We have a tree with n(n < 1e5) nodes and we have a constant k(k < 1e5) can we store the k'th ancestor of all nodes in an array or there is no way to do that??? Thank you for helping :)
We have a tree with n(n < 1e5) nodes and we have a constant k(k < 1e5) can we store the k'th ancestor of all nodes in an array or there is no way to do that??? Thank you for helping :)
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3611 |
| 4 | jiangly | 3583 |
| 5 | strapple | 3515 |
| 6 | tourist | 3470 |
| 7 | dXqwq | 3436 |
| 8 | Radewoosh | 3415 |
| 9 | Otomachi_Una | 3413 |
| 10 | Um_nik | 3376 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 161 |
| 2 | adamant | 150 |
| 3 | Um_nik | 146 |
| 4 | Dominater069 | 144 |
| 5 | errorgorn | 141 |
| 6 | cry | 139 |
| 7 | Proof_by_QED | 136 |
| 8 | YuukiS | 135 |
| 9 | chromate00 | 134 |
| 9 | TheScrasse | 134 |
| Name |
|---|



easiest way is binary lifting to kth parent using sparse table (similar to lca).
space complexity:O(nlogn)
Complexity:O(nlogn)
Hi teja349 how can I do the binary search in O(logn)? I already know it in O(logn*logn) because I know the kth parent in O(logn) plus the binary search
let's say that k has the following binary representation: 0011010
This means that you need to climb up (21 = 2 nodes) + (23 = 8 nodes) + (24 = 16 nodes), the order doesn't matter
To do this you can loop over the bits of k and if the ith bit set, go up 2i nodes
Since we have O(log2(n)) bits, we go up by a power of two and we do O(1) work on the sparse table, the overall complexity is O(log2(n))
You can solve in O(N) by maintaining an explicit dfs stack in a vector as you dfs from the root. From each node you then look at the kth last thing in the vector if it exists.
I think this is actually much easier than using a sparse stable (although sparse tables can be easily extended to non constant k-th ancestor queries)