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 exist an element(say X) which 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
….a. Sum is Odd, implies we cannot find such X, since all entries are integer.
….b. 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++

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// CPP program to check if array has an
// element whose value is half of array
// sum.
#include <bits/stdc++.h>
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;
}

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Java

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// 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..

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Python 3

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# 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

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C#

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// 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

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PHP

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<?php
// PHP program to check if array has an
// element whose value is half of array
// sum.
  
// Function to check if answer exists
function checkForElement(&$array, $n)
{
    // Sum of all array elements
    $sum = 0;
    for ($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
    $start = 0;
    $end = $n - 1;
    while ($start <= $end
    {
        $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
$array = array(1, 2, 3 );
$n = sizeof($array);
if (checkForElement($array, $n))
    echo "Yes";
else
    echo "No";
      
// This code is contributed 
// by Shivi_Aggarwal 
?>

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Output:

Yes

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

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

C++

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// CPP program to check if array has an
// element whose value is half of array
// sum.
#include <bits/stdc++.h>
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<int> 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;
}

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Java

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// 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<Integer> 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

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Python3

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# 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

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Output:

Yes

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



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