Check if it is possible to split given Array into K odd-sum subsets

0 0 4 minutes read

Given an array arr[] of length N, the task is to check if it is possible to split the given array into K non-empty and non-intersecting subsets such that the sum of elements of each subset is odd.
Examples:

Input: K = 4, arr[] = {1, 3, 4, 7, 5, 3, 1}
Output: Yes
Explanation:
[1], [3, 4, 7, 5], [3] and [1] are the possible subsets.
Input: K = 3, arr[] = {2, 3, 4, 7, 2}
Output: No
Explanation:
Given array cannot be split into 3 subset with odd sum.

Approach:
To solve the problem mentioned above we need to observe the following points:

Even numbers don’t change the parity of the sum of subsets, so we can ignore them.
If number of odd integers in the array is less than K, then we cannot split it into K subsets with odd sums since there are not enough odd integers.
Let number of odd integers be cnt. Then, the answer will always be possible only if cnt % 2 = K % 2. This is because we will distribute one odd number in the first K-1 subsets and cnt – K – 1 odd numbers in the last subset. Now since cnt and K have the same parity so cnt – K – 1 will be odd and the sum is also odd.

Hence, to solve the problem, count the number of odd integers present in the array. Let this be cnt. The answer will be ‘Yes’ if cnt is greater than K and cnt % 2 = K % 2. Otherwise, an answer is not possible and we print ‘No’.
Below is the implementation of the above approach:

C++

// C++ implementation to check if it is
// possible to split array into K
// subsets with odd sum
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if array
// can be split in required K
// subsets
bool checkArray(int n, int k, int arr[])
{
    // Store count of
    // odd numbers
    int cnt = 0;
    for (int i = 0; i < n; i++) {
        // Check if element
        // is odd
        if (arr[i] & 1)
            cnt += 1;
    }
 
    // Check if split is possible
    if (cnt >= k && cnt % 2 == k % 2)
        return true;
    else
        return false;
}
 
// Driver Program
int main()
{
    int arr[] = { 1, 3, 4, 7, 5, 3, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    int k = 4;
 
    if (checkArray(n, k, arr))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation to check if it 
// is possible to split array into K
// subsets with odd sum
 
class GFG{
     
// Function to check if array
// can be split in required K
// subsets
static boolean checkArray(int n, int k, 
                          int arr[])
{
     
    // Store count of odd numbers
    int cnt = 0;
    for(int i = 0; i < n; i++)
    {
         
       // Check if element is odd
       if ((arr[i] & 1) != 0)
           cnt += 1;
    }
     
    // Check if split is possible
    if (cnt >= k && cnt % 2 == k % 2)
        return true;
    else
        return false;
}
 
// Driver code
public static void main (String []args)
{
    int arr[] = { 1, 3, 4, 7, 5, 3, 1 };
    int n = arr.length;
    int k = 4;
 
    if (checkArray(n, k, arr))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by chitranayal

Python3

# Python3 implementation to check if 
# it is possible to split array into 
# K subsets with odd sum
 
# Function to check if array
# can be split in required K
# subsets
def checkArray(n, k, arr):
     
    # Store count of
    # odd numbers
    cnt = 0
    for i in range(n):
         
        # Check if element
        # is odd
        if (arr[i] & 1):
            cnt += 1
 
    # Check if split is possible
    if (cnt >= k and cnt % 2 == k % 2):
        return True
    else:
        return False
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, 3, 4, 7, 5, 3, 1 ]
    n = len(arr)
    k = 4
 
    if (checkArray(n, k, arr)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by mohit kumar 29

C#

// C# implementation to check if it 
// is possible to split array into K
// subsets with odd sum
using System;
class GFG{
 
// Function to check if array
// can be split in required K
// subsets
static bool checkArray(int n, int k, 
                       int []arr)
{
     
    // Store count of odd numbers
    int cnt = 0;
    for(int i = 0; i < n; i++)
    {
         
        // Check if element is odd
        if ((arr[i] & 1) != 0)
            cnt += 1;
    }
     
    // Check if split is possible
    if (cnt >= k && cnt % 2 == k % 2)
        return true;
    else
        return false;
}
 
// Driver code
public static void Main (string []args)
{
    int []arr = { 1, 3, 4, 7, 5, 3, 1 };
    int n = arr.Length;
    int k = 4;
 
    if (checkArray(n, k, arr))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by AnkitRai01

Javascript

<script>
// javascript implementation to check if it 
// is possible to split array into K
// subsets with odd sum
    // Function to check if array
    // can be split in required K
    // subsets
    function checkArray(n , k , arr) {
 
        // Store count of odd numbers
        var cnt = 0;
        for (i = 0; i < n; i++) {
 
            // Check if element is odd
            if ((arr[i] & 1) != 0)
                cnt += 1;
        }
 
        // Check if split is possible
        if (cnt >= k && cnt % 2 == k % 2)
            return true;
        else
            return false;
    }
 
    // Driver code
     
        var arr = [ 1, 3, 4, 7, 5, 3, 1 ];
        var n = arr.length;
        var k = 4;
 
        if (checkArray(n, k, arr))
            document.write("Yes");
        else
            document.write("No");
 
// This code contributed by gauravrajput1
</script>

Output:

Yes

Time Complexity: O(N)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!