# Find pairs in array whose sums already exist in array

Given an array of n distinct and positive elements, the task is to find pair whose sum already exists in given array.

Examples :

```Input : arr[] = {2, 8, 7, 1, 5};
Output : 2 5
7 1

Input : arr[] = {7, 8, 5, 9, 11};
Output : Not Exist
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A Naive Approach is to run three loops to find pair whose sum exists in array.

## C++

 `// A simple C++ program to find pair whose sum ` `// already exists in array ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find pair whose sum exists in arr[] ` `void` `findPair(``int` `arr[], ``int` `n) ` `{ ` `    ``bool` `found = ``false``; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` `            ``for` `(``int` `k = 0; k < n; k++) { ` `                ``if` `(arr[i] + arr[j] == arr[k]) { ` `                    ``cout << arr[i] << ``" "` `<< arr[j] << endl; ` `                    ``found = ``true``; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``if` `(found == ``false``) ` `        ``cout << ``"Not exist"` `<< endl; ` `} ` ` `  `// Driven code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 4, 8, 13, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``findPair(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// A simple Java program to find  ` `// pair whose sum already exists ` `// in array ` `import` `java.io.*; ` ` `  `public` `class` `GFG { ` ` `  `    ``// Function to find pair whose ` `    ``// sum exists in arr[] ` `    ``static` `void` `findPair(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``boolean` `found = ``false``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``for` `(``int` `j = i + ``1``; j < n; j++) { ` `                ``for` `(``int` `k = ``0``; k < n; k++) { ` `                    ``if` `(arr[i] + arr[j] == arr[k]) { ` `                        ``System.out.println(arr[i] +  ` `                                      ``" "` `+ arr[j]); ` `                        ``found = ``true``; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``if` `(found == ``false``) ` `            ``System.out.println(``"Not exist"``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``static` `public` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] arr = {``10``, ``4``, ``8``, ``13``, ``5``}; ` `        ``int` `n = arr.length; ` `        ``findPair(arr, n); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## Python3

 `# A simple python program to find pair  ` `# whose sum already exists in array ` ` `  `# Function to find pair whose sum ` `# exists in arr[] ` `def` `findPair(arr, n): ` `    ``found ``=` `False` `    ``for` `i ``in` `range``(``0``, n): ` `        ``for` `j ``in` `range``(i ``+` `1``, n): ` `            ``for` `k ``in` `range``(``0``, n): ` `                ``if` `(arr[i] ``+` `arr[j] ``=``=` `arr[k]): ` `                    ``print``(arr[i], arr[j]) ` `                    ``found ``=` `True` ` `  `    ``if` `(found ``=``=` `False``): ` `        ``print``(``"Not exist"``) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[ ``10``, ``4``, ``8``, ``13``, ``5` `] ` `    ``n ``=` `len``(arr) ` `    ``findPair(arr, n) ` `     `  `# This code contributed by 29AjayKumar `

## C#

 `// A simple C# program to find  ` `// pair whose sum already exists ` `// in array ` `using` `System; ` ` `  `public` `class` `GFG { ` ` `  `    ``// Function to find pair whose ` `    ``// sum exists in arr[] ` `    ``static` `void` `findPair(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``bool` `found = ``false``; ` `        ``for` `(``int` `i = 0; i < n; i++) { ` `            ``for` `(``int` `j = i + 1; j < n; j++) { ` `                ``for` `(``int` `k = 0; k < n; k++) { ` `                    ``if` `(arr[i] + arr[j] == arr[k]) { ` `                        ``Console.WriteLine(arr[i] +  ` `                                      ``" "` `+ arr[j]); ` `                        ``found = ``true``; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``if` `(found == ``false``) ` `            ``Console.WriteLine(``"Not exist"``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``static` `public` `void` `Main(String []args) ` `    ``{ ` `        ``int``[] arr = {10, 4, 8, 13, 5}; ` `        ``int` `n = arr.Length; ` `        ``findPair(arr, n); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output :

`8 5`

An Efficient solution is to store all element in an hash table (unordered_set in C++) and check one by one all pairs and check its sum exists in set or not. If it exists in set then print pair. If no pair found in array then print not exists .

