查找数组是否有一个元素的值是数组总和的一半

给定一个已排序的数组（具有唯一条目），我们必须找到是否存在一个元素（比如 X），它恰好是包括 X 在内的数组所有元素之和的一半。

例子：

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. 计算数组所有元素的总和。
2.可以有两种情况
…。一种。 Sum 是奇数，意味着我们找不到这样的 X，因为所有条目都是整数。
….b. Sum 是 Even，如果 sum 值的一半存在于数组中，则答案为 YES 否则为 NO。
3. 我们可以使用二分查找来查找数组中是否存在 sum/2（因为它没有重复的条目）

下面是上述方法的实现：

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

// 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

输出：

Yes

时间复杂度： O(n)
辅助空间： O(1)

另一个适用于未排序数组的有效解决方案
这个想法是使用散列。

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

蟒蛇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
    # 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

 
// 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
    // and storing in a hash table
    let s = new Set();
    let sum = 0;
     
    for(let 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.has(sum / 2))
    {
        return true;
    }
    else
    {
        return false;
    }
}
 
// Driver code
let array = [ 1, 2, 3 ];
let n = array.length;
 
document.write(
    checkForElement(array, n) ? "Yes" : "No");
 
// This code is contributed by rag2127
 

输出：

Yes

时间复杂度： O(n)
辅助空间： O(n)

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