# Split array into K disjoint subarrays such that sum of each subarray is odd.

Given an array arr[] containing N elements, the task is to divide the array into K(1 ≤ K ≤ N) subarrays and such that the sum of elements of each subarray is odd. Print the starting index (1 based indexing) of each subarray after dividing the array and -1 if no such subarray exists.

Note: For all subarrays S1, S2, S3, …, SK:

• The intersection of S1, S2, S3, …, SK should be NULL.
• The union of S1, S2, S3, …, SK should be equal to the array.

Examples:

Input: N = 5, arr[] = {7, 2, 11, 4, 19}, K = 3
Output: 1 3 5
Explanation:
When the given array arr[] is divided into K = 3 parts, the possible subarrays are: {7, 2}, {11, 4} and {19}

Input: N = 5, arr[] = {2, 4, 6, 8, 10}, K = 3
Output: -1
Explanation:
It is impossible to divide the array arr[] into K = 3 subarrays as all the elements are even and the sum of every subarray is even.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: It can be easily observed that for any subarray to have odd sum:

1. Since only odd values can lead to odd sum, hence we can ignore the even values.
2. The number of odd values must also be odd.
3. So we need at least K odd values in the array for K subarrays. If K is greater than the number of odd elements then the answer is always -1.

Below is the implementation of the above approach:

 // C++ program to split the array into K // disjoint subarrays so that the sum of // each subarray is odd.    #include using namespace std;    // Function to split the array into K // disjoint subarrays so that the sum of // each subarray is odd. void split(int a[], int n, int k) {     // Number of odd elements     int odd_ele = 0;        // Loop to store the number     // of odd elements in the array     for (int i = 0; i < n; i++)         if (a[i] % 2)             odd_ele++;        // If the count of odd elements is < K     // then the answer doesnt exist     if (odd_ele < k)         cout << -1;        // If the number of odd elements is     // greater than K and the extra     // odd elements are odd, then the     // answer doesn't exist     else if (odd_ele > k && (odd_ele - k) % 2)         cout << -1;        else {         for (int i = 0; i < n; i++) {             if (a[i] % 2) {                    // Printing the position of                 // odd elements                 cout << i + 1 << " ";                    // Decrementing K as we need positions                 // of only first k odd numbers                 k--;             }                // When the positions of the first K             // odd numbers are printed             if (k == 0)                 break;         }     } }    // Driver code int main() {     int n = 5;     int arr[] = { 7, 2, 11, 4, 19 };     int k = 3;        split(arr, n, k); }

 // Java program to split the array into K // disjoint subarrays so that the sum of // each subarray is odd. class GFG{     // Function to split the array into K // disjoint subarrays so that the sum of // each subarray is odd. static void split(int a[], int n, int k) {     // Number of odd elements     int odd_ele = 0;         // Loop to store the number     // of odd elements in the array     for (int i = 0; i < n; i++)         if (a[i] % 2==1)             odd_ele++;         // If the count of odd elements is < K     // then the answer doesnt exist     if (odd_ele < k)         System.out.print(-1);         // If the number of odd elements is     // greater than K and the extra     // odd elements are odd, then the     // answer doesn't exist     else if (odd_ele > k && (odd_ele - k) % 2==1)         System.out.print(-1);         else {         for (int i = 0; i < n; i++) {             if (a[i] % 2==1) {                     // Printing the position of                 // odd elements                 System.out.print(i + 1+ " ");                     // Decrementing K as we need positions                 // of only first k odd numbers                 k--;             }                 // When the positions of the first K             // odd numbers are printed             if (k == 0)                 break;         }     } }     // Driver code public static void main(String[] args) {     int n = 5;     int arr[] = { 7, 2, 11, 4, 19 };     int k = 3;         split(arr, n, k); } }    // This code is contributed by 29AjayKumar

 # Python3 program to split the array into K # disjoint subarrays so that the sum of # each subarray is odd.    # Function to split the array into K # disjoint subarrays so that the sum of # each subarray is odd. def split(a, n, k) :        # Number of odd elements     odd_ele = 0;        # Loop to store the number     # of odd elements in the array     for i in range(n) :         if (a[i] % 2) :             odd_ele += 1;        # If the count of odd elements is < K     # then the answer doesnt exist     if (odd_ele < k) :         print(-1);        # If the number of odd elements is     # greater than K and the extra     # odd elements are odd, then the     # answer doesn't exist     elif (odd_ele > k and (odd_ele - k) % 2) :         print(-1);        else :         for i in range(n) :             if (a[i] % 2) :                    # Printing the position of                 # odd elements                 print(i + 1 ,end= " ");                    # Decrementing K as we need positions                 # of only first k odd numbers                 k -= 1;                # When the positions of the first K             # odd numbers are printed             if (k == 0) :                 break;    # Driver code if __name__ == "__main__" :        n = 5;     arr = [ 7, 2, 11, 4, 19 ];     k = 3;        split(arr, n, k);            # This code is contributed by AnkitRai01

 // C# program to split the array into K // disjoint subarrays so that the sum of // each subarray is odd. using System;    class GFG{     // Function to split the array into K // disjoint subarrays so that the sum of // each subarray is odd. static void split(int []a, int n, int k) {     // Number of odd elements     int odd_ele = 0;         // Loop to store the number     // of odd elements in the array     for (int i = 0; i < n; i++)         if (a[i] % 2 == 1)             odd_ele++;         // If the count of odd elements is < K     // then the answer doesnt exist     if (odd_ele < k)         Console.Write(-1);         // If the number of odd elements is     // greater than K and the extra     // odd elements are odd, then the     // answer doesn't exist     else if (odd_ele > k && (odd_ele - k) % 2 == 1)         Console.Write(-1);         else {         for (int i = 0; i < n; i++) {             if (a[i] % 2 == 1) {                     // Printing the position of                 // odd elements                 Console.Write(i + 1 + " ");                     // Decrementing K as we need positions                 // of only first k odd numbers                 k--;             }                 // When the positions of the first K             // odd numbers are printed             if (k == 0)                 break;         }     } }     // Driver code public static void Main(string[] args) {     int n = 5;     int []arr = { 7, 2, 11, 4, 19 };     int k = 3;         split(arr, n, k); } }    // This code is contributed by AnkitRai01

Output:
1 3 5

Time Complexity: O(N)

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : 29AjayKumar, AnkitRai01

Article Tags :
Practice Tags :