# 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++

 `// 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

 `// 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

 `# 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#

 `// 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)

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Improved By : 29AjayKumar, AnkitRai01

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