Given an integer n. There’s a full binary tree with 2n – 1 nodes. The foundation of that tree is the node with the worth 1, and each node with a worth x has two kids the place the left node has the worth
2*x and the proper node has the worth 2*x + 1, you’re given Ok queries of kind (ai, bi), and the duty is to return the LCA for the node pair ai and bi for all Ok queries.
Examples:
Enter: n = 5, queries = [ { 17, 21 }, { 23, 5 }, { 15, 7 }, { 3, 21 }, { 31, 9 }, { 5, 15 }, { 11, 2 }, { 19, 7 } ]
Full binary tree for given enter n=5
Output: [ 2, 5, 7, 1, 1, 1, 2, 1 ]
Enter: n = 3, queries = [ {2, 5}, {3, 6}, {4, 1}, {7, 3} ]
Full binary tree for given enter n=3
Output: [2, 3, 1, 3]
Method: The issue may be solved based mostly on the next concept:
As all values on a degree are smaller than values on the subsequent degree. Verify which node is having better worth in a question, and divide it by 2 to achieve its dad or mum node. Repeat this step till we get frequent ingredient.
Comply with the steps to unravel the issue:
- In a question, we’re having 2 nodes a and b, whose lowest frequent ancestor we’ve to search out.
- By dividing the worth of the node by 2, we’ll at all times get the dad or mum node worth.
- From a and b whichever node is having better worth divide by 2. So, as to maneuver in direction of the foundation of the foundation.
- When a and b turns into equal, the frequent ancestor between them is obtained and returned.
Beneath is the implementation for the strategy mentioned:
C++
// C++ code for the above strategy
#embrace <bits/stdc++.h>
utilizing namespace std;
// Operate to search out lca for
// given two nodes in tree
int helper(int a, int b)
{
whereas (a != b) {
if (a > b)
a = a / 2;
else
b = b / 2;
}
return a;
}
// Driver code
int foremost()
{
// 2^n - 1 nodes in full
// binary tree
int n = 5;
// Queries enter vector
vector<vector<int> > queries
= { { 17, 21 }, { 23, 5 }, { 15, 7 }, { 3, 21 }, { 31, 9 }, { 5, 15 }, { 11, 2 }, { 19, 7 } };
// Processing every question in
// queries vector
for (auto e : queries) {
// Operate name
int lca = helper(e[0], e[1]);
cout << lca << ' ';
}
return 0;
}
Time Complexity: O(n)
Auxiliary Area: O(1)
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