# Find if array has an element whose value is half of array sum

Given a sorted array (with unique entries), we have to find whether there exists an element(say X) that is exactly half the sum of all the elements of the array including X.

Examples:

Input : A = {1, 2, 3}
Output : YES
Sum of all the elements is 6 = 3*2;

Input : A = {2, 4}
Output : NO
Sum of all the elements is 6, and 3 is not present in the array.
1. Calculate the sum of all the elements of the array.
2. There can be two cases
1. Sum is Odd, implies we cannot find such X, since all entries are integer.
2. Sum is Even, if half the value of sum exist in array then answer is YES else NO.
3. We can use Binary Search to find if sum/2 exist in array or not (Since it does not have duplicate entries)

Below is the implementation of above approach:

## C++

 // CPP program to check if array has an // element whose value is half of array // sum. #include using namespace std;   // Function to check if answer exists bool checkForElement(int array[], int n) {     // Sum of all array elements     int sum = 0;     for (int i = 0; i < n; i++)         sum += array[i];       // If sum is odd     if (sum % 2)        return false;       sum /= 2; // If sum is Even       // Do binary search for the required element     int start = 0;     int end = n - 1;     while (start <= end)     {         int mid = start + (end - start) / 2;         if (array[mid] == sum)             return true;                else if (array[mid] > sum)             end = mid - 1;                else             start = mid + 1;     }       return false; }   // Driver code int main() {     int array[] = { 1, 2, 3 };     int n = sizeof(array) / sizeof(array[0]);     if (checkForElement(array, n))       cout << "Yes";     else       cout << "No";     return 0; }

## Java

 // Java program to check if array has an // element whose value is half of array // sum.   import java.io.*;   class GFG {   // Function to check if answer exists static boolean checkForElement(int array[], int n) {     // Sum of all array elements     int sum = 0;     for (int i = 0; i < n; i++)         sum += array[i];       // If sum is odd     if (sum % 2>0)     return false;       sum /= 2; // If sum is Even       // Do binary search for the required element     int start = 0;     int end = n - 1;     while (start <= end)     {         int mid = start + (end - start) / 2;         if (array[mid] == sum)             return true;             else if (array[mid] > sum)             end = mid - 1;             else             start = mid + 1;     }       return false; }   // Driver code       public static void main (String[] args) {     int array[] = { 1, 2, 3 };     int n = array.length;     if (checkForElement(array, n))     System.out.println( "Yes");     else     System.out.println( "No");     } } // This code is contributed by anuj_67..

## Python3

 # Python 3 program to check if array # has an element whose value is half # of array sum.   # Function to check if answer exists def checkForElement(array, n):           # Sum of all array elements     sum = 0     for i in range(n):         sum += array[i]       # If sum is odd     if (sum % 2):         return False       sum //= 2     # If sum is Even       # Do binary search for the     # required element     start = 0     end = n - 1     while (start <= end) :         mid = start + (end - start) // 2         if (array[mid] == sum):             return True            elif (array[mid] > sum) :             end = mid - 1;             else:             start = mid + 1       return False   # Driver code if __name__ == "__main__":     array = [ 1, 2, 3 ]     n = len(array)     if (checkForElement(array, n)):         print("Yes")     else:         print("No")   # This code is contributed # by ChitraNayal

## C#

 // C# program to check if array has // an element whose value is half // of array sum. using System;   class GFG { // Function to check if answer exists static bool checkForElement(int[] array,                             int n) {     // Sum of all array elements     int sum = 0;     for (int i = 0; i < n; i++)         sum += array[i];       // If sum is odd     if (sum % 2 > 0)     return false;       sum /= 2; // If sum is Even       // Do binary search for the     // required element     int start = 0;     int end = n - 1;     while (start <= end)     {         int mid = start + (end - start) / 2;         if (array[mid] == sum)             return true;             else if (array[mid] > sum)             end = mid - 1;             else             start = mid + 1;     }       return false; }   // Driver Code static void Main() {     int []array = { 1, 2, 3 };     int n = array.Length;     if (checkForElement(array, n))         Console.WriteLine("Yes");     else         Console.WriteLine("No"); } }   // This code is contributed by ANKITRAI1

