检查数组是否有一个元素等于所有剩余元素的总和

给定一个包含 N 个元素的数组，任务是检查该数组是否有一个元素等于所有剩余元素的总和。
例子：

Input: a[] = {5, 1, 2, 2}
Output: Yes
we can write 5=(1+2+2)

Input: a[] = {2, 1, 2, 4, 3}
Output: No

方法：假设数组中的总元素为N 。现在，如果存在任何这样的元素，使得该元素等于剩余元素的总和，那么可以说该数组可以分为具有相等总和的两半，这样一半只有一个值为 sum/2 的元素.
此外，由于两半的总和相等，数组的总和必须是偶数，正如我们所知：

• 奇数 + 奇数 = 偶数
• 偶数 + 偶数 = 偶数

算法：

• 迭代数组，并计算所有元素的出现并存储在地图中。还要对数组元素求和。
• 问题中给出的条件只有在满足以下条件时才有可能。
  1. 数组的总和是偶数
  2. 数组中出现的 sum/2 应至少等于 1。
• 如果不满足上述条件，则无法删除任何此类元素。

下面是上述方法的实现：

C++

// C++ program to Check if the array
// has an element which is equal to sum
// of all the remaining elements
 
#include 
using namespace std;
 
// Function to check if such element exists or not
bool isExists(int a[], int n)
{
    // Storing frequency in map
    unordered_map freq;
 
    // Stores the sum
    int sum = 0;
 
    // Traverse the array and count the
    // array elements
    for (int i = 0; i < n; i++) {
        freq[a[i]]++;
        sum += a[i];
    }
 
    // Only possible if sum is even
    if (sum % 2 == 0) {
        // If half sum is available
        if (freq[sum / 2])
            return true;
    }
 
    return false;
}
 
// Driver code
int main()
{
    int a[] = { 5, 1, 2, 2 };
 
    int n = sizeof(a) / sizeof(a[0]);
 
    if (isExists(a, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java program to Check if the array
// has an element which is equal to sum
// of all the remaining elements
import java.util.*;
class Solution{
 
// Function to check if such element exists or not
static boolean isExists(int a[], int n)
{
    // Storing frequency in map
    Map freq= new HashMap();
   
    // Stores the sum
    int sum = 0;
   
    // Traverse the array and count the
    // array elements
    for (int i = 0; i < n; i++) {
        freq.put(a[i],freq.get(a[i])==null?0:freq.get(a[i])+1);
        sum += a[i];
    }
   
    // Only possible if sum is even
    if (sum % 2 == 0) {
        // If half sum is available
        if (freq.get(sum / 2)!=null)
            return true;
    }
   
    return false;
}
   
// Driver code
public static void main(String args[])
{
    int a[] = { 5, 1, 2, 2 };
   
    int n = a.length;
   
    if (isExists(a, n))
        System.out.println( "Yes");
    else
        System.out.println( "No");
   
}
}
//contributed by Arnab Kundu

Python3

# Python3 code to check if the array has
# an element which is equal to sum of all
# the remaining elements
 
# function to check if such element
# exists or not
def isExists(a, n):
     
    # storing frequency in dict
    freq = {i : 0 for i in a}
     
    #stores the sum
    Sum = 0
     
    # traverse the array and count
    # the array element
    for i in range(n):
        freq[a[i]] += 1
        Sum += a[i]
     
    # Only possible if sum is even
    if Sum % 2 == 0:
         
        #if half sum is available
        if freq[Sum // 2]:
            return True
    return False
     
# Driver code
a = [5, 1, 2, 2]
 
n = len(a)
 
if isExists(a, n):
    print("Yes")
else:
    print("No")
 
# This code is contributed
# by Mohit Kumar

C#

// C# program to Check if the array
// has an element which is equal to sum
// of all the remaining elements
using System;
using System.Collections.Generic;
     
class Solution
{
 
// Function to check if such element exists or not
static Boolean isExists(int []arr, int n)
{
    // Storing frequency in map
    Dictionary m = new Dictionary();
     
    // Stores the sum
    int sum = 0;
     
    // Traverse the array and count the
    // array elements
    for (int i = 0; i < n; i++)
    {
        if(m.ContainsKey(arr[i]))
        {
            var val = m[arr[i]];
            m.Remove(arr[i]);
            m.Add(arr[i], val + 1);
        }
        else
        {
            m.Add(arr[i], 1);
        }
        sum += arr[i];
    }
     
    // Only possible if sum is even
    if (sum % 2 == 0)
    {
        // If half sum is available
        if (m[sum / 2] != 0)
            return true;
    }
     
    return false;
}
     
// Driver code
public static void Main()
{
    int []a = { 5, 1, 2, 2 };
     
    int n = a.Length;
     
    if (isExists(a, n))
        Console.WriteLine( "Yes");
    else
        Console.WriteLine( "No");
}
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

 
// JavaScript program to Check if the array
// has an element which is equal to sum
// of all the remaining elements
 
// Function to check if such element exists or not
function isExists(a, n)
{
    // Storing frequency in map
    let freq= new Map();
   
    // Stores the sum
    let sum = 0;
   
    // Traverse the array and count the
    // array elements
    for (let i = 0; i < n; i++) {
        freq.set(a[i],freq.get(a[i])==null?0:freq.get(a[i])+1);
        sum += a[i];
    }
   
    // Only possible if sum is even
    if (sum % 2 == 0) {
        // If half sum is available
        if (freq.get(sum / 2)!=null)
            return true;
    }
   
    return false;
}
     
// Driver Code
 
           let a = [ 5, 1, 2, 2 ];
   
    let n = a.length;
   
    if (isExists(a, n))
        document.write( "Yes");
    else
        document.write( "No");
         

输出：

Yes

时间复杂度： O(N * log N)
辅助空间： O(N)
另一种方法：计算总数 = 数组中所有元素的总和。然后运行一个 FOR 循环来检查每个元素是否 * 2 == 总数。如果找到任何此类元素，则在循环结束时返回 True，否则返回 False。时间复杂度 = O(N)，空间复杂度 = O(1)。