# Check if the array has an element which is equal to sum of all the remaining elements

• Difficulty Level : Easy
• Last Updated : 13 Jan, 2023

Given an array of N elements, the task is to check if the array has an element that is equal to the sum of all the remaining elements.

Examples

```Input: a[] = {5, 1, 2, 2}
Output: Yes
we can write 5=(1+2+2)

Input: a[] = {2, 1, 2, 4, 3}
Output: No```

Approach: Suppose that the total elements in the array is N. Now, if there exists any such element such that the element is equal to the sum of remaining elements then it can be said that the array can be divided into two halves with equal sum such that one half has only one element with value sum/2.

Also, since both halves have equal sum, the overall sum of the array must be even as we know that:

• ODD + ODD = EVEN
• EVEN + EVEN = EVEN

Algorithm

• Iterate over the array, and count the occurrence of all the elements and store in a map. Also summate the array elements.
• The condition given in the problem is only possible when the below conditions are met.
1. Total Sum of the array is even
2. sum/2 occurrence in the array should be equal to atleast 1.
• If the above conditions are not met, hence it is not possible to remove any such element.

Below is the implementation of the above approach:

## C++

 `// C++ program to Check if the array``// has an element which is equal to sum``// of all the remaining elements` `#include ``using` `namespace` `std;` `// Function to check if such element exists or not``bool` `isExists(``int` `a[], ``int` `n)``{``    ``// Storing frequency in map``    ``unordered_map<``int``, ``int``> freq;` `    ``// Stores the sum``    ``int` `sum = 0;` `    ``// Traverse the array and count the``    ``// array elements``    ``for` `(``int` `i = 0; i < n; i++) {``        ``freq[a[i]]++;``        ``sum += a[i];``    ``}` `    ``// Only possible if sum is even``    ``if` `(sum % 2 == 0) {``        ``// If half sum is available``        ``if` `(freq[sum / 2])``            ``return` `true``;``    ``}` `    ``return` `false``;``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 5, 1, 2, 2 };` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);` `    ``if` `(isExists(a, n))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java program to Check if the array``// has an element which is equal to sum``// of all the remaining elements``import` `java.util.*;``class` `Solution{` `// Function to check if such element exists or not``static` `boolean` `isExists(``int` `a[], ``int` `n)``{``    ``// Storing frequency in map``    ``Map freq= ``new` `HashMap();``  ` `    ``// Stores the sum``    ``int` `sum = ``0``;``  ` `    ``// Traverse the array and count the``    ``// array elements``    ``for` `(``int` `i = ``0``; i < n; i++) {``        ``freq.put(a[i],freq.get(a[i])==``null``?``0``:freq.get(a[i])+``1``);``        ``sum += a[i];``    ``}``  ` `    ``// Only possible if sum is even``    ``if` `(sum % ``2` `== ``0``) {``        ``// If half sum is available``        ``if` `(freq.get(sum / ``2``)!=``null``)``            ``return` `true``;``    ``}``  ` `    ``return` `false``;``}``  ` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `a[] = { ``5``, ``1``, ``2``, ``2` `};``  ` `    ``int` `n = a.length;``  ` `    ``if` `(isExists(a, n))``        ``System.out.println( ``"Yes"``);``    ``else``        ``System.out.println( ``"No"``);``  ` `}``}``//contributed by Arnab Kundu`

## Python3

 `# Python3 code to check if the array has``# an element which is equal to sum of all``# the remaining elements` `# function to check if such element``# exists or not``def` `isExists(a, n):``    ` `    ``# storing frequency in dict``    ``freq ``=` `{i : ``0` `for` `i ``in` `a}``    ` `    ``#stores the sum``    ``Sum` `=` `0``    ` `    ``# traverse the array and count``    ``# the array element``    ``for` `i ``in` `range``(n):``        ``freq[a[i]] ``+``=` `1``        ``Sum` `+``=` `a[i]``    ` `    ``# Only possible if sum is even``    ``if` `Sum` `%` `2` `=``=` `0``:``        ` `        ``#if half sum is available``        ``if` `freq[``Sum` `/``/` `2``]:``            ``return` `True``    ``return` `False``    ` `# Driver code``a ``=` `[``5``, ``1``, ``2``, ``2``]` `n ``=` `len``(a)` `if` `isExists(a, n):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)` `# This code is contributed``# by Mohit Kumar`

## C#

 `// C# program to Check if the array``// has an element which is equal to sum``// of all the remaining elements``using` `System;``using` `System.Collections.Generic;``    ` `class` `Solution``{` `// Function to check if such element exists or not``static` `Boolean isExists(``int` `[]arr, ``int` `n)``{``    ``// Storing frequency in map``    ``Dictionary<``int``, ``int``> m = ``new` `Dictionary<``int``, ``int``>();``    ` `    ``// Stores the sum``    ``int` `sum = 0;``    ` `    ``// Traverse the array and count the``    ``// array elements``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if``(m.ContainsKey(arr[i]))``        ``{``            ``var` `val = m[arr[i]];``            ``m.Remove(arr[i]);``            ``m.Add(arr[i], val + 1);``        ``}``        ``else``        ``{``            ``m.Add(arr[i], 1);``        ``}``        ``sum += arr[i];``    ``}``    ` `    ``// Only possible if sum is even``    ``if` `(sum % 2 == 0)``    ``{``        ``// If half sum is available``        ``if` `(m[sum / 2] != 0)``            ``return` `true``;``    ``}``    ` `    ``return` `false``;``}``    ` `// Driver code``public` `static` `void` `Main()``{``    ``int` `[]a = { 5, 1, 2, 2 };``    ` `    ``int` `n = a.Length;``    ` `    ``if` `(isExists(a, n))``        ``Console.WriteLine( ``"Yes"``);``    ``else``        ``Console.WriteLine( ``"No"``);``}``}` `/* This code contributed by PrinciRaj1992 */`

