Check if the array has an element which is equal to sum of all the remaining elements
Last Updated :
11 Jul, 2025
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.
- Total Sum of the array is even
- 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 <bits/stdc++.h>
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 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<Integer, Integer> freq= new HashMap<Integer, Integer>();
// 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 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# 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 */
<script>
// 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(a, n)
{
// Storing frequency in map
let freq= new Map();
// Stores the sum
let sum = 0;
// Traverse the array and count the
// array elements
for (let i = 0; i < n; i++) {
freq.set(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
let a = [ 5, 1, 2, 2 ];
let n = a.length;
if (isExists(a, n))
document.write( "Yes");
else
document.write( "No");
</script>
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.
- 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 <bits/stdc++.h>
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 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
# 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# 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 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.
Time Complexity: O(N), where N is the length of the given array.
Auxiliary Space: O(1)
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