Two elements whose sum is closest to zero

Question: An Array of integers is given, both +ve and -ve. You need to find the two elements such that their sum is closest to zero.

For the below array, program should print -80 and 85.

METHOD 1 (Simple) 
For each element, find the sum of it with every other element in the array and compare sums. Finally, return the minimum sum.

Implementation

# include <stdio.h> # include <stdlib.h> /* for abs() */ # include <math.h> void minAbsSumPair(int arr[], int arr_size) {   int inv_count = 0;   int l, r, min_sum, sum, min_l, min_r;     /* Array should have at least two elements*/   if(arr_size < 2)   {     printf("Invalid Input");     return;   }     /* Initialization of values */   min_l = 0;   min_r = 1;   min_sum = arr[0] + arr[1];     for(l = 0; l < arr_size - 1; l++)   {     for(r = l+1; r < arr_size; r++)     {       sum = arr[l] + arr[r];       if(abs(min_sum) > abs(sum))       {         min_sum = sum;         min_l = l;         min_r = r;       }     }   }     printf(" The two elements whose sum is minimum are %d and %d",           arr[min_l], arr[min_r]); }   /* Driver program to test above function */ int main() {   int arr[] = {1, 60, -10, 70, -80, 85};   minAbsSumPair(arr, 6);   getchar();   return 0; }

Time complexity: O(n^2)

METHOD 2 (Use Sorting) 
Thanks to baskin for suggesting this approach. We recommend to read this post for background of this approach.

Algorithm 
1) Sort all the elements of the input array.
2) Use two index variables l and r to traverse from left and right ends respectively. Initialize l as 0 and r as n-1.
3) sum = a[l] + a[r]
4) If sum is -ve, then l++
5) If sum is +ve, then r–
6) Keep track of abs min sum.
7) Repeat steps 3, 4, 5 and 6 while l < r Implementation

# include <stdio.h> # include <math.h> # include <limits.h>   void quickSort(int *, int, int);   /* Function to print pair of elements having minimum sum */ void minAbsSumPair(int arr[], int n) {   // Variables to keep track of current sum and minimum sum   int sum, min_sum = INT_MAX;     // left and right index variables   int l = 0, r = n-1;     // variable to keep track of the left and right pair for min_sum   int min_l = l, min_r = n-1;     /* Array should have at least two elements*/   if(n < 2)   {     printf("Invalid Input");     return;   }     /* Sort the elements */   quickSort(arr, l, r);     while(l < r)   {     sum = arr[l] + arr[r];       /*If abs(sum) is less then update the result items*/     if(abs(sum) < abs(min_sum))     {       min_sum = sum;       min_l = l;       min_r = r;     }     if(sum < 0)       l++;     else       r--;   }     printf(" The two elements whose sum is minimum are %d and %d",           arr[min_l], arr[min_r]); }   /* Driver program to test above function */ int main() {   int arr[] = {1, 60, -10, 70, -80, 85};   int n = sizeof(arr)/sizeof(arr[0]);   minAbsSumPair(arr, n);   getchar();   return 0; }   /* FOLLOWING FUNCTIONS ARE ONLY FOR SORTING     PURPOSE */ void exchange(int *a, int *b) {   int temp;   temp = *a;   *a   = *b;   *b   = temp; }   int partition(int arr[], int si, int ei) {   int x = arr[ei];   int i = (si - 1);   int j;     for (j = si; j <= ei - 1; j++)   {     if(arr[j] <= x)     {       i++;       exchange(&arr[i], &arr[j]);     }   }     exchange (&arr[i + 1], &arr[ei]);   return (i + 1); }   /* Implementation of Quick Sort arr[] --> Array to be sorted si  --> Starting index ei  --> Ending index */ void quickSort(int arr[], int si, int ei) {   int pi;    /* Partitioning index */   if(si < ei)   {     pi = partition(arr, si, ei);     quickSort(arr, si, pi - 1);     quickSort(arr, pi + 1, ei);   } }

Time Complexity: complexity to sort + complexity of finding the optimum pair = O(nlogn) + O(n) = O(nlogn)

Asked by Vineet

Please write comments if you find any bug in the above program/algorithm or other ways to solve the same problem.