## C++

 `// C++ program to find pair whose sum already ` `// exists in array ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find pair whose sum exists in arr[] ` `void` `findPair(``int` `arr[], ``int` `n) ` `{ ` `    ``// Hash to store all element of array ` `    ``unordered_set<``int``> s; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``s.insert(arr[i]); ` ` `  `    ``bool` `found = ``false``; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` `            ``// Check sum already exists or not ` `            ``if` `(s.find(arr[i] + arr[j]) != s.end()) { ` `                ``cout << arr[i] << ``" "` `<< arr[j] << endl; ` `                ``found = ``true``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``if` `(found == ``false``) ` `        ``cout << ``"Not exist"` `<< endl; ` `} ` ` `  `// Driven code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 4, 8, 13, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``findPair(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find pair whose sum ` `// already exists in array ` `import` `java.util.*; ` `import` `java.lang.*; ` `import` `java.io.*; ` ` `  `class` `Getpairs { ` `    ``// Function to find pair whose sum ` `    ``// exists in arr[] ` `    ``public` `static` `void` `findPair(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``/* store all the array elements as a  ` `        ``Hash value*/` `        ``HashSet s = ``new` `HashSet(); ` ` `  `        ``for` `(Integer i : arr) { ` `            ``s.add(i); ` `        ``} ` ` `  `        ``/* Run two loop and check for the sum  ` `    ``in the Hashset*/` `        ``/* if not a single pair exist then found  ` `    ``will be false else true*/` `        ``boolean` `found = ``false``; ` ` `  `        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) { ` `            ``for` `(``int` `j = i + ``1``; j < n; j++) { ` `                ``int` `sum = arr[i] + arr[j]; ` `                ``if` `(s.contains(sum)) { ` `                    ``/* if the sum is present in  ` `                 ``hashset then found become ` `                ``true*/` `                    ``found = ``true``; ` ` `  `                    ``System.out.println(arr[i] + ``" "` `                                       ``+ arr[j]); ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``if` `(found == ``false``) ` `            ``System.out.println(``"Not Exist "``); ` `    ``} ` ` `  `    ``// driver function ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int``[] arr = { ``10``, ``4``, ``8``, ``13``, ``5` `}; ` `        ``int` `n = arr.length; ` `        ``findPair(arr, n); ` `    ``} ` `} ` ` `  `// This code is contributed by Smarak Chopdar `

## Python3

 `# Python3 program to find pair whose  ` `# sum already exist in arrar ` ` `  `# Function to find pair whose  ` `# sum sxists in arr[] ` `def` `findPair(arr, n): ` `     `  `    ``# hash to store all element of array ` `    ``s ``=` `{i : ``1` `for` `i ``in` `arr} ` `     `  `    ``found ``=` `False` `     `  `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(i ``+` `1``, n): ` `             `  `            ``# check if sum already exists or not ` `            ``if` `arr[i] ``+` `arr[j] ``in` `s.keys(): ` `                ``print``(arr[i], arr[j]) ` `                ``found ``=` `True` `    ``if` `found ``=``=` `False``: ` `        ``print``(``"Not exist"``) ` `         `  `# Driver code ` `arr ``=` `[``10``, ``4``, ``8``, ``13``, ``5``] ` ` `  `n ``=` `len``(arr) ` ` `  `findPair(arr, n) ` `     `  `# This code is contributed  ` `# by Mohit Kumar `

## C#

 `// C# program to find pair whose sum ` `// already exists in array ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `Getpairs ` `{ ` `    ``// Function to find pair whose sum ` `    ``// exists in arr[] ` `    ``public` `static` `void` `findPair(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``/* store all the array elements as a  ` `        ``Hash value*/` `        ``HashSet<``int``> s = ``new` `HashSet<``int``>(); ` ` `  `        ``foreach` `(``int` `i ``in` `arr) ` `        ``{ ` `            ``s.Add(i); ` `        ``} ` ` `  `        ``/* Run two loop and check for the sum  ` `    ``in the Hashset*/` `        ``/* if not a single pair exist then found  ` `    ``will be false else true*/` `        ``bool` `found = ``false``; ` ` `  `        ``for` `(``int` `i = 0; i < n - 1; i++) ` `        ``{ ` `            ``for` `(``int` `j = i + 1; j < n; j++)  ` `            ``{ ` `                ``int` `sum = arr[i] + arr[j]; ` `                ``if` `(s.Contains(sum)) ` `                ``{ ` `                    ``/* if the sum is present in  ` `                    ``hashset then found become ` `                    ``true*/` `                    ``found = ``true``; ` ` `  `                    ``Console.WriteLine(arr[i] + ``" "` `                                    ``+ arr[j]); ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``if` `(found == ``false``) ` `            ``Console.WriteLine(``"Not Exist "``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = { 10, 4, 8, 13, 5 }; ` `        ``int` `n = arr.Length; ` `        ``findPair(arr, n); ` `    ``} ` `} ` ` `  `// This code contributed by Rajput-Ji `

Output:

```8 5
```

Time Complexity : O(n2)

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