Given a tree, and the weights of all the nodes and an integer X, the task is to count all the nodes i such that (weight[i] + X) is a Fibonacci Number.
First few fibonacci numbers are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, …
X = 5
Only the nodes 3 and 5 give a fibonacci number when 5 is added to them.
i.e. (3 + 5) = 8 and (16 + 5) = 21 are both Fibonacci numbers.
Approach: Perform dfs on the tree and count all the nodes sum of whose weight with x is a fibonacci number.
Below is the implementation of the above approach:
- Count of cells in a matrix which give a Fibonacci number when the count of adjacent cells is added
- Given a n-ary tree, count number of nodes which have more number of children than parents
- Count the number of non-reachable nodes
- Count all pairs of adjacent nodes whose XOR is an odd number
- Count the number of nodes at given level in a tree using BFS.
- Count the number of nodes at a given level in a tree using DFS
- Check if a M-th fibonacci number divides N-th fibonacci number
- Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes
- Print levels with odd number of nodes and even number of nodes
- Count Fibonacci numbers in given range in O(Log n) time and O(1) space
- Number of ways to represent a number as sum of k fibonacci numbers
- Finding number of digits in n'th Fibonacci number
- Count BST nodes that lie in a given range
- Nth Even Fibonacci Number
- Count the nodes in the given tree whose weight is even
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