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 the nodes in the given Tree whose weight is a Perfect Number
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- Count the nodes whose weight is a perfect square
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- Query to find the maximum and minimum weight between two nodes in the given tree using LCA.
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- Count all pairs of adjacent nodes whose XOR is an odd number
- Count of n digit numbers whose sum of digits equals to given sum
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