Given an array arr[] of integers, the task is to check if it is possible to obtain a subsequence of K elements from the array such that their sum is odd. If it is possible, print Yes. Otherwise, print No.
Examples:
Input: arr[] = { 2, 5, 6, 7, 4 }, K = 3
Output: Yes
Explanation:
Subsequences {2, 5, 6}, {2, 6, 7} and {2, 7, 4} have odd sumInput: arr[] = { 1, 5, 7, 11 }, K = 4
Output: No
Explanation:
Only subsequence of length 4 is {1, 5, 7, 11} which has even sum (24). Hence no such subsequence exists.
Naive Approach:
The simplest method to solve the problem is to generate all subsequences of length K and check if any of these subsequences have an odd sum. The time complexity for such an approach will be exponential and thus inefficient.
Efficient Approach:
The efficient method to approach the above problem will be to count the number of odd elements in the array and then, simply checking for all the edge cases when it will not be possible to find a subsequence with odd sum.
The edge cases to be considered when such a subsequence cannot be generated are as follows:
- If there are no odd elements in the array, any subsequence will contain only even elements and an even sum will be obtained. So, it is not possible to generate a subsequence with odd sum.
- If K is even and there are no even elements present in the array, a subsequence with odd sum is not possible.
For all other cases, it will be possible to generate a subsequence with an odd sum.
Below is the implementation of above approach:
C++
// C++ program to check if a // subsequence of length K // with odd sum exists in the // given array #include <bits/stdc++.h> using namespace std; // Function to check if any required // subsequence exists or not bool isSubseqPossible(int arr[], int N, int K) { int i; // Store count of odd and // even elements in the array int odd = 0, even = 0; // Calculate the count of // odd and even elements for (i = 0; i < N; i++) { if (arr[i] % 2 == 1) odd++; else even++; } // If no odd elements exists // or no even elements exists // when K is even if (odd == 0 || (even == 0 && K % 2 == 0)) // Subsequence is not possible return false; // Possible otherwise return true; } // Driver Code int main() { int arr[] = { 2, 3, 5, 7, 4 }; int N = sizeof(arr) / sizeof(arr[0]); int K = 3; cout << (isSubseqPossible(arr, N, K) ? "Yes" : "No"); return 0; } |
Java
// Java program to check if a // subsequence of length K // with odd sum exists in the // given array class GFG{ // Function to check if any required // subsequence exists or not static boolean isSubseqPossible(int []arr, int N, int K) { int i; // Store count of odd and // even elements in the array int odd = 0, even = 0; // Calculate the count of // odd and even elements for (i = 0; i < N; i++) { if (arr[i] % 2 == 1) odd++; else even++; } // If no odd elements exists // or no even elements exists // when K is even if (odd == 0 || (even == 0 && K % 2 == 0)) // Subsequence is not possible return false; // Possible otherwise return true; } // Driver Code public static void main(String args[]) { int []arr = { 2, 3, 5, 7, 4 }; int N = arr.length; int K = 3; System.out.print(isSubseqPossible(arr, N, K) ? "Yes" : "No"); } } // This code is contributed by Code_Mech |
Python3
# Python3 program to check if a subsequence # of length K with odd sum exists in the # given array # Function to check if any required # subsequence exists or not def isSubseqPossible(arr, N, K): i = 0 # Store count of odd and # even elements in the array odd = 0 even = 0 # Calculate the count of # odd and even elements for i in range(N): if (arr[i] % 2 == 1): odd += 1 else: even += 1 # If no odd element exists or no # even element exists when K even if (odd == 0 or (even == 0 and K % 2 == 0)): # Subsequence is not possible return False # Otherwise possible return True # Driver code if __name__ == '__main__': arr = [ 2, 3, 5, 7, 4 ] N = len(arr) K = 3 print("Yes" if isSubseqPossible(arr, N, K) else "No") # This code is contributed by himanshu77 |
C#
// C# program to check if a // subsequence of length K // with odd sum exists in the // given array using System; class GFG{ // Function to check if any required // subsequence exists or not static bool isSubseqPossible(int []arr, int N, int K) { int i; // Store count of odd and // even elements in the array int odd = 0, even = 0; // Calculate the count of // odd and even elements for (i = 0; i < N; i++) { if (arr[i] % 2 == 1) odd++; else even++; } // If no odd elements exists // or no even elements exists // when K is even if (odd == 0 || (even == 0 && K % 2 == 0)) // Subsequence is not possible return false; // Possible otherwise return true; } // Driver Code public static void Main() { int []arr = { 2, 3, 5, 7, 4 }; int N = arr.Length; int K = 3; Console.Write(isSubseqPossible(arr, N, K) ? "Yes" : "No"); } } // This code is contributed by Code_Mech |
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
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