# 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:

## C++

 `# include ` `# include /* for abs() */ ` `# include ` ` `  `using` `namespace` `std; ` `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) ` `    ``{ ` `        ``cout << ``"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; ` `        ``} ` `        ``} ` `    ``} ` `     `  `    ``cout << ``"The two elements whose sum is minimum are "` `         ``<< arr[min_l] << ``" and "` `<< arr[min_r]; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = {1, 60, -10, 70, -80, 85}; ` `    ``minAbsSumPair(arr, 6); ` `    ``return` `0; ` `} ` ` `  `// This code is contributed ` `// by Akanksha Rai(Abby_akku) `

## C

 `# include ` `# include /* for abs() */ ` `# include ` `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; ` `} `

## Java

 `import` `java.util.*; ` `import` `java.lang.*; ` `class` `Main ` `{ ` `    ``static` `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``) ` `      ``{ ` `        ``System.out.println(``"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``(Math.abs(min_sum) > Math.abs(sum)) ` `          ``{ ` `            ``min_sum = sum; ` `            ``min_l = l; ` `            ``min_r = r; ` `          ``} ` `        ``} ` `      ``} ` `      `  `      ``System.out.println(``" The two elements whose "``+ ` `                              ``"sum is minimum are "``+ ` `                        ``arr[min_l]+ ``" and "``+arr[min_r]); ` `    ``} ` `     `  `    ``// main function ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `arr[] = {``1``, ``60``, -``10``, ``70``, -``80``, ``85``}; ` `        ``minAbsSumPair(arr, ``6``); ` `    ``} ` `     `  `} `

## Python3

 `# Python3 code to find Two elements ` `# whose sum is closest to zero ` ` `  `def` `minAbsSumPair(arr,arr_size): ` `    ``inv_count ``=` `0` ` `  `    ``# Array should have at least ` `    ``# two elements ` `    ``if` `arr_size < ``2``: ` `        ``print``(``"Invalid Input"``) ` `        ``return` ` `  `    ``# Initialization of values  ` `    ``min_l ``=` `0` `    ``min_r ``=` `1` `    ``min_sum ``=` `arr[``0``] ``+` `arr[``1``] ` `    ``for` `l ``in` `range` `(``0``, arr_size ``-` `1``): ` `        ``for` `r ``in` `range` `(l ``+` `1``, arr_size): ` `            ``sum` `=` `arr[l] ``+` `arr[r]                  ` `            ``if` `abs``(min_sum) > ``abs``(``sum``):          ` `                ``min_sum ``=` `sum` `                ``min_l ``=` `l ` `                ``min_r ``=` `r ` ` `  `    ``print``(``"The two elements whose sum is minimum are"``,  ` `            ``arr[min_l], ``"and "``, arr[min_r]) ` ` `  `# Driver program to test above function  ` `arr ``=` `[``1``, ``60``, ``-``10``, ``70``, ``-``80``, ``85``] ` ` `  `minAbsSumPair(arr, ``6``); ` ` `  `# This code is contributed by Smitha Dinesh Semwal `

## C#

 `// C# code to find Two elements ` `// whose sum is closest to zero ` `using` `System; ` ` `  `class` `GFG ` `{ ` `static` `void` `minAbsSumPair(``int` `[]arr, ` `                        ``int` `arr_size) ` `    ``{ ` `     `  `    ``int` `l, r, min_sum, sum, min_l, min_r; ` `     `  `    ``/* Array should have at least two elements*/` `    ``if` `(arr_size < 2) ` `    ``{ ` `        ``Console.Write(``"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` `(Math.Abs(min_sum) > Math.Abs(sum)) ` `            ``{ ` `                ``min_sum = sum; ` `                ``min_l = l; ` `                ``min_r = r; ` `            ``} ` `        ``} ` `    ``} ` `     `  `    ``Console.Write(``" The two elements whose "``+ ` `                        ``"sum is minimum are "``+ ` `                    ``arr[min_l]+ ``" and "``+arr[min_r]); ` `    ``} ` `     `  `    ``// main function ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``int` `[]arr = {1, 60, -10, 70, -80, 85}; ` `     `  `        ``minAbsSumPair(arr, 6); ` `    ``} ` `     `  `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 ` ``abs``(``\$sum``)) ` `        ``{ ` `            ``\$min_sum` `= ``\$sum``; ` `            ``\$min_l` `= ``\$l``; ` `            ``\$min_r` `= ``\$r``; ` `        ``} ` `        ``} ` `    ``} ` `     `  `    ``echo` `"The two elements whose sum is minimum are "` `            ``.``\$arr``[``\$min_l``].``" and "``. ``\$arr``[``\$min_r``]; ` `             `  `} ` ` `  `// Driver Code ` `\$arr` `= ``array``(1, 60, -10, 70, -80, 85); ` `minAbsSumPair(``\$arr``, 6); ` ` `  `// This code is contributed by Sam007 ` `?> `

Output:

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

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

## C

 `# include ` `# include ` `# include ` ` `  `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); ` `  ``} ` `} `

