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

Given an array of N elements, the task is to check if the array has an element which 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 <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

// 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

# 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 */

Output:

Yes

Time Complexity: O(N * log N)
Auxiliary Space: O(N)

Another Approach: Calculate 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. Time complexity = O(N), space complexity = O(1).

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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