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++ 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 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 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# 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 |
<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> |
3 5 6
Time Complexity: O(N2)
Auxiliary Space: O(N)