Given an array arr[] with N elements, the task is to find whether all the elements of the given array can be made 0 by given operations. Only 2 types of operations can be performed on this array:
- Split an element B into 2 elements C and D such that B = C + D.
- Merge 2 elements P and Q as one element R such that R = P^Q i.e. (XOR of P and Q).
Examples:
Input: arr = [9, 17]
Output: Yes
Explanation: Following is one possible sequence of operations –
1) Merge i.e 9 XOR 17 = 24
2) Split 24 into two parts each of size 12
3) Merge i.e 12 XOR 12 = 0
As there is only 1 element i.e 0. So it is possible.Input: arr = [1]
Output: No
Explanation: There is no possible way to make it 0.
Approach :
- If any element in the array is even then it can be made 0. Split that element in two equal parts of arr[i]/2 and arr[i]/2. XOR of two equal numbers is zero. Therefore this strategy makes an element 0.
- If any element is odd. Split it in two parts: 1 and arr[i]-1. Since arr[i]-1 is even, it can be made 0 by the above strategy. Therefore an odd element can reduce its size to 1. Two odd elements can, therefore, be made 0 by following above strategy and finally XOR them (i.e. 1) as 1 XOR 1 = 0. Therefore if the number of odd elements in the array is even, then the answer is possible.Otherwise, an element of value 1 will be left and it is not possible to satisfy the condition.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function that finds if it is // possible to make the array // contain only 1 element i.e. 0 string solve(vector<int>& A) { int i, ctr = 0; for (i = 0; i < A.size(); i++) { // Check if element is odd if (A[i] % 2) { ctr++; } } // According to the logic // in above approach if (ctr % 2) { return "No"; } else { return "Yes"; } } // Driver code int main() { vector<int> arr = { 9, 17 }; cout << solve(arr) << endl; return 0; } |
Java
// Java program for the above approach class GFG{ // Function that finds if it is // possible to make the array // contain only 1 element i.e. 0 public static String solve(int[] A) { int i, ctr = 0; for(i = 0; i < A.length; i++) { // Check if element is odd if (A[i] % 2 == 1) { ctr++; } } // According to the logic // in above approach if (ctr % 2 == 1) { return "No"; } else { return "Yes"; } } // Driver code public static void main(String[] args) { int[] arr = { 9, 17 }; System.out.println(solve(arr)); } } // This code is contributed by divyeshrabadiya07 |
Python3
# Python3 program for the above approach # Function that finds if it is # possible to make the array # contain only 1 element i.e. 0 def solve(A): ctr = 0 for i in range(len(A)): # Check if element is odd if A[i] % 2 == 1: ctr += 1 # According to the logic # in above approach if ctr % 2 == 1: return 'No' else : return 'Yes' # Driver code if __name__=='__main__': arr = [9, 17] print(solve(arr)) # This code is contributed by rutvik_56 |
C#
// C# program for the above approach using System; class GFG{ // Function that finds if it is // possible to make the array // contain only 1 element i.e. 0 public static string solve(int[] A) { int i, ctr = 0; for(i = 0; i < A.Length; i++) { // Check if element is odd if (A[i] % 2 == 1) { ctr++; } } // According to the logic // in above approach if (ctr % 2 == 1) { return "No"; } else { return "Yes"; } } // Driver code public static void Main() { int[] arr = { 9, 17 }; Console.Write(solve(arr)); } } // This code is contributed by chitranayal |
Yes
Time Complexity: O(N)
Auxiliary Space Complexity: O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
