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++
// 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 codeint 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 codepublic 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 codeif __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 codepublic 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 codelet 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
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