairs(A, ``len``(A), n) ` ` `  `# This code is contributed by __Devesh Agrawal__ `

## 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) `

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).
As a hash map has been used to store array elements.

Note: If the range of numbers includes negative numbers then also it will work fine.

Related Problems:

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

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