Find two elements whose sum is closest to zero

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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:

import java.util.*;
public class SumCloseToZero {
public int findSum(int [] a){
Arrays.sort(a);
int i=0;
int j = a.length–1;
int positiveClose = Integer.MAX_VALUE;
int negativeClose = Integer.MIN_VALUE;
while(i<j){
int temp = a[i] + a[j];
//check if temp is greater than 0
if(temp>0){
//check if positiveClose needs to be updated
if(positiveClose>temp){
positiveClose = temp;
}
//decrement j, in order to reduce the difference, close to 0
j—;
}else if(temp<0){
//check if negative needs to be updated
if(negativeClose<temp){
negativeClose = temp;
}
//increment i, in order to reduce the difference, close to 0
i++;
}else{
//means temp is 0, that is the closest to zero we can get, return 0
return 0;
}
}
//check the least absolute value in postiveClose and negativeClose
return Math.min(Math.abs(positiveClose), Math.abs(negativeClose));
}
public static void main(String[] args) {
int a [] = {1, 4, –5, 3, –2, 10, –6, 20};
int closestSum = new SumCloseToZero().findSum(a);
System.out.println("Sum close to zero in the given array is : " + closestSum);
}
}

view raw
SumCloseToZero.java
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Output:
Sum close to zero in the given array is : 1

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