# Only integer with positive value in positive negative value in array

Given an array of N integers. In the given, for each positive element ‘x’ there exist a negative value ‘-x’, except one integer whose negative value is not present. That integer may occur multiple number of time. The task is find that integer.

Examples:

```Input : arr[] = { 1, 8, -6, -1, 6, 8, 8 }
Output : 8
All positive elements have an equal negative
value except 8.

Input : arr[] = { 15, 6, -9, 4, 15, 9,
-6, -4, 15, 15 }
Output : 15
```

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

Method 1: (hashing) The idea is to create a hash table, initialize with zero value. Whenever we encounter a positive value, we add +1 on corresponding index position in hash. And whenever we encounter negative value we add -1. Finally we traverse the whole hash. After traversing, the index with non-zero value is the only integer with only value without negative pair.

Below is C++ implementation of this approach:

 `// A hashing based solution to find the only ` `// element that doesn't have negative value. ` `#include ` `using` `namespace` `std; ` ` `  `// Return the integer with have no negative value. ` `int` `findInteger(``int` `arr[], ``int` `n) ` `{ ` `    ``unorder_map<``int``, ``int``> hash; ` `    ``int` `maximum = 0; ` ` `  `    ``// Traversing the array. ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// If negative, then subtract 1 in hash array. ` `        ``if` `(arr[i] < 0) ` `            ``hash[``abs``(arr[i])] -= 1; ` ` `  `        ``// Else add 1 in hash array. ` `        ``else` `            ``hash[arr[i]] += 1; ` `    ``} ` ` `  `    ``// Traverse the hash array. ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``if` `(hash[arr[i]] == 0) ` `            ``return` `i; ` ` `  `    ``return` `-1; ` `} ` ` `  `// Driven Program ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 8, -6, -1, 6, 8 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``cout << findInteger(arr, n) << endl; ` `    ``return` `0; ` `} `

Output:

```8
```

Method 2: (efficient) The idea is to find the sum of each element of array and also count the number of even and odd numbers in the array. Finally, divide the sum by absolute difference of number of odd element and even element.

Below is the implementation of this approach:

## C++

 `// An efficient solution to find the only ` `// element that doesn't have negative value. ` `#include ` `using` `namespace` `std; ` ` `  `// Return the integer with have no negative value. ` `int` `findInteger(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `neg = 0, pos = 0; ` `    ``int` `sum = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``sum += arr[i]; ` ` `  `        ``// If negative, then increment neg count. ` `        ``if` `(arr[i] < 0)  ` `            ``neg++;         ` ` `  `        ``// Else increment pos count.. ` `        ``else`  `            ``pos++;         ` `    ``} ` ` `  `    ``return` `(sum / ``abs``(neg - pos)); ` `} ` ` `  `// Driven Program ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 8, -6, -1, 6, 8 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``cout << findInteger(arr, n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// An efficient solution to find the ` `// only element that doesn't have ` `// negative value. ` `import` `java.lang.*; ` ` `  `class` `GFG { ` `     `  `    ``// Return the integer with have ` `    ``// no negative value. ` `    ``static` `int` `findInteger(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``int` `neg = ``0``, pos = ``0``; ` `        ``int` `sum = ``0``; ` `     `  `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``sum += arr[i]; ` `     `  `            ``// If negative, then increment ` `            ``// neg count. ` `            ``if` `(arr[i] < ``0``)  ` `                ``neg++;  ` `     `  `            ``// Else increment pos count.. ` `            ``else` `                ``pos++;  ` `        ``} ` `     `  `        ``return` `(sum / Math.abs(neg - pos)); ` `    ``} ` `     `  `    ``// Driven Program ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `         `  `        ``int` `arr[] = ``new` `int``[]{ ``1``, ``8``, -``6``, -``1``, ``6``, ``8` `}; ` `        ``int` `n = arr.length; ` `         `  `        ``System.out.println(findInteger(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Smitha. `

## Python 3

 `# An efficient solution to find the ` `# only element that doesn't have ` `# negative value. ` ` `  `# Return the integer with have no ` `# negative value. ` `def` `findInteger(arr, n): ` ` `  `    ``neg ``=` `0` `    ``pos ``=` `0` `    ``sum` `=` `0` ` `  `    ``for` `i ``in` `range``(``0``, n): ` `        ``sum` `+``=` `arr[i] ` ` `  `        ``# If negative, then increment ` `        ``# neg count. ` `        ``if` `(arr[i] < ``0``): ` `            ``neg ``+``=` `1` ` `  `        ``# Else increment pos count.. ` `        ``else``: ` `            ``pos ``+``=` `1` `     `  `    ``return` `(``sum` `/` `abs``(neg ``-` `pos)) ` ` `  `# Driven Program ` `arr ``=` `[``1``, ``8``, ``-``6``, ``-``1``, ``6``, ``8` `] ` `n ``=` `len``(arr) ` `print``(``int``(findInteger(arr, n))) ` ` `  `# This code is contributed by Smitha. `

## C#

 `// An efficient solution to find the ` `// only element that doesn't have ` `// negative value. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Return the integer with have ` `    ``// no negative value. ` `    ``static` `int` `findInteger(``int` `[] arr, ``int` `n) ` `    ``{ ` `        ``int` `neg = 0, pos = 0; ` `        ``int` `sum = 0; ` `     `  `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``sum += arr[i]; ` `     `  `            ``// If negative, then increment ` `            ``// neg count. ` `            ``if` `(arr[i] < 0)  ` `                ``neg++;  ` `     `  `            ``// Else increment pos count.. ` `            ``else` `                ``pos++;  ` `        ``} ` `     `  `        ``return` `(sum / Math.Abs(neg - pos)); ` `    ``} ` `     `  `    ``// Driven Program ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[] arr = ``new` `int``[]{ 1, 8, -6, ` `                                ``-1, 6, 8 }; ` `        ``int` `n = arr.Length; ` `         `  `        ``Console.Write(findInteger(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Smitha. `

## PHP

 ` `

Output:

```8
```

## tags:

Arrays Hash cpp-unordered_map Arrays Hash