Given a tree and the weights of all the nodes, the task is to count the number of nodes whose weights are even parity i.e. whether the count of set bits in them is even.
Weight Binary Representation Parity 5 0101 Even 10 1010 Even 11 1011 Odd 8 1000 Odd 6 0110 Even
Approach: Perform dfs on the tree and for every node, check if it’s weight is even parity or not. If yes then increment count.
Below is the implementation of the above approach:
- Time Complexity: O(N).
In DFS, every node of the tree is processed once and hence the complexity due to the DFS is O(N) for N nodes in the tree. Therefore, the time complexity is O(N).
- Auxiliary Space: O(1).
Any extra space is not required, so the space complexity is constant.
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