Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weights contain a vowel.
Only the strings of the nodes 1 and 5 contain vowels.
Approach: Perform dfs on the tree and for every node, check if it’s string contains vowels, If yes then increment the count.
Below is the implementation of the above approach:
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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