Given an integer Q representing the number of queries and an array where each query has an integer N. Our task is to iterate through each query and find the number of paths such that bitwise AND of all nodes on that path is odd.
Binary Tree is constructed with N vertices numbered 1 through N. For each i, from 2 to N, there is an edge between vertex i and vertex i / 2 (rounded off).
Input: Q = 2, [5, 2] Output: 1 0 Explanation: For first query the binary tree will be 1 / \ 2 3 / \ 4 5 The path which satisfies the condition is 1 -> 3. Hence only 1 path. For the second query, the binary tree will be 1 / 2 There no such path that satisfies the condition. Input: Q = 3, [3, 7, 13] Output: 1 3 4
Approach: The idea is to use Dynamic Programming and precomputation of answers for all values from 1 to maximum value of N in the query.
- Firstly, observe that if a bitwise AND of a path is odd then none elements of that path can be even. Hence required path should have odd elements.
- We know that for an ith node (except for node 1) the parent node will be i/2 (rounded off). Maintain a dp array for storing the answer for the ith node. Another array will be used to store the number of odd elements from the current node until the parents are odd.
- While computing the dp array the first condition is if the ith node value is even then dp[i] = dp[i – 1] because the ith node will not contribute in the answer so dp[i] will be the answer of (i-1)th node. Secondly, if ith node is odd then dp[i] = dp[i-1] + nC2 -(n-1)C2. On simplification dp[i] = dp[i-1] + (number of odd elements till now) – 1.
Below is the implementation of the above approach:
Time Complexity: O( Nmax + Q*(1) ), where Nmax is the maximum value of N. Q*(1) since we are precomputing each queries.
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