Split array into K disjoint subarrays such that sum of each subarray is odd.
Last Updated :
06 Jan, 2023
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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, ..., 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:
- 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++ 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 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
<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.
Related Topic: Subarrays, Subsequences, and Subsets in Array