## Javascript

 ``

Output

`Yes`

Complexity Analysis:

• Time Complexity: O(N)
• Auxiliary Space: O(N)

Another Approach: Calculate the total sum of all the elements in the array. Then run a FOR-loop to check if each element * 2 == total. If any such element is found, return True, else False at the end of the loop.

• Traverse the array and calculate the total sum of array elements
• Check if the array has an element that is equal to the sum of all the remaining elements.
• If true, return true,
• Return false.

Below is the implementation of the above approach:

## C++

 `// C++ program to Check if the array``// has an element which is equal to sum``// of all the remaining elements` `#include ``using` `namespace` `std;` `// Function to check if such element exists or not``bool` `isExists(``int` `arr[], ``int` `n)``{``    ``// Stores the sum``    ``int` `sum = 0;` `    ``// Traverse the array and calculate the sum``    ``// of array elements``    ``for` `(``int` `i = 0; i < n; i++) {``        ``sum += arr[i];``    ``}` `    ``// Check if the array has an element that is equal to``    ``// the sum of all the remaining elements.``    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(sum - arr[i] == arr[i])``            ``return` `true``;``    ``}` `    ``return` `false``;``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 5, 1, 2, 2 };` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);` `    ``if` `(isExists(a, n))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java program to Check if the array``// has an element which is equal to sum``// of all the remaining elements` `import` `java.util.*;` `public` `class` `GFG {` `    ``// Function to check if such element exists or not``    ``static` `boolean` `isExists(``int` `arr[], ``int` `n)``    ``{``        ``// Stores the sum``        ``int` `sum = ``0``;` `        ``// Traverse the array and calculate the sum``        ``// of array elements``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``sum += arr[i];``        ``}` `        ``// Check if the array has an element that is equal``        ``// to the sum of all the remaining elements.``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(sum - arr[i] == arr[i])``                ``return` `true``;``        ``}` `        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a[] = { ``5``, ``1``, ``2``, ``2` `};` `        ``int` `n = a.length;` `        ``if` `(isExists(a, n))``            ``System.out.println(``"Yes"``);``        ``else``            ``System.out.println(``"No"``);``    ``}``}` `// This code is contributed by Karandeep1234`

## Python3

 `# Python program to Check if the array``# has an element which is equal to sum``# of all the remaining elements` `# Function to check if such element exists or not``def` `isExists(arr, n):``    ``# Stores the sum``    ``sum` `=` `0` `    ``# Traverse the array and calculate the sum``    ``# of array elements``    ``for` `i ``in` `range``(n):``        ``sum` `+``=` `arr[i]` `    ``# Check if the array has an element that is equal to``    ``# the sum of all the remaining elements.``    ``for` `i ``in` `range``(n):``        ``if` `sum``-``arr[i] ``=``=` `arr[i]: ``return` `True` `    ``return` `False` `# Driver code``a ``=` `[``5``, ``1``, ``2``, ``2``]``n ``=` `len``(a)``if` `isExists(a,n): ``print``(``'Yes'``)``else``: ``print``(``'No'``)` `# This code is contributed by hardikkushwaha.`

## C#

 `// C# program to Check if the array``// has an element which is equal to sum``// of all the remaining elements` `using` `System;` `public` `class` `GFG {` `    ``// Function to check if such element exists or not``    ``static` `bool` `isExists(``int``[] arr, ``int` `n)``    ``{``        ``// Stores the sum``        ``int` `sum = 0;` `        ``// Traverse the array and calculate the sum``        ``// of array elements``        ``for` `(``int` `i = 0; i < n; i++) {``            ``sum += arr[i];``        ``}` `        ``// Check if the array has an element that is equal``        ``// to the sum of all the remaining elements.``        ``for` `(``int` `i = 0; i < n; i++) {``            ``if` `(sum - arr[i] == arr[i])``                ``return` `true``;``        ``}` `        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``int``[] a = { 5, 1, 2, 2 };` `        ``int` `n = a.Length;` `        ``if` `(isExists(a, n))``            ``Console.WriteLine(``"Yes"``);``        ``else``            ``Console.WriteLine(``"No"``);``    ``}``}` `// This code is contributed by Karandeep1234`

## Javascript

 `// Javascript program to Check if the array``// has an element which is equal to sum``// of all the remaining elements` `// Function to check if such element exists or not``function` `isExists(arr, n)``{``    ``// Stores the sum``    ``let sum = 0;` `    ``// Traverse the array and calculate the sum``    ``// of array elements``    ``for` `(let i = 0; i < n; i++) {``        ``sum += arr[i];``    ``}` `    ``// Check if the array has an element that is equal to``    ``// the sum of all the remaining elements.``    ``for` `(let i = 0; i < n; i++) {``        ``if` `(sum - arr[i] == arr[i])``            ``return` `true``;``    ``}` `    ``return` `false``;``}` `// Driver code``let a = [ 5, 1, 2, 2 ];``let n = a.length;` `if` `(isExists(a, n))``    ``console.log(``"Yes"``);``else``    ``console.log(``"No"``);` `// This code is contributed by poojaagarwals.`

Output

`Yes`

Time Complexity: O(N), where N is the length of the given array.
Auxiliary Space: O(1)

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