Split the array into equal sum parts according to given conditions

Given an integer array arr[], the task is to check if the input array can be split in two sub-arrays such that:

If possible then print Yes else print No.

Examples:



Input: arr[] = {1, 2}
Output: No
The elements cannot be divided in groups such that there sum is equal.

Input: arr[] = {1, 4, 3}
Output: Yes
{1, 3} and {4} are the groups satisfying the given condition.

Approach: We can use a recursive approach by keeping left sum and right sum to maintain two different groups. Left sum is for multiples of 5 and right sum is for multiples of 3 (which are not multiples of 5) and the elements which are neither divisible by 5 nor by 3 can lie in any group satisfying the equal sum rule (include them in left sum and right sum one by one).

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Recursive function that returns true if the array
// can be divided into two sub-arrays
// satisfying the given condition
bool helper(int* arr, int n, int start, int lsum, int rsum)
{
  
    // If reached the end
    if (start == n)
        return lsum == rsum;
  
    // If divisible by 5 then add to the left sum
    if (arr[start] % 5 == 0)
        lsum += arr[start];
  
    // If divisible by 3 but not by 5
    // then add to the right sum
    else if (arr[start] % 3 == 0)
        rsum += arr[start];
  
    // Else it can be added to any of the sub-arrays
    else
  
        // Try adding in both the sub-arrays (one by one)
        // and check whether the condition satisfies
        return helper(arr, n, start + 1, lsum + arr[start], rsum)
           || helper(arr, n, start + 1, lsum, rsum + arr[start]);
  
    // For cases when element is multiple of 3 or 5.
    return helper(arr, n, start + 1, lsum, rsum);
}
  
// Function to start the recursive calls
bool splitArray(int* arr, int n)
{
    // Initially start, lsum and rsum will all be 0
    return helper(arr, n, 0, 0, 0);
}
  
// Driver code
int main()
{
    int arr[] = { 1, 4, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    if (splitArray(arr, n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}
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// Java implementation of the approach
class Solution
{
  
// Recursive function that returns true if the array
// can be divided into two sub-arrays
// satisfying the given condition
static boolean helper(int arr[], int n,
                    int start, int lsum, int rsum)
{
  
    // If reached the end
    if (start == n)
        return lsum == rsum;
  
    // If divisible by 5 then add to the left sum
    if (arr[start] % 5 == 0)
        lsum += arr[start];
  
    // If divisible by 3 but not by 5
    // then add to the right sum
    else if (arr[start] % 3 == 0)
        rsum += arr[start];
  
    // Else it can be added to any of the sub-arrays
    else
  
        // Try adding in both the sub-arrays (one by one)
        // and check whether the condition satisfies
        return helper(arr, n, start + 1, lsum + arr[start], rsum)
        || helper(arr, n, start + 1, lsum, rsum + arr[start]);
  
    // For cases when element is multiple of 3 or 5.
    return helper(arr, n, start + 1, lsum, rsum);
}
  
// Function to start the recursive calls
static boolean splitArray(int arr[], int n)
{
    // Initially start, lsum and rsum will all be 0
    return helper(arr, n, 0, 0, 0);
}
  
// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 4, 3 };
    int n = arr.length;
  
    if (splitArray(arr, n))
        System.out.println( "Yes");
    else
        System.out.println( "No");
}
}
  
// This code is contributed by Arnab Kundu
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# Python 3 implementation of the approach
  
# Recursive function that returns true if 
# the array can be divided into two sub-arrays
# satisfying the given condition
def helper(arr, n, start, lsum, rsum):
  
    # If reached the end
    if (start == n):
        return lsum == rsum
  
    # If divisible by 5 then add 
    # to the left sum
    if (arr[start] % 5 == 0):
        lsum += arr[start]
  
    # If divisible by 3 but not by 5
    # then add to the right sum
    elif (arr[start] % 3 == 0):
        rsum += arr[start]
  
