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# Check if there exist two elements in an array whose sum is equal to the sum of rest of the array

We have an array of integers and we have to find two such elements in the array such that sum of these two elements is equal to the sum of rest of elements in array.

Examples:

```Input  : arr[] = {2, 11, 5, 1, 4, 7}
Output : Elements are 4 and 11
Note that 4 + 11 = 2 + 5 + 1 + 7

Input  : arr[] = {2, 4, 2, 1, 11, 15}
Output : Elements do not exist ```

A simple solution is to consider every pair one by one, find its sum and compare the sum with sum of rest of the elements. If we find a pair whose sum is equal to rest of elements, we print the pair and return true. Time complexity of this solution is O(n3)

An efficient solution is to find sum of all array elements. Let this sum be “sum”. Now the task reduces to finding a pair with sum equals to sum/2.

Another optimization is, a pair can exist only if the sum of whole array is even because we are basically dividing it into two parts with equal sum.

1. Find the sum of whole array. Let this sum be “sum”
2. If sum is odd, return false.
3. Find a pair with sum equals to “sum/2” using hashing based method discussed here as method 2. If a pair is found, print it and return true.
4. If no pair exists, return false.

Below is the implementation of above steps.

## C++

 `// C++ program to find whether two elements exist``// whose sum is equal to sum of rest of the elements.``#include ``using` `namespace` `std;` `// Function to check whether two elements exist``// whose sum is equal to sum of rest of the elements.``bool` `checkPair(``int` `arr[], ``int` `n)``{``    ``// Find sum of whole array``    ``int` `sum = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``sum += arr[i];` `    ``// If sum of array is not even then we can not``    ``// divide it into two part``    ``if` `(sum % 2 != 0)``        ``return` `false``;` `    ``sum = sum / 2;` `    ``// For each element arr[i], see if there is``    ``// another element with value sum - arr[i]``    ``unordered_set<``int``> s;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``int` `val = sum - arr[i];` `        ``// If element exist than return the pair``        ``if` `(s.find(val) != s.end()) {``            ``printf``(``"Pair elements are %d and %d\n"``, arr[i],``                   ``val);``            ``return` `true``;``        ``}` `        ``s.insert(arr[i]);``    ``}` `    ``return` `false``;``}` `// Driver program.``int` `main()``{``    ``int` `arr[] = { 2, 11, 5, 1, 4, 7 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``if` `(checkPair(arr, n) == ``false``)``        ``printf``(``"No pair found"``);``    ``return` `0;``}`

## Java

 `// Java program to find whether two elements exist``// whose sum is equal to sum of rest of the elements.``import` `java.util.*;` `class` `GFG {` `    ``// Function to check whether two elements exist``    ``// whose sum is equal to sum of rest of the elements.``    ``static` `boolean` `checkPair(``int` `arr[], ``int` `n)``    ``{``        ``// Find sum of whole array``        ``int` `sum = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``sum += arr[i];``        ``}` `        ``// If sum of array is not even then we can not``        ``// divide it into two part``        ``if` `(sum % ``2` `!= ``0``) {``            ``return` `false``;``        ``}` `        ``sum = sum / ``2``;` `        ``// For each element arr[i], see if there is``        ``// another element with value sum - arr[i]``        ``HashSet s = ``new` `HashSet();``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``int` `val = sum - arr[i];` `            ``// If element exist than return the pair``            ``if` `(s.contains(val)``                ``&& val == (``int``)s.toArray()[s.size() - ``1``]) {``                ``System.out.printf(``                    ``"Pair elements are %d and %d\n"``, arr[i],``                    ``val);``                ``return` `true``;``            ``}``            ``s.add(arr[i]);``        ``}``        ``return` `false``;``    ``}` `    ``// Driver program.