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# Find pairs in array whose sum does not exist in Array

• Difficulty Level : Easy
• Last Updated : 18 May, 2021

Given an array arr[] consisting of N positive integers, the task is to print all pairs of array elements whose sum does not exist in the given array. If no such pair exists, print “-1”.
Examples:

Input: arr[] = {2, 4, 2, 6}
Output:
(2, 6)
(4, 6)
(2, 6)
Explanation:
All possible pairs in the array are (2, 4), (2, 2), (2, 6), (4, 2), (4, 6) and (2, 6).
Among these, the pairs (2, 6), (4, 6) and (2, 6) have sums {8, 10, 8} respectively which are not present in the array.
Input: arr[] = {1, 1, 2, 3}
Output:
(2, 3)
Explanation:
All possible pairs in the array are (1, 1), (1, 2), (1, 3), (1, 2), (1, 3) and (2, 3).
Among these, the only pair whose sum is not present in the array is (2, 3).

Naive Approach:
The simplest approach to solve the problem is to generate all possible pairs one by one, and for every pair check if its sum exists in the array by traversing the array. If any pair is found with its sum existing in the array, print the pair. Otherwise, move to the next pair.
Time Complexity: O(N3
Auxiliary Space: O(N)
Efficient Approach:
The problem can be solved using a HashSet. Follow the steps below to solve the problem:

• Store the elements in the of the array in a HashSet.
• Now, traverse over all the array elements and generate all possible pairs.
• For each pair, check if the sum of that pair is present in the HashSet or not. If so, print the pair. Otherwise, move to the next pair.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to print all pairs``// with sum not present in the array``void` `findPair(``int` `arr[], ``int` `n)``{``    ``int` `i, j;` `    ``// Corner Case``    ``if` `(n < 2)``    ``{``        ``cout << ``"-1"` `<< endl;``    ``}` `    ``// Stores the distinct array``    ``// elements``    ``set <``int``> hashMap;` `    ``for``(``int` `k = 0; k < n; k++)``    ``{``        ``hashMap.insert(arr[k]);``    ``}` `    ``// Generate all possible pairs``    ``for``(i = 0; i < n - 1; i++)``    ``{``        ``for``(j = i + 1; j < n; j++)``        ``{``            ` `            ``// Calculate sum of current pair``            ``int` `sum = arr[i] + arr[j];` `            ``// Check if the sum exists in``            ``// the HashSet or not``            ``if` `(hashMap.find(sum) == hashMap.end())``            ``{``    ` `                ``cout << ``"("` `<< arr[i] << ``", "``                    ``<< arr[j] << ``")"` `<< endl;``            ``}``        ``}``    ``}``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 2, 4, 2, 6 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``findPair(arr, n);``    ``return` `0;``}` `// This code is contributed by divyeshrabadiya07`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG {` `    ``// Function to print all pairs``    ``// with sum not present in the array``    ``public` `static` `void` `findPair(``        ``int``[] arr, ``int` `n)``    ``{``        ``int` `i, j;` `        ``// Corner Case``        ``if` `(n < ``2``) {``            ``System.out.println(``"-1"``);``        ``}` `        ``// Stores the distinct array``        ``// elements``        ``HashSet hashMap``            ``= ``new` `HashSet();` `        ``for` `(Integer k : arr) {``            ``hashMap.add(k);``        ``}` `        ``// Generate all possible pairs``        ``for` `(i = ``0``; i < n - ``1``; i++) {` `            ``for` `(j = i + ``1``; j < n; j++) {` `                ``// Calculate sum of current pair``                ``int` `sum = arr[i] + arr[j];` `                ``// Check if the sum exists in``                ``// the HashSet or not``                ``if` `(!hashMap.contains(sum)) {` `                    ``System.out.println(``                        ``"("` `+ arr[i] + ``", "``                        ``+ arr[j] + ``")"``);``                ``}``            ``}``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = { ``2``, ``4``, ``2``, ``6` `};``        ``int` `n = arr.length;` `        ``findPair(arr, n);``    ``}``}`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to print all pairs``# with sum not present in the array``def` `findPair(arr, n):` `    ``# Corner Case``    ``if` `(n < ``2``):``        ``print``(``"-1"``)` `    ``# Stores the distinct array``    ``# elements``    ``hashMap ``=` `[]` `    ``for` `k ``in` `arr:``        ``hashMap.append(k)` `    ``# Generate all possible pairs``    ``for` `i ``in` `range` `(n ``-` `1``):``        ``for` `j ``in` `range` `(i ``+` `1``, n):``            ``# Calculate sum of current pair``            ``sum` `=` `arr[i] ``+` `arr[j]` `            ``# Check if the sum exists in``            ``# the HashSet or not``            ``if` `sum` `not` `in` `hashMap:``                ``print``(``"("``, arr[i] , ``", "``,``                        ``arr[j] , ``")"``)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``arr ``=` `[ ``2``, ``4``, ``2``, ``6` `]``    ``n ``=` `len``(arr)``        ` `    ``findPair(arr, n)` `# This code is contributed by ChitraNayal`

## C#

 `// C# program to implement``// the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to print all pairs``// with sum not present in the array``public` `static` `void` `findPair(``int``[] arr, ``int` `n)``{``    ``int` `i, j;` `    ``// Corner Case``    ``if` `(n < 2)``    ``{``        ``Console.Write(``"-1"``);``    ``}` `    ``// Stores the distinct array``    ``// elements``    ``HashSet<``int``> hashMap = ``new` `HashSet<``int``>();` `    ``foreach``(``int` `k ``in` `arr)``    ``{``        ``hashMap.Add(k);``    ``}` `    ``// Generate all possible pairs``    ``for``(i = 0; i < n - 1; i++)``    ``{``        ``for``(j = i + 1; j < n; j++)``        ``{``            ` `            ``// Calculate sum of current pair``            ``int` `sum = arr[i] + arr[j];` `            ``// Check if the sum exists in``            ``// the HashSet or not``            ``if` `(!hashMap.Contains(sum))``            ``{``                ``Console.Write(``"("` `+ arr[i] +``                             ``", "` `+ arr[j] +``                             ``")\n"``);``            ``}``        ``}``    ``}``}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``    ``int``[] arr = { 2, 4, 2, 6 };``    ``int` `n = arr.Length;` `    ``findPair(arr, n);``}``}` `// This code is contributed by rutvik_56`

## Javascript

 ``
Output:
```(2, 6)
(4, 6)
(2, 6)```

Time Complexity: O(N2
Auxiliary Space: O(N)

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