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Find if array has an element whose value is half of array sum
  • Difficulty Level : Basic
  • Last Updated : 09 Jun, 2021

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++




// 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;
}

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..

Python 3




# 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




<?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
?>

Javascript




<script>
// Javascript 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
    let sum = 0;
    for (let i = 0; i < n; i++)
        sum += array[i];
 
    // If sum is odd
    if (sum % 2)
    return false;
 
    sum = Math.floor(sum / 2); // If sum is Even
 
    // Do binary search for the
    // required element
    let start = 0;
    let end = n - 1;
    while (start <= end)
    {
        let mid = Math.floor(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
let array = new Array(1, 2, 3 );
let n = array.length;
if (checkForElement(array, n))
    document.write("Yes");
else
    document.write("No");
     
// This code is contributed by _saurabh_jaiswal
</script>
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++




// 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;
}

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<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

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<int> s = new HashSet<int>();
        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
Output: 
Yes

 

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

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