# Pairs of Positive Negative values in an array

Given an array of distinct integers, print all the pairs having positive value and negative value of a number that exists in the array. We need to print pairs in order of their occurrences. A pair whose any element appears first should be printed first.

Examples:

```Input  :  arr[] = { 1, -3, 2, 3, 6, -1 }
Output : -1 1 -3 3

Input  :  arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 }
Output : -1 1 -4 4 -8 8 -9 9
```

Method 1 (Simple : O(n2))
The idea is to use two nested loop. For each element arr[i], find negative of arr[i] from index i + 1 to n – 1 and store it in another array. For output, sort the stored element and print negative positive value of the stored element.

Below is C++ implementation of this approach:

`// Simple CPP program to find pairs of positive`
`// and negative values present in an array.`
`#include <bits/stdc++.h>`
`using` `namespace` `std;`
`// Print pair with negative and positive value`
`void` `printPairs(``int` `arr[], ``int` `n)`
`{`
`    ``vector<``int``> v;`
`    ``// For each element of array.`
`    ``for` `(``int` `i = 0; i < n; i++) `
`        ``// Try to find the negative value of `
`        ``// arr[i] from i + 1 to n`
`        ``for` `(``int` `j = i + 1; j < n; j++) `
`            ``// If absolute values are equal print pair.`
`            ``if` `(``abs``(arr[i]) == ``abs``(arr[j])) `
`                ``v.push_back(``abs``(arr[i]));      `
`    ``// If size of vector is 0, therefore there is no `
`    ``// element with positive negative value, print "0"`
`    ``if` `(v.size() == 0)`
`       ``return``;`
`    ``// Sort the vector`
`    ``sort(v.begin(), v.end());`
`    ``// Print the pair with negative positive value.`
`    ``for` `(``int` `i = 0; i < v.size(); i++)`
`        ``cout << -v[i] << ``" "` `<< v[i];    `
`}`
`// Driven Program`
`int` `main()`
`{`
`    ``int` `arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };`
`    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`
`    ``printPairs(arr, n);`
`    ``return` `0;`
`}`

Output:

```4 4 -8 8 -9 9 -1 1
```

Method 2 (Hahing)
The idea is to use hashing. Traverse the given array, increase the count at absolute value of hash table. If count becomes 2, store its absolute value in another vector. And finally sort the vector. If the size of the vector is 0, print “0”, else for each term in vector print first its negative value and the the positive value.

Below is C++ implementation of this approach:

`// Efficient CPP program to find pairs of `
`// positive and negative values present in`
`// an array.`
`#include <bits/stdc++.h>`
`using` `namespace` `std;`
`// Print pair with negative and positive value`
`void` `printPairs(``int` `arr[], ``int` `n)`
`{`
`    ``vector<``int``> v;`
`    ``unordered_map<``int``, ``bool``> cnt;`
`    ``// For each element of array.`
`    ``for` `(``int` `i = 0; i < n; i++) {`
`        ``// If element has not encounter early,`
`        ``// mark it on cnt array.`
`        ``if` `(cnt[``abs``(arr[i])] == 0)`
`            ``cnt[``abs``(arr[i])] = 1;`
`        ``// If seen before, push it in vector (`
`        ``// given that elements are distinct)`
`        ``else` `{`
`            ``v.push_back(``abs``(arr[i]));`
`            ``cnt[``abs``(arr[i])] = 0;`
`        ``}`
`    ``}`
`    ``if` `(v.size() == 0)`
`        ``return``;`
`    ``sort(v.begin(), v.end());`
`    ``for` `(``int` `i = 0; i < v.size(); i++)`
`        ``cout << -v[i] << ``" "` `<< v[i] << ``" "``;`
`}`
`// Driven Program`
`int` `main()`
`{`
`    ``int` `arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };`
`    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`
`    ``printPairs(arr, n);`
`    ``return` `0;`
`}`

Output:

```4 4 -8 8 -9 9 -1 1

```

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
We have built the accelerating growth-oriented website for budding engineers and aspiring job holders of technology companies such as Google, Facebook, and Amazon
If you would like to study our free courses you can join us at