``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``2``, ``11``, ``5``, ``1``, ``4``, ``7` `};``        ``int` `n = arr.length;``        ``if` `(checkPair(arr, n) == ``false``) {``            ``System.out.printf(``"No pair found"``);``        ``}``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 program to find whether``# two elements exist whose sum is``# equal to sum of rest of the elements.` `# Function to check whether two``# elements exist whose sum is equal``# to sum of rest of the elements.``def` `checkPair(arr, n):``    ``s ``=` `set``()``    ``sum` `=` `0` `    ``# Find sum of whole array``    ``for` `i ``in` `range``(n):``        ``sum` `+``=` `arr[i]``    ` `    ``# / If sum of array is not``    ``# even then we can not``    ``# divide it into two part``    ``if` `sum` `%` `2` `!``=` `0``:``        ``return` `False``    ``sum` `=` `sum` `/` `2` `    ``# For each element arr[i], see if``    ``# there is another element with``    ``# value sum - arr[i]``    ``for` `i ``in` `range``(n):``        ``val ``=` `sum` `-` `arr[i]``        ``if` `arr[i] ``not` `in` `s:``            ``s.add(arr[i])``            ` `        ``# If element exist than``        ``# return the pair``        ``if` `val ``in` `s:``            ``print``(``"Pair elements are"``,``                   ``arr[i], ``"and"``, ``int``(val))` `# Driver Code``arr ``=` `[``2``, ``11``, ``5``, ``1``, ``4``, ``7``]``n ``=` `len``(arr)``if` `checkPair(arr, n) ``=``=` `False``:``    ``print``(``"No pair found"``)` `# This code is contributed``# by Shrikant13`

## C#

 `// C# program to find whether two elements exist``// whose sum is equal to sum of rest of the elements.``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ` `    ``// Function to check whether two elements exist``    ``// whose sum is equal to sum of rest of the elements.``    ``static` `bool` `checkPair(``int` `[]arr, ``int` `n)``    ``{``        ``// Find sum of whole array``        ``int` `sum = 0;``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``sum += arr[i];``        ``}` `        ``// If sum of array is not even then we can not``        ``// divide it into two part``        ``if` `(sum % 2 != 0)``        ``{``            ``return` `false``;``        ``}` `        ``sum = sum / 2;` `        ``// For each element arr[i], see if there is``        ``// another element with value sum - arr[i]``        ``HashSet<``int``> s = ``new` `HashSet<``int``>();``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``int` `val = sum - arr[i];` `            ``// If element exist than return the pair``            ``if` `(s.Contains(val))``            ``{``                ``Console.Write(``"Pair elements are {0} and {1}\n"``,``                        ``arr[i], val);``                ``return` `true``;``            ``}``            ``s.Add(arr[i]);``        ``}``        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `[]arr = {2, 11, 5, 1, 4, 7};``        ``int` `n = arr.Length;``        ``if` `(checkPair(arr, n) == ``false``)``        ``{``            ``Console.Write(``"No pair found"``);``        ``}``    ``}``}` `// This code contributed by Rajput-Ji`

## PHP

 ``

## Javascript

 ``

Output

`Pair elements are 4 and 11`

Time complexity : O(n). unordered_set is implemented using hashing. Time complexity hash search and insert is assumed as O(1) here.
Auxiliary Space: O(n)

Another Efficient Approach ( Space Optimization): First we will sort the array for Binary search . Then we will iterate whole array and check if a index exist in the array that pair with i such that arr[index] + a[i] == Rest sum of the array . We can use Binary search  to  find a index in the array by modifying the Binary search program . If a pair exist then print that pair . else print no pair exists.