    # Else it can be added to any of
    # the sub-arrays
    else:
  
        # Try adding in both the sub-arrays 
        # (one by one) and check whether
        # the condition satisfies
        return (helper(arr, n, start + 1
                lsum + arr[start], rsum) or
                helper(arr, n, start + 1
                lsum, rsum + arr[start]));
  
    # For cases when element is multiple of 3 or 5.
    return helper(arr, n, start + 1, lsum, rsum)
  
# Function to start the recursive calls
def splitArray(arr, n):
      
    # Initially start, lsum and rsum 
    # will all be 0
    return helper(arr, n, 0, 0, 0)
  
# Driver code
if __name__ == "__main__":
      
    arr = [ 1, 4, 3 ]
    n = len(arr)
  
    if (splitArray(arr, n)):
        print("Yes")
    else:
        print("No")
  
# This code is contributed by ita_c
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// C# implementation of the approach 
using System;
  
class GFG 
  
    // Recursive function that returns true if the array 
    // can be divided into two sub-arrays 
    // satisfying the given condition 
    static bool helper(int []arr, int n, 
                        int start, int lsum, int rsum) 
    
      
        // If reached the end 
        if (start == n) 
            return lsum == rsum; 
      
        // If divisible by 5 then add to the left sum 
        if (arr[start] % 5 == 0) 
            lsum += arr[start]; 
      
        // If divisible by 3 but not by 5 
        // then add to the right sum 
        else if (arr[start] % 3 == 0) 
            rsum += arr[start]; 
      
        // Else it can be added to any of the sub-arrays 
        else
      
            // Try adding in both the sub-arrays (one by one) 
            // and check whether the condition satisfies 
            return helper(arr, n, start + 1, lsum + arr[start], rsum) 
            || helper(arr, n, start + 1, lsum, rsum + arr[start]); 
      
        // For cases when element is multiple of 3 or 5. 
        return helper(arr, n, start + 1, lsum, rsum); 
    
      
    // Function to start the recursive calls 
    static bool splitArray(int []arr, int n) 
    
        // Initially start, lsum and rsum will all be 0 
        return helper(arr, n, 0, 0, 0); 
    
      
    // Driver code 
    public static void Main() 
    
        int []arr = { 1, 4, 3 }; 
        int n = arr.Length; 
      
        if (splitArray(arr, n)) 
            Console.WriteLine( "Yes"); 
        else
            Console.WriteLine( "No"); 
    }
  
// This code is contributed by Ryuga
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<?php
// PHP implementation of the approach
  
// Recursive function that returns true 
// if the array can be divided into two
// sub-arrays satisfying the given condition
function helper(&$arr, $n, $start
                    $lsum, $rsum)
{
  
    // If reached the end
    if ($start == $n)
        return $lsum == $rsum;
  
    // If divisible by 5 then 
    // add to the left sum
    if ($arr[$start] % 5 == 0)
        $lsum += $arr[$start];
  
    // If divisible by 3 but not by 5
    // then add to the right sum
    else if ($arr[$start] % 3 == 0)
        $rsum += $arr[$start];
  
    // Else it can be added to any 
    // of the sub-arrays
    else
  
        // Try adding in both the sub-arrays (one by one)
        // and check whether the condition satisfies
        return helper($arr, $n, $start + 1, 
                      $lsum + $arr[$start], $rsum) || 
               helper($arr, $n, $start + 1, 
                      $lsum, $rsum + $arr[$start]);
  
    // For cases when element is 
    // multiple of 3 or 5.
    return helper($arr, $n, $start + 1,
                        $lsum, $rsum);
}
  
// Function to start the recursive calls
function splitArray($arr, $n)
{
    // Initially start, lsum and r
    // sum will all be 0
    return helper($arr, $n, 0, 0, 0);
}
  
// Driver code
$arr = array( 1, 4, 3 );
$n = count($arr);
  
if (splitArray($arr, $n))
    print("Yes");
else
    print("No");
  
// This code is contributed by mits
?>
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Output:
Yes

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