Find array elements equal to sum of any subarray of at least size 2

Given an array arr[], the task is to find the elements from the array which are equal to the sum of any sub-array of size greater than 1.
Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6} 
Output: 3, 5, 6 
Explanation:
The elements 3, 5, 6 are equal to sum of subarrays {1, 2},{2, 3} and {1, 2, 3} respectively.
Input: arr[] = {5, 6, 10, 1, 3, 4, 8, 16} 
Output: 4, 8, 16 
Explanation:
The elements 4, 8, 16 are equal to the sum of subarrays {1, 3}, {1, 3, 4}, {1, 3, 4, 8} respectively 

Approach: The idea is to use a prefix sum array to solve the given problem. Create a prefix[] array that stores the prefix sum of all its preceding elements for every index. Store the sum of all subarrays in a map and search if any array element is present in the map or not.
Below is the implementation of the above approach: 

C++

// C++ implementation to find array
// elements equal to the sum
// of any subarray of at least size 2

#include <bits/stdc++.h>
using namespace std;

// Function to find all elements
void findElements(int* arr, int n)
{
    if (n == 1)
        return;

    int pre[n];

    // Create a prefix array
    pre[0] = arr[0];

    for (int i = 1; i < n; i++)
        pre[i] = arr[i] + pre[i - 1];

    unordered_map<int, bool> mp;

    // Mark the sum of all sub arrays as true
    for (int i = 1; i < n - 1; i++) {
        mp[pre[i]] = true;

        for (int j = i + 1; j < n; j++) {
            mp[pre[j] - pre[i - 1]] = true;
        }
    }

    // Check all elements
    // which are marked
    // true in the map
    for (int i = 0; i < n; i++) {
        if (mp[arr[i]]) {
            cout << arr[i] << " ";
        }
    }
    cout << endl;
}

// Driver Code
int main()
{
    int arr1[] = { 1, 2, 3, 4, 5, 6 };

    int n1 = sizeof(arr1) / sizeof(arr1[0]);

    findElements(arr1, n1);

    return 0;
}

Java

// Java implementation to find array
// elements equal to the sum of any
// subarray of at least size 2
import java.util.*;

class GFG{

// Function to find all elements
static void findElements(int[] arr, int n)
{
    if (n == 1)
        return;

    int pre[] = new int[n];

    // Create a prefix array
    pre[0] = arr[0];

    for(int i = 1; i < n; i++)
        pre[i] = arr[i] + pre[i - 1];

    HashMap<Integer, Boolean> mp = new HashMap<>();

    // Mark the sum of all sub arrays as true
    for(int i = 1; i < n - 1; i++)
    {
        mp.put(pre[i], true);

        for(int j = i + 1; j < n; j++)
        {
            mp.put(pre[j] - pre[i - 1], true);
        }
    }

    // Check all elements
    // which are marked
    // true in the map
    for(int i = 0; i < n; i++)
    {
        if (mp.get(arr[i]) != null)
        {
            System.out.print(arr[i] + " ");
        }
    }
    System.out.println();
}

// Driver Code
public static void main(String[] args)
{
    int arr1[] = { 1, 2, 3, 4, 5, 6 };

    int n1 = arr1.length;

    findElements(arr1, n1);
}
}

// This code is contributed by jrishabh99

Python3

# Python3 implementation to find array
# elements equal to the sum of any
# subarray of at least size 2

# Function to find all elements
def findElements(arr, n):

    if (n == 1):
        return;

    pre = [0] * n;

    # Create a prefix array
    pre[0] = arr[0];

    for i in range(1, n):
        pre[i] = arr[i] + pre[i - 1];

    mp = {};

    # Mark the sum of all sub arrays as true
    for i in range(1, n - 1):
        mp[pre[i]] = True;

        for j in range(i + 1, n):
            mp[pre[j] - pre[i - 1]] = True;

    # Check all elements which
    # are marked true in the map
    for i in range(n):
        if arr[i] in mp:
            print(arr[i], end = " ");
        else:
            pass

    print()

# Driver Code
if __name__ == "__main__":

    arr1 = [ 1, 2, 3, 4, 5, 6 ];
    n1 = len(arr1);

    findElements(arr1, n1);

# This code is contributed by AnkitRai01

C#

// C# implementation to find array
// elements equal to the sum of any
// subarray of at least size 2
using System;
using System.Collections.Generic;
class GFG{

// Function to find all elements
static void findElements(int[] arr,
                         int n)
{
  if (n == 1)
    return;

  int []pre = new int[n];

  // Create a prefix array
  pre[0] = arr[0];

  for(int i = 1; i < n; i++)
    pre[i] = arr[i] + pre[i - 1];

  Dictionary<int,
             Boolean> mp = new Dictionary<int,
                                          Boolean>();

  // Mark the sum of all sub arrays as true
  for(int i = 1; i < n - 1; i++)
  {
    if(!mp.ContainsKey(pre[i]))
      mp.Add(pre[i], true);

    for(int j = i + 1; j < n; j++)
    {
      if(!mp.ContainsKey(pre[j] - pre[i - 1]))
        mp.Add(pre[j] - pre[i - 1], true);
    }
  }

  // Check all elements
  // which are marked
  // true in the map
  for(int i = 0; i < n; i++)
  {
    if (mp.ContainsKey(arr[i]))
    {
      Console.Write(arr[i] + " ");
    }
  }
  Console.WriteLine();
}

// Driver Code
public static void Main(String[] args)
{
  int []arr1 = {1, 2, 3, 4, 5, 6};
  int n1 = arr1.Length;
  findElements(arr1, n1);
}
}

// This code is contributed by gauravrajput1

Javascript

<script>

// Javascript implementation to find array
// elements equal to the sum of any
// subarray of at least size 2

// Function to find all elements
function findElements(arr, n)
{
    if (n == 1)
        return;

    let pre = Array.from({length: n}, (_, i) => 0);

    // Create a prefix array
    pre[0] = arr[0];

    for(let i = 1; i < n; i++)
        pre[i] = arr[i] + pre[i - 1];

    let mp = new Map();

    // Mark the sum of all sub arrays as true
    for(let i = 1; i < n - 1; i++)
    {
        mp.set(pre[i], true);

        for(let j = i + 1; j < n; j++)
        {
            mp.set(pre[j] - pre[i - 1], true);
        }
    }

    // Check all elements
    // which are marked
    // true in the map
    for(let i = 0; i < n; i++)
    {
        if (mp.get(arr[i]) != null)
        {
            document.write(arr[i] + " ");
        }
    }
    document.write("<br/>");
}

// Driver code

      let arr1 = [ 1, 2, 3, 4, 5, 6 ];

    let n1 = arr1.length;

    findElements(arr1, n1);

 // This code is contributed by souravghosh0416.
</script>

Output:

3 5 6

Time Complexity: O(N²)

Auxiliary Space: O(N)