Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose sum of digits of weights is odd.
Node 1: digitSum(144) = 1 + 4 + 4 = 9
Node 2: digitSum(1234) = 1 + 2 + 3 + 4 = 10
Node 3: digitSum(21) = 2 + 1 = 3
Node 4: digitSum(5) = 5
Node 5: digitSum(77) = 7 + 7 = 14
Only the sum of digits of the weights of nodes 1, 3 and 4 are odd.
Approach: Perform dfs on the tree and for every node, check if the sum of the digits of its weight is odd. If yes then increment the 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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- Count the nodes in the given tree whose weight is even parity
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- Count nodes in the given tree whose weight is a fibonacci number
- Count the nodes in the given tree whose weight is a powerful number
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- Count numbers from given range having odd digits at odd places and even digits at even places
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- Count of integers in a range which have even number of odd digits and odd number of even digits
- Query to find the maximum and minimum weight between two nodes in the given tree using LCA.
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- Check if there is a cycle with odd weight sum in an undirected graph
- Sum of all odd nodes in the path connecting two given nodes
- Count of n digit numbers whose sum of digits equals to given sum
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