Below is the implementation of the above approach :

## C++

 `// C++ program for the above approach` `#include``using` `namespace` `std;` `// Function to Find if a index exist in array such that``// arr[index] + a[i] == Rest sum of the array``int` `binarysearch(``int` `arr[], ``int` `n, ``int` `i ,``int` `Totalsum)``{``     ``int` `l = 0, r = n - 1 , index = -1;``//initialize as -1` `     ``while` `(l <= r)``     ``{   ``int` `mid = (l + r) / 2;``     ` `         ``int` `Pairsum = arr[mid] + arr[i];``//pair sum``         ``int` `Restsum = Totalsum - Pairsum;``//Rest sum``        ` `        ``if` `( Pairsum == Restsum )``         ``{``            ``if``( index != i )``// checking a pair has same position or not``            ``{  index = mid; }``//Then update index -1 to mid``              ` `            ``// Checking for adjacent element``            ``else` `if``(index == i && mid>0 && arr[mid-1]==arr[i])``              ``{  index = mid-1; }``//Then update index -1 to mid-1``          ` `            ``else` `if``(index == i && mid Restsum)``        ``{ ``// If pair sum is greater than rest sum , our index will``          ``// be in the Range [mid+1,R]``            ``l = mid + 1;``        ``}``        ``else` `{``          ``// If pair sum is smaller than rest sum , our index will``          ``// be in the Range [L,mid-1]``            ``r = mid - 1;``        ``}``    ``}``    ``// return index=-1 if a pair not exist with arr[i]``    ``// else return index such that arr[i]+arr[index] == sum of rest of arr``    ``return` `index;``}``// Function to check if a pair exist such their sum``// equal to rest of the array or not``bool` `checkPair(``int` `arr[],``int` `n)``{   ``int` `Totalsum=0;``    ``sort(arr , arr + n);``//sort arr for Binary search`` ` `    ``for``(``int` `i=0;i

## Java

 `// Java program for the above approach` `import` `java.util.*;` `class` `GFG {``    ``// Function to Find if a index exist in array such that``    ``// arr[index] + a[i] == Rest sum of the array``    ``static` `int` `binarysearch(``int` `arr[], ``int` `n, ``int` `i,``                            ``int` `Totalsum)``    ``{``        ``int` `l = ``0``, r = n - ``1``, index = -``1``; ``// initialize as -1` `        ``while` `(l <= r) {``            ``int` `mid = (l + r) / ``2``;` `            ``int` `Pairsum = arr[mid] + arr[i]; ``// pair sum``            ``int` `Restsum = Totalsum - Pairsum; ``// Rest sum` `            ``if` `(Pairsum == Restsum) {``                ``if` `(index != i) ``// checking a pair has same``                                ``// position or not``                ``{``                    ``index = mid;``                ``} ``// Then update index -1 to mid` `                ``// Checking for adjacent element``                ``else` `if` `(index == i && mid > ``0``                         ``&& arr[mid - ``1``] == arr[i]) {``                    ``index = mid - ``1``;``                ``} ``// Then update index -1 to mid-1` `                ``else` `if` `(index == i && mid < n - ``1``                         ``&& arr[mid + ``1``] == arr[i]) {``                    ``index = mid + ``1``;``                ``} ``// Then update index-1 to mid+1``                ``break``;``            ``}``            ``else` `if` `(Pairsum > Restsum) {``                ``// If pair sum is greater than rest sum ,``                ``// our index will be in the Range [mid+1,R]``                ``l = mid + ``1``;``            ``}``            ``else` `{``                ``// If pair sum is smaller than rest sum ,``                ``// our index will be in the Range [L,mid-1]``                ``r = mid - ``1``;``            ``}``        ``}``        ``// return index=-1 if a pair not exist with arr[i]``        ``// else return index such that arr[i]+arr[index] ==``        ``// sum of rest of arr``        ``return` `index;``    ``}` `    ``// Function to check if a pair exist such their sum``    ``// equal to rest of the array or not``    ``static` `boolean` `checkPair(``int` `arr[], ``int` `n)``    ``{``        ``int` `Totalsum = ``0``;``        ``Arrays.sort(arr); ``// sort arr for Binary