## Java

 `import` `java.util.*; ` `import` `java.lang.*; ` `class` `Main ` `{ ` `    ``static` `void` `minAbsSumPair(``int` `arr[], ``int` `n) ` `    ``{ ` `      ``// Variables to keep track of current sum and minimum sum ` `      ``int` `sum, min_sum = ``999999``; ` `      `  `      ``// 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``) ` `      ``{ ` `        ``System.out.println(``"Invalid Input"``); ` `        ``return``; ` `      ``} ` `      `  `      ``/* Sort the elements */` `      ``sort(arr, l, r); ` `      `  `      ``while``(l < r) ` `      ``{ ` `        ``sum = arr[l] + arr[r]; ` `      `  `        ``/*If abs(sum) is less then update the result items*/` `        ``if``(Math.abs(sum) < Math.abs(min_sum)) ` `        ``{ ` `          ``min_sum = sum; ` `          ``min_l = l; ` `          ``min_r = r; ` `        ``} ` `        ``if``(sum < ``0``) ` `          ``l++; ` `        ``else` `          ``r--; ` `      ``} ` `      `  `       `  `      ``System.out.println(``" The two elements whose "``+ ` `                              ``"sum is minimum are "``+ ` `                        ``arr[min_l]+ ``" and "``+arr[min_r]); ` `    ``} ` `      `  `    ``// main function ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `arr[] = {``1``, ``60``, -``10``, ``70``, -``80``, ``85``}; ` `        ``int` `n = arr.length; ` `        ``minAbsSumPair(arr, n); ` `    ``} ` `     `  `    ``/* Functions for QuickSort */` `     `  `    ``/* This function takes last element as pivot, ` `       ``places the pivot element at its correct ` `       ``position in sorted array, and places all ` `       ``smaller (smaller than pivot) to left of ` `       ``pivot and all greater elements to right ` `       ``of pivot */` `    ``static` `int` `partition(``int` `arr[], ``int` `low, ``int` `high) ` `    ``{ ` `        ``int` `pivot = arr[high];  ` `        ``int` `i = (low-``1``); ``// index of smaller element ` `        ``for` `(``int` `j=low; j Array to be sorted, ` `      ``low  --> Starting index, ` `      ``high  --> Ending index */` `    ``static` `void` `sort(``int` `arr[], ``int` `low, ``int` `high) ` `    ``{ ` `        ``if` `(low < high) ` `        ``{ ` `            ``/* pi is partitioning index, arr[pi] is  ` `              ``now at right place */` `            ``int` `pi = partition(arr, low, high); ` ` `  `            ``// Recursively sort elements before ` `            ``// partition and after partition ` `            ``sort(arr, low, pi-``1``); ` `            ``sort(arr, pi+``1``, high); ` `        ``} ` `    ``} ` `} `

## C#

 `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``static` `void` `minAbsSumPair(``int` `[]arr ,``int` `n) ` `    ``{ ` `        ``// Variables to keep track  ` `        ``// of current sum and minimum sum ` `        ``int` `sum, min_sum = 999999; ` `         `  `        ``// 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) ` `        ``{ ` `            ``Console.Write(``"Invalid Input"``); ` `            ``return``; ` `        ``} ` `         `  `        ``/* Sort the elements */` `        ``sort(arr, l, r); ` `         `  `        ``while``(l < r) ` `        ``{ ` `            ``sum = arr[l] + arr[r]; ` `         `  `            ``/*If abs(sum) is less then update the result items*/` `            ``if` `(Math.Abs(sum) < Math.Abs(min_sum)) ` `            ``{ ` `                ``min_sum = sum; ` `                ``min_l = l; ` `                ``min_r = r; ` `            ``} ` `            ``if` `(sum < 0) ` `                ``l++; ` `            ``else` `                ``r--; ` `        ``} ` `         `  `        ``Console.Write(``" The two elements whose "` `+  ` `                                ``"sum is minimum are "` `+ ` `                            ``arr[min_l]+ ``" and "` `+ arr[min_r]); ` `    ``} ` `     `  `    ``// driver code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``int` `[]arr = {1, 60, -10, 70, -80, 85}; ` `        ``int` `n = arr.Length; ` `         `  `        ``minAbsSumPair(arr, n); ` `    ``} ` `     `  `    ``/* Functions for QuickSort */` `     `  `    ``/* This function takes last element as pivot, ` `    ``places the pivot element at its correct ` `    ``position in sorted array, and places all ` `    ``smaller (smaller than pivot) to left of ` `    ``pivot and all greater elements to right ` `    ``of pivot */` `    ``static` `int` `partition(``int` `[]arr, ``int` `low, ``int` `high) ` `    ``{ ` `        ``int` `pivot = arr[high];  ` `        ``int` `i = (low-1); ``// index of smaller element ` `         `  `        ``for` `(``int` `j = low; j < high; j++) ` `        ``{ ` `            ``// If current element is smaller than or ` `            ``// equal to pivot ` `            ``if` `(arr[j] <= pivot) ` `            ``{ ` `                ``i++; ` ` `  `                ``// swap arr[i] and arr[j] ` `                ``int` `temp = arr[i]; ` `                ``arr[i] = arr[j]; ` `                ``arr[j] = temp; ` `            ``} ` `        ``} ` ` `  `        ``// swap arr[i+1] and arr[high] (or pivot) ` `        ``int` `temp1 = arr[i+1]; ` `        ``arr[i+1] = arr[high]; ` `        ``arr[high] = temp1; ` ` `  `        ``return` `i+1; ` `    ``} ` ` `  ` `  `    ``/* The main function that implements QuickSort() ` `    ``arr[] --> Array to be sorted, ` `    ``low --> Starting index, ` `    ``high --> Ending index */` `    ``static` `void` `sort(``int` `[]arr, ``int` `low, ``int` `high) ` `    ``{ ` `        ``if` `(low < high) ` `        ``{ ` `            ``/* pi is partitioning index, arr[pi] is  ` `            ``now at right place */` `            ``int` `pi = partition(arr, low, high); ` ` `  `            ``// Recursively sort elements before ` `            ``// partition and after partition ` `            ``sort(arr, low, pi-1); ` `            ``sort(arr, pi+1, high); ` `        ``} ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

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)

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