检查数组中是否存在两个元素的总和等于数组其余部分的总和

我们有一个整数数组，我们必须在数组中找到两个这样的元素，以使这两个元素的总和等于数组中其余元素的总和。
例子：

Input  : arr[] = {2, 11, 5, 1, 4, 7}
Output : Elements are 4 and 11
Note that 4 + 11 = 2 + 5 + 1 + 7

Input  : arr[] = {2, 4, 2, 1, 11, 15}
Output : Elements do not exist

一个简单的解决方案是逐一考虑每对，找到其总和，然后将总和与其余元素的总和进行比较。如果找到一对总和等于其余元素的对，则打印该对并返回true。该解决方案的时间复杂度为O(n ³ )
一个有效的解决方案是找到所有数组元素的总和。将该总和设为“总和”。现在，任务简化为找到总和等于sum / 2的一对。
另一个优化是，只有在整个数组的和为偶数的情况下，一对配对才能存在，因为我们基本上将其分为相等和的两个部分。
1-找到整个数组的总和。让这个总和为“总和”
2-如果sum为奇数，则返回false。
3-使用此处讨论的基于散列的方法找到对，其总和等于“ sum / 2”的对作为方法2。如果找到对，则将其打印并返回true。
4-如果不存在任何对，则返回false。
下面是上述步骤的实现。

C++

// C++ program to find whether two elements exist
// whose sum is equal to sum of rest of the elements.
#include
using namespace std;
 
// Function to check whether two elements exist
// whose sum is equal to sum of rest of the elements.
bool checkPair(int arr[],int n)
{
    // Find sum of whole array
    int sum = 0;
    for (int i=0; i s;
    for (int i=0; i

---

Java

// Java program to find whether two elements exist
// whose sum is equal to sum of rest of the elements.
import java.util.*;
 
class GFG
{
     
    // Function to check whether two elements exist
    // whose sum is equal to sum of rest of the elements.
    static boolean checkPair(int arr[], int n)
    {
        // Find sum of whole array
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            sum += arr[i];
        }
 
        // If sum of array is not even than we can not
        // divide it into two part
        if (sum % 2 != 0)
        {
            return false;
        }
 
        sum = sum / 2;
 
        // For each element arr[i], see if there is
        // another element with vaalue sum - arr[i]
        HashSet s = new HashSet();
        for (int i = 0; i < n; i++)
        {
            int val = sum - arr[i];
 
            // If element exist than return the pair
            if (s.contains(val) &&
                val == (int) s.toArray()[s.size() - 1])
            {
                System.out.printf("Pair elements are %d and %d\n",
                        arr[i], val);
                return true;
            }
            s.add(arr[i]);
        }
        return false;
    }
 
    // Driver program.
    public static void main(String[] args)
    {
        int arr[] = {2, 11, 5, 1, 4, 7};
        int n = arr.length;
        if (checkPair(arr, n) == false)
        {
            System.out.printf("No pair found");
        }
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 program to find whether
# two elements exist whose sum is
# equal to sum of rest of the elements.
 
# Function to check whether two
# elements exist whose sum is equal
# to sum of rest of the elements.
def checkPair(arr, n):
    s = set()
    sum = 0
 
    # Find sum of whole array
    for i in range(n):
        sum += arr[i]
     
    # / If sum of array is not
    # even than we can not
    # divide it into two part
    if sum % 2 != 0:
        return False
    sum = sum / 2
 
    # For each element arr[i], see if
    # there is another element with
    # value sum - arr[i]
    for i in range(n):
        val = sum - arr[i]
        if arr[i] not in s:
            s.add(arr[i])
             
        # If element exist than
        # return the pair
        if val in s:
            print("Pair elements are",
                   arr[i], "and", int(val))
 
# Driver Code
arr = [2, 11, 5, 1, 4, 7]
n = len(arr)
if checkPair(arr, n) == False:
    print("No pair found")
 
# This code is contributed
# by Shrikant13

C#

// C# program to find whether two elements exist
// whose sum is equal to sum of rest of the elements.
using System;
using System.Collections.Generic;
 
class GFG
{
     
    // Function to check whether two elements exist
    // whose sum is equal to sum of rest of the elements.
    static bool checkPair(int []arr, int n)
    {
        // Find sum of whole array
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            sum += arr[i];
        }
 
        // If sum of array is not even than we can not
        // divide it into two part
        if (sum % 2 != 0)
        {
            return false;
        }
 
        sum = sum / 2;
 
        // For each element arr[i], see if there is
        // another element with vaalue sum - arr[i]
        HashSet s = new HashSet();
        for (int i = 0; i < n; i++)
        {
            int val = sum - arr[i];
 
            // If element exist than return the pair
            if (s.Contains(val))
            {
                Console.Write("Pair elements are {0} and {1}\n",
                        arr[i], val);
                return true;
            }
            s.Add(arr[i]);
        }
        return false;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {2, 11, 5, 1, 4, 7};
        int n = arr.Length;
        if (checkPair(arr, n) == false)
        {
            Console.Write("No pair found");
        }
    }
}
 
// This code contributed by Rajput-Ji

Javascript

 
// Javascript program to find
// whether two elements exist
// whose sum is equal to sum of rest
// of the elements.   
     
    // Function to check whether
    // two elements exist
    // whose sum is equal to sum of
    // rest of the elements.
    function checkPair(arr,n)
    {
        // Find sum of whole array
        let sum = 0;
        for (let i = 0; i < n; i++)
        {
            sum += arr[i];
        }
   
        // If sum of array is not even than we can not
        // divide it into two part
        if (sum % 2 != 0)
        {
            return false;
        }
   
        sum = Math.floor(sum / 2);
   
        // For each element arr[i], see if there is
        // another element with vaalue sum - arr[i]
        let s = new Set();
        for (let i = 0; i < n; i++)
        {
            let val = sum - arr[i];
   
            // If element exist than return the pair
             
            if(!s.has(arr[i]))
            {
                s.add(arr[i])
            }
             
            if (s.has(val) )
            {
                document.write("Pair elements are "+
                        arr[i]+" and "+ val+"<br>");
                return true;
            }
            s.add(arr[i]);
        }
        return false;
    }
    // Driver program.
    let arr=[2, 11, 5, 1, 4, 7];
    let n = arr.length;
    if (checkPair(arr, n) == false)
    {
        document.write("No pair found");
    }
     
    // This code is contributed by rag2127
     

输出：

Pair elements are 4 and 11