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Split array into K disjoint subarrays such that sum of each subarray is odd.
  • Difficulty Level : Easy
  • Last Updated : 09 May, 2020

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.



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 <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


Output:

1 3 5

Time Complexity: O(N)

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