## PHP

 \$sum)             \$end = \$mid - 1;             else             \$start = \$mid + 1;     }       return false; }   // Driver code \$array = array(1, 2, 3 ); \$n = sizeof(\$array); if (checkForElement(\$array, \$n))     echo "Yes"; else     echo "No";       // This code is contributed // by Shivi_Aggarwal ?>

## Javascript



Output

Yes

Complexity Analysis:

• Time Complexity: O(n)
• Auxiliary Space: O(1)

Another efficient solution that works for unsorted arrays also
Implementation: The idea is to use hashing.

## C++

 // CPP program to check if array has an // element whose value is half of array // sum. #include using namespace std;   // Function to check if answer exists bool checkForElement(int array[], int n) {     // Sum of all array elements     // and storing in a hash table     unordered_set s;     int sum = 0;     for (int i = 0; i < n; i++) {         sum += array[i];         s.insert(array[i]);     }         // If sum/2 is present in hash table     if (sum % 2 == 0 && s.find(sum/2) != s.end())         return true;     else         return false; }   // Driver code int main() {     int array[] = { 1, 2, 3 };     int n = sizeof(array) / sizeof(array[0]);     if (checkForElement(array, n))       cout << "Yes";     else       cout << "No";     return 0; }

## Java

 // Java program to check if array has an // element whose value is half of array // sum.   import java.util.*;   class GFG {   // Function to check if answer exists     static boolean checkForElement(int array[], int n) {         // Sum of all array elements         // and storing in a hash table         Set s = new LinkedHashSet<>();         int sum = 0;         for (int i = 0; i < n; i++) {             sum += array[i];             s.add(array[i]);         }         // If sum/2 is present in hash table         if (sum % 2 == 0 && s.contains(sum / 2)                 && (sum / 2 )== s.stream().skip(s.size() - 1).findFirst().get()) {             return true;         } else {             return false;         }     }   // Driver code     public static void main(String[] args) {         int array[] = {1, 2, 3};         int n = array.length;         System.out.println(checkForElement(array, n) ? "Yes" : "No");     } } // This code is contributed by 29AjayKumar

## Python3

 # Python 3 program to check if array has an # element whose value is half of array # sum.   # Function to check if answer exists def checkForElement(array, n):     # Sum of all array elements     # and storing in a hash table     s = set()     sum = 0     for i in range(n):         sum += array[i]         s.add(array[i])       # If sum/2 is present in hash table     f = int(sum / 2)     if (sum % 2 == 0 and  f in s):         return True     else:         return False   # Driver code if __name__ == '__main__':     array = [1, 2, 3]     n = len(array)     if (checkForElement(array, n)):         print("Yes")     else:         print("No")   # This code is contributed by # Surendra_Gangwar

## C#

 // C# program to check if array has an // element whose value is half of array // sum. using System; using System.Collections.Generic;   class GFG {       // Function to check if answer exists     static Boolean checkForElement(int []array, int n)     {         // Sum of all array elements         // and storing in a hash table         HashSet s = new HashSet();         int sum = 0;         for (int i = 0; i < n; i++)         {             sum += array[i];             s.Add(array[i]);         }                   // If sum/2 is present in hash table         if (sum % 2 == 0 && s.Contains(sum / 2))         {             return true;         }         else         {             return false;         }     }       // Driver code     public static void Main(String[] args)     {         int []array = {1, 2, 3};         int n = array.Length;         Console.WriteLine(checkForElement(array, n) ? "Yes" : "No");     } }   // This code is contributed by Princi Singh

## Javascript



Output

Yes

Complexity Analysis:

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

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