Dare2Solve
The problem requires connecting two trees such that the resulting tree has the minimum possible diameter. The diameter of a tree is defined as the longest path between any two nodes in the tree. To minimize the diameter after connecting two trees, we need to strategically choose nodes from each tree to connect.
Calculate Tree Diameters: First, compute the diameters of both trees:
Connecting Trees: After obtaining the diameters (d1
for the first tree and d2
for the second tree), the possible diameters of the resulting tree are:
d1 + d2 + 1
: This accounts for the worst-case scenario where the longest path in the resulting tree goes through the connecting edge.Choose Connecting Nodes: To achieve the minimum diameter, connect nodes that minimize the impact on the existing diameters. A strategic choice is often connecting the centroid or a node close to the centroid of each tree.
Compute Minimum Diameter: Calculate the possible diameters resulting from connecting different pairs of nodes from both trees and choose the minimum diameter.
Implementation Details: Implement DFS to calculate tree diameters and carefully choose nodes for connection to minimize the resulting diameter. Edge cases include trees with minimal nodes (like single-node trees).
Time Complexity:
Space Complexity:
This approach ensures that we find the optimal way to connect the two trees while minimizing the diameter of the resulting tree efficiently.
var minimumDiameterAfterMerge = function (edges1, edges2) {
minimumDiameterAfterMerge
that takes two arguments, edges1
and edges2
. var getDiameter = (edges) => {
getDiameter
that calculates the diameter of a tree given its edges.let map = Array(edges.length + 1)
.fill()
.map(() => []);
map
to represent the graph. It initializes an array of empty arrays.for (let [i, j] of edges) {
map[i].push(j);
map[j].push(i);
}
[i, j]
and adding j
to the list of neighbors of i
and vice versa.let res = 0;
res
to store the maximum diameter found. var dfs = (node, parent) => {
let maxDepth = 1;
dfs
that calculates the depth of the tree and updates the diameter. It initializes maxDepth
to 1. for (let neighbor of map[node]) {
if (neighbor === parent) continue;
let depth = dfs(neighbor, node);
res = Math.max(res, maxDepth + depth);
maxDepth = Math.max(maxDepth, 1 + depth);
}
return maxDepth;
};
node
. If the neighbor is the parent
, it continues to the next iteration. It calculates the depth
by recursively calling dfs
on the neighbor. Updates res
with the maximum diameter found so far. Updates maxDepth
with the maximum depth found so far. Returns maxDepth
.dfs(0, -1);
dfs
function starting from node 0
with -1
as the parent. return res;
};
let d1 = getDiameter(edges1);
let d2 = getDiameter(edges2);
d1
and d2
. return Math.max(d1 - 1, d2 - 1, Math.floor(d1 / 2) + Math.floor(d2 / 2) + 1);
};
class Solution {
public:
int minimumDiameterAfterMerge(vector<vector<int>>& edges1, vector<vector<int>>& edges2) {
int d1 = getDiameter(edges1);
int d2 = getDiameter(edges2);
return max(max(d1 - 1, d2 - 1), d1 / 2 + d2 / 2 + 1);
}
private:
int getDiameter(vector<vector<int>>& edges) {
unordered_map<int, vector<int>> map;
for (auto& edge : edges) {
int i = edge[0];
int j = edge[1];
map[i].push_back(j);
map[j].push_back(i);
}
int res = 0;
dfs(0, -1, map, res);
return res;
}
int dfs(int node, int parent, unordered_map<int, vector<int>>& map, int& res) {
int maxDepth = 1;
for (int neighbor : map[node]) {
if (neighbor == parent)
continue;
int depth = dfs(neighbor, node, map, res);
res = max(res, maxDepth + depth);
maxDepth = max(maxDepth, 1 + depth);
}
return maxDepth;
}
};
class Solution:
def minimumDiameterAfterMerge(self, edges1: List[List[int]], edges2: List[List[int]]) -> int:
def getDiameter(edges):
map = defaultdict(list)
for i, j in edges:
map[i].append(j)
map[j].append(i)
res = [0]
def dfs(node, parent):
maxDepth = 1
for neighbor in map[node]:
if neighbor == parent:
continue
depth = dfs(neighbor, node)
res[0] = max(res[0], maxDepth + depth)
maxDepth = max(maxDepth, 1 + depth)
return maxDepth
dfs(0, -1)
return res[0]
d1 = getDiameter(edges1)
d2 = getDiameter(edges2)
return max(d1 - 1, d2 - 1, d1 // 2 + d2 // 2 + 1)
class Solution {
public int minimumDiameterAfterMerge(int[][] edges1, int[][] edges2) {
int d1 = getDiameter(edges1);
int d2 = getDiameter(edges2);
return Math.max(Math.max(d1 - 1, d2 - 1), d1 / 2 + d2 / 2 + 1);
}
private int getDiameter(int[][] edges) {
List<List<Integer>> map = new ArrayList<>();
for (int i = 0; i < edges.length + 1; i++) {
map.add(new ArrayList<>());
}
for (int[] edge : edges) {
int i = edge[0];
int j = edge[1];
map.get(i).add(j);
map.get(j).add(i);
}
int[] res = { 0 };
dfs(0, -1, map, res);
return res[0];
}
private int dfs(int node, int parent, List<List<Integer>> map, int[] res) {
int maxDepth = 1;
for (int neighbor : map.get(node)) {
if (neighbor == parent)
continue;
int depth = dfs(neighbor, node, map, res);
res[0] = Math.max(res[0], maxDepth + depth);
maxDepth = Math.max(maxDepth, 1 + depth);
}
return maxDepth;
}
}
/**
* @param {number[][]} edges1
* @param {number[][]} edges2
* @return {number}
*/
var minimumDiameterAfterMerge = function (edges1, edges2) {
var getDiameter = (edges) => {
let map = Array(edges.length + 1).fill().map(() => []);
for (let [i, j] of edges) {
map[i].push(j);
map[j].push(i);
}
let res = 0;
var dfs = (node, parent) => {
let maxDepth = 1;
for (let neighbor of map[node]) {
if (neighbor === parent) continue;
let depth = dfs(neighbor, node);
res = Math.max(res, maxDepth + depth);
maxDepth = Math.max(maxDepth, 1 + depth);
}
return maxDepth;
};
dfs(0, -1);
return res;
};
let d1 = getDiameter(edges1);
let d2 = getDiameter(edges2);
return Math.max(d1 - 1, d2 - 1, Math.floor(d1 / 2) + Math.floor(d2 / 2) + 1);
};