Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weights are palindrome.
Input: Output: 3 Only the weights of the nodes 2, 3 and 5 are palindromes.
Approach: Perform dfs on the tree and for every node, check if it’s string is palindrome or not. If yes then increment the count.
- Time Complexity: O(N*Len) where Len is the maximum length of the weighted string of a node in the given tree.
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. Also, processing of every node involves traversing the weighted string of that node once, thus adding a complexity of O(Len) where Len is the length of the weighted string. Therefore, the total time complexity is O(N*Len).
- Auxiliary Space: O(1).
Any extra space is not required, so the space complexity is constant.
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