Two elements whose sum is closest to zero

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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;
}

Output:

The two elements whose sum is minimum are -80 and 85

Time complexity: O(n^2)

METHOD 2 (Use Sorting) 
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);
  }
}

Output:

The two elements whose sum is minimum are -80 and 85


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

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits:http://www.geeksforgeeks.org/two-elements-whose-sum-is-closest-to-zero/
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rakesh

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