# 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:**

inta [] = {1, 4, -5, 3, -2, 10, -6, 20};Output: Sum close to zero in the given array is : 1inta [] = {-5, 5, 8};Output: Sum close to zero in the given array is : 0inta [] = {5, 8};Output: Sum close to zero in the given array is : 13inta [] = {-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: **

- Sort the array.
- This will bring all the negative elements at the front and positive elements at the end of the array.
- 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.
- 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.
- Add A[i] + A[j], if sum > 0 then reduce j, j– and if sum < 0 then increase i ++.
- If sum > 0 compare it with positiveClose, if positiveClose>sum, do positiveClose = sum.
- If sum < 0 compare it with negativeClose, if negativeClose<sum, do negativeClose = sum.
- Repeat the step 5, 6, 7 till i<j.
- Return the minimum of absolute values of positiveClose and negativeClose, this will be the answer.
- At any point if sum = 0 then return 0 since this will be the closest to the 0 is possible.
- 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

The solution does not return negative closest sum.

Arrays.sort(array);

int i = 0;

int j = array.length – 1;

int leftIndex = 0;

int rightIndex = array.length – 1;

int minimumSum = Integer.MAX_VALUE;

int sum;

if (array.length < 2) { //For one element return that element

return array[0];

}

while(i < j) {

sum = array[i] + array[j];

if (Math.abs(sum) < Math.abs(minimumSum)) {

minimumSum = sum;

leftIndex = i;

rightIndex = j;

}

if (sum < 0) {

i++;

} else {

j–;

}

}

System.out.println("The two elements whose sum is mimimum is " + array[leftIndex] + ", " + array[rightIndex]);

return minimumSum;

It should return two elements, not closest sum.