Hard
You are given an integer n
. There is a complete binary tree with 2n - 1
nodes. The root of that tree is the node with the value 1
, and every node with a value val
in the range [1, 2n - 1 - 1]
has two children where:
2 * val
, and2 * val + 1
.You are also given a 2D integer array queries
of length m
, where queries[i] = [ai, bi]
. For each query, solve the following problem:
ai
and bi
.ai
and bi
.Note that:
Return an array answer
of length m
where answer[i]
is the answer to the ith
query.
Example 1:
Input: n = 3, queries = [[5,3],[4,7],[2,3]]
Output: [4,5,3]
Explanation: The diagrams above show the tree of 23 - 1 nodes. Nodes colored in red describe the nodes in the cycle after adding the edge.
After adding the edge between nodes 3 and 5, the graph contains a cycle of nodes [5,2,1,3]. Thus answer to the first query is 4. We delete the added edge and process the next query.
After adding the edge between nodes 4 and 7, the graph contains a cycle of nodes [4,2,1,3,7]. Thus answer to the second query is 5. We delete the added edge and process the next query.
After adding the edge between nodes 2 and 3, the graph contains a cycle of nodes [2,1,3]. Thus answer to the third query is 3. We delete the added edge.
Example 2:
Input: n = 2, queries = [[1,2]]
Output: [2]
Explanation: The diagram above shows the tree of 22 - 1 nodes. Nodes colored in red describe the nodes in the cycle after adding the edge.
Constraints:
2 <= n <= 30
m == queries.length
1 <= m <= 105
queries[i].length == 2
1 <= ai, bi <= 2n - 1
ai != bi
@Suppress("UNUSED_PARAMETER")
class Solution {
fun cycleLengthQueries(n: Int, queries: Array<IntArray>): IntArray {
val m = queries.size
val res = IntArray(m)
for (i in 0 until m) {
var a = queries[i][0]
var b = queries[i][1]
var count = 1
while (a != b) {
if (a > b) {
a = a shr 1
} else {
b = b shr 1
}
count++
}
res[i] = count
}
return res
}
}