Question 17 : Find a Pair Whose Sum is Closest to zero in Array

Given array of +ve and -ve integers ,we need to find a pair whose sum is closed to Zero in Array.

For example::

array[]={1,3,-5,7,8,20,-40,6};
The pair whose sum is closest to zero :  -5 and 6

Solution :

Solution 1:

You can check each and every pair of numbers and find the minimum sum.
Java code:

public static void findPairWithMinSumBruteForce(int arr[])
{
    if(arr.length<2)
        return;
    // Suppose 1st two element has minimum sum
    int minimumSum=arr[0]+arr[1];
    int pair1stIndex=0;
    int pair2ndIndex=1;
    for (int i = 0; i < arr.length; i++) {
        for (int j = i+1; j < arr.length; j++) {
            int tempSum=arr[i]+arr[j];
            if(Math.abs(tempSum) < Math.abs(minimumSum))
            {
                pair1stIndex=i;
                pair2ndIndex=j;
                minimumSum=tempSum;
            }
        }
    }
    System.out.println(" The pair whose sum is closest to zero : "+arr[pair1stIndex]+" "+ arr[pair2ndIndex]);
}

 Solution 2:

• Sort the array.
• We will maintain two indexes one at beginning (l=0) and one at end (r=n-1)
• iterate until l <  r
• Calculate sum of arr[l] + arr[r]
• if abs (sum) < abs (minSum), then update the minimum sum and pair.
• If sum is less than 0, this means if we want to find sum close to 0, do r–
• If sum is greater than 0,this means if we want to find sum close to 0 , do l++

Java code::

public static void findPairWithMinSum(int arr[]) {
 
        // Sort the array, you can use any sorting algorithm to sort it
        Arrays.sort(arr);
        int sum=0; 
        int minimumSum = Integer.MAX_VALUE;
        int n=arr.length;
        if(n<0)
            return;
        // left and right index variables
        int l = 0, r = n-1;
 
        // variables to keep track of the left and right index pair for minimumSum
        int minLeft = l, minRight = n-1;
 
        while(l < r)
        {
            sum = arr[l] + arr[r];
 
            /*If abs(sum) is less than min sum, we need to update sum and pair */
            if(Math.abs(sum) < Math.abs(minimumSum))
            {
                minimumSum = sum;
                minLeft = l;
                minRight = r;
            }
            if(sum < 0)
                l++;
            else
                r--;
        }
 
        System.out.println(" The pair whose sum is minimun : "+arr[minLeft]+" "+ arr[minRight]);
    }

Time complexity: O(NLogN)

Java program to find a pair whose sum is closest to zero::

import java.util.Arrays;
 
public class findPairClosestToZero {
 
    public static void main(String[] args)
    {
        int array[]={1,30,-5,70,-8,20,-40,60};
        findPairWithMinSumBruteForce(array);
        findPairWithMinSum(array);
    }
    public static void  findPairWithMinSumBruteForce(int arr[])
    {
        if(arr.length<2)
            return;
        // Suppose 1st two element has minimum sum
        int minimumSum=arr[0]+arr[1];
        int pair1stIndex=0;
        int pair2ndIndex=1;
        for (int i = 0; i < arr.length; i++) {
            for (int j = i+1; j < arr.length; j++) {
                int tempSum=arr[i]+arr[j];
                if(Math.abs(tempSum) < Math.abs(minimumSum))
                {
                    pair1stIndex=i;
                    pair2ndIndex=j;
                    minimumSum=tempSum;
                }
            }
        }
        System.out.println(" The pair whose sum is closest to zero using brute force method: "+arr[pair1stIndex]+" "+ arr[pair2ndIndex]);
    }
 
    public static void findPairWithMinSum(int arr[]) {
 
        // Sort the array, you can use any sorting algorithm to sort it
        Arrays.sort(arr);
        int sum=0; 
        int minimumSum = Integer.MAX_VALUE;
        int n=arr.length;
        if(n<0)
            return;
        // left and right index variables
        int l = 0, r = n-1;
 
        // variables to keep track of the left and right index pair for minimumSum
        int minLeft = l, minRight = n-1;
 
        while(l < r)
        {
            sum = arr[l] + arr[r];
 
            /*If abs(sum) is less than min sum, we need to update sum and pair */
            if(Math.abs(sum) < Math.abs(minimumSum))
            {
                minimumSum = sum;
                minLeft = l;
                minRight = r;
            }
            if(sum < 0)
                l++;
            else
                r--;
        }
 
        System.out.println(" The pair whose sum is minimun : "+arr[minLeft]+" "+ arr[minRight]);
    }
}

When you run the program, you will get the below output::

 The pair whose sum is closest to zero using brute force method: 1 -5
 The pair whose sum is closest to zero : -5 1

Don't miss the next article! 
Be the first to be notified when a new article or Kubernetes experiment is published.                            

   

Share This

• Facebook
• Twitter
• Linkedin
• Pinterest