Given an array, arr[] and an integer K. Check whether it is possible to get an odd sum by choosing exactly K elements of the array.
Examples:
Input: arr[] = { 1, 2, 3 }, K = 2
Output: Possible
Explanation: {2, 3} -> 2 + 3 = 5Input: arr[] = { 1, 2, 4, 2 }, K = 4
Output: Not Possible
Explanation: {2, 2, 4, 2} -> 2 + 2 + 4 + 2 = 10
No other possibilities as K is equal to the size of array
Approach:
On observing, it is found that there are three cases.
- First count the number of odd and even elements.
- Case 1: When all elements are even. Then, sum will always be even irrespective of the value of K as even + even = even.
- Case 2: When all elements are odd. Then, the sum will depend only on the value of K.
If K is odd, then the sum will be odd as every odd pair makes the sum even and in the end, one odd element makes the sum odd as even + odd = odd.
If K is even then every odd element get paired and become even, therefore the sum becomes even. - Case 3: When K <= N, then sum depends only on the number of odd elements. If the number of odd elements is even, then the sum will be even as odd + odd = even, which implies every odd pair will become even. And if we add even elements to the sum, the sum will remain even as even + even = even.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function returns true if // it is possible to have // odd sum bool isPossible(int arr[], int N, int K) { int oddCount = 0, evenCount = 0; // counting number of odd // and even elements for (int i = 0; i < N; i++) { if (arr[i] % 2 == 0) evenCount++; else oddCount++; } if (evenCount == N || (oddCount == N && K % 2 == 0) || (K == N && oddCount % 2 == 0)) return false; else return true; } // Driver code int main() { int arr[] = { 1, 2, 3, 4, 5, 8 }; int K = 5; int N = sizeof(arr) / sizeof(arr[0]); if (isPossible(arr, N, K)) cout << "Possible"; else cout << "Not Possible"; return 0; } |
Java
// Java implementation of the above approach class GFG{ // Function returns true if // it is possible to have // odd sum static boolean isPossible(int arr[], int N, int K) { int oddCount = 0, evenCount = 0; // Counting number of odd // and even elements for(int i = 0; i < N; i++) { if (arr[i] % 2 == 0) { evenCount++; } else { oddCount++; } } if (evenCount == N || (oddCount == N && K % 2 == 0) || (K == N && oddCount % 2 == 0)) { return false; } else { return true; } } // Driver code public static void main (String[] args) { int arr[] = { 1, 2, 3, 4, 5, 8 }; int K = 5; int N = arr.length; if (isPossible(arr, N, K)) { System.out.println("Possible"); } else { System.out.println("Not Possible"); } } } // This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the above approach # Function returns true if it # is possible to have odd sum def isPossible(arr, N, K): oddCount = 0 evenCount = 0 # Counting number of odd # and even elements for i in range(N): if (arr[i] % 2 == 0): evenCount += 1 else: oddCount += 1 if (evenCount == N or (oddCount == N and K % 2 == 0) or (K == N and oddCount % 2 == 0)): return False else: return True # Driver code if __name__ == '__main__': arr = [ 1, 2, 3, 4, 5, 8 ] K = 5 N = len(arr) if (isPossible(arr, N, K)): print("Possible") else: print("Not Possible") # This code is contributed by mohit kumar 29 |
C#
// C# implementation of the above approach using System; class GFG{ // Function returns true if // it is possible to have // odd sum static bool isPossible(int []arr, int N, int K) { int oddCount = 0, evenCount = 0; // Counting number of odd // and even elements for(int i = 0; i < N; i++) { if (arr[i] % 2 == 0) { evenCount++; } else { oddCount++; } } if (evenCount == N || (oddCount == N && K % 2 == 0) || (K == N && oddCount % 2 == 0)) { return false; } else { return true; } } // Driver code public static void Main (string[] args) { int []arr = { 1, 2, 3, 4, 5, 8 }; int K = 5; int N = arr.Length; if (isPossible(arr, N, K)) { Console.WriteLine("Possible"); } else { Console.WriteLine("Not Possible"); } } } // This code is contributed by AnkitRai01 |
Possible
Time complexity: O(N)
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