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:
- Sum of both the sub-arrays is equal.
- All the elements which are divisible by 5 should be in the same group.
- All the elements which are divisible by 3 (but not divisible by 5) should be in the other group.
- Elements which are neither divisible by 5 nor by 3 can be put in any group.
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:
C++
// 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; } |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
PHP
<?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 ?> |
Output:
Yes
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