search` `        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``Totalsum += arr[i];``        ``} ``// Finding total sum of the arr` `        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``// If index is -1 , Means arr[i] can't pair with``            ``// any element else arr[i]+a[index] == sum of``            ``// rest of the arr``            ``int` `index = binarysearch(arr, n, i, Totalsum);` `            ``if` `(index != -``1``) {``                ``System.out.println(``"Pair elements are "``                                   ``+ arr[i] + ``" and "``                                   ``+ arr[index]);``                ``return` `true``;``            ``}``        ``}``        ``return` `false``; ``// Return false if a pair not exist``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``2``, ``11``, ``5``, ``1``, ``4``, ``7` `};``        ``int` `n = arr.length;` `        ``// Function call``        ``if` `(checkPair(arr, n) == ``false``) {``            ``System.out.println(``"No pair found"``);``        ``}``    ``}``}`

## Python3

 `# Python program for the above approach` `# Function to find if a index exist in array such that``# arr[index] + a[i] == Rest sum of the array`  `def` `binarysearch(arr, n, i, Totalsum):``    ``l ``=` `0``    ``r ``=` `n ``-` `1``    ``index ``=` `-``1`  `# Initialize as -1` `    ``while` `l <``=` `r:``        ``mid ``=` `(l ``+` `r) ``/``/` `2``        ``Pairsum ``=` `arr[mid] ``+` `arr[i]  ``# Pair sum``        ``Restsum ``=` `Totalsum ``-` `Pairsum  ``# Rest sum` `        ``if` `Pairsum ``=``=` `Restsum:``            ``if` `index !``=` `i:  ``# Checking if a pair has the same position or not``                ``index ``=` `mid  ``# Then update index -1 to mid` `            ``# Checking for adjacent element``            ``elif` `index ``=``=` `i ``and` `mid > ``0` `and` `arr[mid ``-` `1``] ``=``=` `arr[i]:``                ``index ``=` `mid ``-` `1`  `# Then update index -1 to mid-1` `            ``elif` `index ``=``=` `i ``and` `mid < n ``-` `1` `and` `arr[mid ``+` `1``] ``=``=` `arr[i]:``                ``index ``=` `mid ``+` `1`  `# Then update index-1 to mid+1` `            ``break``        ``elif` `Pairsum > Restsum:``            ``# If pair sum is greater than rest sum, our index will``            ``# be in the Range [mid+1,R]``            ``l ``=` `mid ``+` `1``        ``else``:``            ``# If pair sum is smaller than rest sum, our index will``            ``# be in the Range [L,mid-1]``            ``r ``=` `mid ``-` `1` `    ``# Return index=-1 if a pair not exist with arr[i]``    ``# else return index such that arr[i]+arr[index] == sum of rest of arr``    ``return` `index` `# Function to check if a pair exists such that their sum``# equals to rest of the array or not`  `def` `checkPair(arr, n):``    ``Totalsum ``=` `0``    ``arr ``=` `sorted``(arr)  ``# Sort arr for Binary search` `    ``for` `i ``in` `range``(n):``        ``Totalsum ``+``=` `arr[i]  ``# Finding total sum of the arr` `    ``for` `i ``in` `range``(n):``        ``# If index is -1, means arr[i] can't pair with any element``        ``# else arr[i]+a[index] == sum of rest of the arr``        ``index ``=` `binarysearch(arr, n, i, Totalsum)` `        ``if` `index !``=` `-``1``:``            ``print``(``"Pair elements are"``, arr[i], ``"and"``, arr[index])``            ``return` `True` `    ``return` `False`  `# Return false if a pair not exist`  `# Driver Code``arr ``=` `[``2``, ``11``, ``5``, ``1``, ``4``, ``7``]``n ``=` `len``(arr)` `# Function call``if` `checkPair(arr, n) ``=``=` `False``:``    ``print``(``"No pair found"``)`

