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

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.
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

// 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# 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

<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>

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 <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

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

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