Given a tree, and the weights of all the nodes, the task is to count the nodes whose weights are divisible by x.
x = 5
Only the nodes 1 and 2 have weights divisible by 5.
Approach: Perform dfs on the tree and for every node, check if it’s weight is divisible by x or not. If yes then increment the count.
- 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) when there are total 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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