# Find two elements whose sum is closest to zero

Objective: Given an array of positive and negative integers, write a algorithm to find the two elements such their sum is closest to zero.

Example:

```int a [] = {1, 4, -5, 3, -2, 10, -6, 20};
Output: Sum close to zero in the given array is : 1

int a [] = {-5, 5, 8};
Output: Sum close to zero in the given array is : 0

int a [] = {5, 8};
Output: Sum close to zero in the given array is : 13

int a [] = {-5,-5,-8};
Sum close to zero in the given array is : -10
```

Approach:

Naive approach: Use 2 loops. For each element in the array, find the sum with every other element and keep track of the minimum sum found.

Time Complexity : O(n^2) Space Complexity: O(1)

Sorting approach:

1. Sort the array.
2. This will bring all the negative elements at the front and positive elements at the end of the array.
3. Track 2 numbers, one positive number close to 0, let’s call it as positiveClose  and one negative number close to 0, let’s call it as negativeClose.
4. take 2 pointers, one from the beginning of the array and another one from the end of the array. say pointers are i and j.
5. Add A[i] + A[j], if sum > 0 then reduce j, j– and if sum < 0 then increase i ++.
6. If sum > 0 compare it with positiveClose, if positiveClose>sum, do positiveClose = sum.
7. If sum < 0 compare it with negativeClose, if negativeClose<sum, do negativeClose = sum.
8. Repeat the step 5, 6, 7 till i<j.
9. Return the minimum of absolute values of positiveClose and negativeClose, this will be the answer.
10. At any point if sum = 0 then return 0 since this will be the closest to the 0 is possible.
11. We can easily modify the code given below to track the two indexes as well for which the closest sum is possible.

Time Complexity : O(nlogn) Space Complexity: O(n) by using merge sort.

Complete Code:

```Output:
Sum close to zero in the given array is : 1
```

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
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