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

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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, …, S_K:

The intersection of S₁, S₂, S₃, …, S_K should be NULL.
The union of S₁, S₂, S₃, …, S_K 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.

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

Since only odd values can lead to odd sum, hence we can ignore the even values.
The number of odd values must also be odd.
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 <iostream>
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

Javascript

<script>
// Javascript 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.
function split(a, n, k)
{
    // Number of odd elements
    let odd_ele = 0;
 
    // Loop to store the number
    // of odd elements in the array
    for (let 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)
        document.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)
        document.write(1);
 
    else {
        for (let i = 0; i < n; i++) {
            if (a[i] % 2) {
 
                // Printing the position of
                // odd elements
                document.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
 
let n = 5;
let arr = [ 7, 2, 11, 4, 19 ];
let k = 3;
 
split(arr, n, k);
 
 
// This code is contributed by gfgking
</script>

Output:

1 3 5

Time Complexity: O(N)
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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