## C#

 `using` `System;` `class` `GFG``{``  ``// Function to Find if a index exist in array such that``  ``// arr[index] + a[i] == Rest sum of the array``  ``static` `int` `BinarySearch(``int``[] arr, ``int` `n, ``int` `i, ``int` `totalSum)``  ``{``    ``int` `l = 0, r = n - 1, index = -1; ``// initialize as -1` `    ``while` `(l <= r)``    ``{``      ``int` `mid = (l + r) / 2;` `      ``int` `pairSum = arr[mid] + arr[i]; ``// pair sum``      ``int` `restSum = totalSum - pairSum; ``// rest sum` `      ``if` `(pairSum == restSum)``      ``{``        ``if` `(index != i) ``// checking a pair has same``          ``// position or not``        ``{``          ``index = mid;``        ``} ``// Then update index -1 to mid` `        ``// Checking for adjacent element``        ``else` `if` `(index == i && mid > 0``                 ``&& arr[mid - 1] == arr[i])``        ``{``          ``index = mid - 1;``        ``} ``// Then update index -1 to mid-1` `        ``else` `if` `(index == i && mid < n - 1``                 ``&& arr[mid + 1] == arr[i])``        ``{``          ``index = mid + 1;``        ``} ``// Then update index-1 to mid+1``        ``break``;``      ``}``      ``else` `if` `(pairSum > restSum)``      ``{``        ``// If pair sum is greater than rest sum ,``        ``// our index will be in the Range [mid+1,R]``        ``l = mid + 1;``      ``}``      ``else``      ``{``        ``// If pair sum is smaller than rest sum ,``        ``// our index will be in the Range [L,mid-1]``        ``r = mid - 1;``      ``}``    ``}``    ``// return index=-1 if a pair not exist with arr[i]``    ``// else return index such that arr[i]+arr[index] ==``    ``// sum of rest of arr``    ``return` `index;``  ``}` `  ``// Function to check if a pair exist such their sum``  ``// equal to rest of the array or not``  ``static` `bool` `CheckPair(``int``[] arr, ``int` `n)``  ``{``    ``int` `totalSum = 0;``    ``Array.Sort(arr); ``// sort arr for Binary search` `    ``for` `(``int` `i = 0; i < n; i++)``    ``{``      ``totalSum += arr[i];``    ``} ``// Finding total sum of the arr` `    ``for` `(``int` `i = 0; i < n; i++)``    ``{``      ``// If index is -1 , Means arr[i] can't pair with``      ``// any element else arr[i]+a[index] == sum of``      ``// rest of the arr``      ``int` `index = BinarySearch(arr, n, i, totalSum);` `      ``if` `(index != -1)``      ``{``        ``Console.WriteLine(``"Pair elements are "` `+ arr[i] + ``" and "` `+ arr[index]);``        ``return` `true``;``      ``}``    ``}``    ``return` `false``; ``// Return false if a pair not exist``  ``}` `  ``// Driver Code``  ``static` `void` `Main(``string``[] args)``  ``{``    ``int``[] arr = { 2, 11, 5, 1, 4, 7 };``    ``int` `n = arr.Length;` `    ``// Function call``    ``if` `(!CheckPair(arr, n))``    ``{``      ``Console.WriteLine(``"No pair found"``);``    ``}``  ``}``}`

## Javascript

 `// JavaScript program for the above approach``// function to find if a index exist in array such that``// arr[index] + a[i] == Rest sum of the array` `function` `binarysearch(arr, n, i, TotalSum){``    ``let l = 0;``    ``let r = n-1;``    ``let index = -1;``    ` `    ``while``(l <= r){``        ``let mid = parseInt((l+r)/2);``        ``let Pairsum = arr[mid] + arr[i];``        ``let Restsum = TotalSum - Pairsum;``        ` `        ``if` `( Pairsum == Restsum )``        ``{``            ``if``( index != i )``// checking a pair has same position or not``            ``{  index = mid; }``//Then update index -1 to mid``               ` `            ``// Checking for adjacent element``            ``else` `if``(index == i && mid>0 && arr[mid-1]==arr[i])``            ``{  index = mid-1; }``//Then update index -1 to mid-1``            ` `            ``else` `if``(index == i && mid Restsum)``        ``{ ``// If pair sum is greater than rest sum , our index will``          ``// be in the Range [mid+1,R]``            ``l = mid + 1;``        ``}``        ``else` `{``          ``// If pair sum is smaller than rest sum , our index will``          ``// be in the Range [L,mid-1]``            ``r = mid - 1;``        ``}``    ``}``    ``// return index=-1 if a pair not exist with arr[i]``    ``// else return index such that arr[i]+arr[index] == sum of rest of arr``    ``return` `index;``}` `// Function to check if a pair exist such their sum``// equal to rest of the array or not``function` `checkPair(arr, n){``    ``let Totalsum = 0;``    ``arr.sort(``function``(a, b){``return` `a - b});``    ` `    ``for``(let i=0;i

Output

`Pair elements are 11 and 4`

Time complexity : O(n * logn)
Auxiliary Space: O(1)

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