Minimum of the Maximum distances from any node to all other nodes of given Tree
Last Updated :
05 Dec, 2022
Given a tree with N vertices and N-1 edges represented by a 2D array edges[], the task is to find the minimum value among the maximum distances from any node to all other nodes of the tree.
Examples:
Input: N = 4, edges[] = { {1, 2}, {2, 3}, {2, 4} }
Output: 1
Explanation: The Tree looks like the following.
2
/ | \
1 3 4
Maximum distance from house number 1 to any other node is 2.
Maximum distance from house number 2 to any other node is 1.
Maximum distance from house number 3 to any other node is 2.
Maximum distance from house number 4 to any other node is 2.
The minimum among these is 1.
Input: N = 10, edges[] = { {1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 7}, {7, 8}, {8, 9}, {9, 10} }
Output: 5
Approach: This problem can be solved using Depth First Search based on the following idea:
For each node find the farthest node and the distance to that node. Then find the minimum among those values.
Follow the steps mentioned below to implement the idea:
- Create a distance array (say d[]) where d[i] stores the maximum distance to all other nodes from the ith node.
- For every node present in the tree consider each of them as the source one by one:
- Mark the distance of the source from the source as zero.
- Find the maximum distance of all other nodes from the source.
- Find the minimum value among these maximum distances.
Below is the implementation of the above approach:
C++
// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to run DFS
void dfs(vector<int>& d, int node,
vector<vector<int> >& adj, int dist)
{
d[node] = dist;
// DFS call for all its neighbours
for (auto child : adj[node]) {
if (dist + 1 < d[child])
dfs(d, child, adj, dist + 1);
}
}
// Function to find the minimum distance
int minDist(int N, vector<vector<int> >& edges)
{
int ans = INT_MAX;
// Creation of the adjacency matrix
vector<vector<int> > adj(N + 1);
for (auto u : edges) {
adj[u[0]].push_back(u[1]);
adj[u[1]].push_back(u[0]);
}
// Consider ith node as source
// in each iteration
for (int i = 1; i <= N; i++) {
// Distance array to store
// distance of all the nodes
// from source
vector<int> d(N + 1, INT_MAX);
// DFS traversal of the tree and store
// distance of all nodes from source
dfs(d, i, adj, 0);
int dist = 0;
// Find max distance from distance array
for (int j = 1; j <= N; j++)
dist = max(dist, d[j]);
// If distance found is smaller than ans
// then make ans equal to distance
ans = min(ans, dist);
}
// Return the minimum value
return ans;
}
// Driver Code
int main()
{
int N = 4;
vector<vector<int> > edges
= { { 1, 2 }, { 2, 3 }, { 2, 4 } };
// Function call
cout << minDist(N, edges);
return 0;
}
// Java code to implement the above approach
import java.util.ArrayList;
public class GFG {
// Function to run DFS
static void dfs(int[] d, int node,
ArrayList<Integer>[] adj, int dist)
{
d[node] = dist;
// DFS call for all its neighbours
for (int child : adj[node]) {
if (dist + 1 < d[child])
dfs(d, child, adj, dist + 1);
}
}
// Function to find the minimum distance
@SuppressWarnings("unchecked")
static int minDist(int N, int[][] edges)
{
int ans = Integer.MAX_VALUE;
// Creation of the adjacency matrix
ArrayList<Integer>[] adj = new ArrayList[N + 1];
for (int i = 0; i <= N; i++)
adj[i] = new ArrayList<Integer>();
for (int[] u : edges) {
adj[u[0]].add(u[1]);
adj[u[1]].add(u[0]);
}
// Consider ith node as source
// in each iteration
for (int i = 1; i <= N; i++) {
// Distance array to store
// distance of all the nodes
// from source
int[] d = new int[N + 1];
for (int j = 0; j <= N; j++)
d[j] = Integer.MAX_VALUE;
// DFS traversal of the tree and store
// distance of all nodes from source
dfs(d, i, adj, 0);
int dist = 0;
// Find max distance from distance array
for (int j = 1; j <= N; j++)
dist = Math.max(dist, d[j]);
// If distance found is smaller than ans
// then make ans equal to distance
ans = Math.min(ans, dist);
}
// Return the minimum value
return ans;
}
// Driver Code
public static void main(String[] args)
{
int N = 4;
int[][] edges = { { 1, 2 }, { 2, 3 }, { 2, 4 } };
// Function call
System.out.println(minDist(N, edges));
}
}
// This code is contributed by Lovely Jain
# Python code to implement the above approach
# Function to run DFS
import sys
def dfs(d, node, adj, dist):
d[node] = dist
# DFS call for all its neighbours
for child in adj[node]:
if (dist + 1 < d[child]):
dfs(d, child, adj, dist + 1)
# Function to find the minimum distance
def minDist(N, edges):
ans = sys.maxsize
# Creation of the adjacency matrix
adj = [[] for i in range(N+1)]
for u in edges:
adj[u[0]].append(u[1])
adj[u[1]].append(u[0])
# Consider ith node as source
# in each iteration
for i in range(1,N+1):
# Distance array to store
# distance of all the nodes
# from source
d = [sys.maxsize for i in range(N+1)]
# DFS traversal of the tree and store
# distance of all nodes from source
dfs(d, i, adj, 0)
dist = 0
# Find max distance from distance array
for j in range(1,N+1):
dist = max(dist, d[j])
# If distance found is smaller than ans
# then make ans equal to distance
ans = min(ans, dist)
# Return the minimum value
return ans
# Driver Code
N = 4
edges = [[1, 2], [2, 3], [2, 4]]
# Function call
print(minDist(N, edges))
# This code is contributed by shinjanpatra
// C# code to implement the above approach
using System;
using System.Collections;
using System.Collections.Generic;
public class GFG {
// Function to run DFS
static void dfs(int[] d, int node, List<int>[] adj,
int dist)
{
d[node] = dist;
// DFS call for all its neighbours
foreach(int child in adj[node])
{
if (dist + 1 < d[child])
dfs(d, child, adj, dist + 1);
}
}
// Function to find the minimum distance
static int minDist(int N, int[, ] edges)
{
int ans = Int32.MaxValue;
// Creation of the adjacency matrix
List<int>[] adj = new List<int>[ N + 1 ];
for (int i = 0; i <= N; i++)
adj[i] = new List<int>();
for (int i = 0; i < edges.GetLength(0); i++) {
int[] u = { edges[i, 0], edges[i, 1] };
adj[u[0]].Add(u[1]);
adj[u[1]].Add(u[0]);
}
// Consider ith node as source
// in each iteration
for (int i = 1; i <= N; i++) {
// Distance array to store
// distance of all the nodes
// from source
int[] d = new int[N + 1];
for (int j = 0; j <= N; j++)
d[j] = Int32.MaxValue;
// DFS traversal of the tree and store
// distance of all nodes from source
dfs(d, i, adj, 0);
int dist = 0;
// Find max distance from distance array
for (int j = 1; j <= N; j++)
dist = Math.Max(dist, d[j]);
// If distance found is smaller than ans
// then make ans equal to distance
ans = Math.Min(ans, dist);
}
// Return the minimum value
return ans;
}
// Driver Code
public static void Main(string[] args)
{
int N = 4;
int[, ] edges = { { 1, 2 }, { 2, 3 }, { 2, 4 } };
// Function call
Console.WriteLine(minDist(N, edges));
}
}
// This code is contributed by karandeep1234.
<script>
// Javascript code to implement the above approach
// Function to run DFS
function dfs(d, node, adj, dist) {
d[node] = dist;
// DFS call for all its neighbours
for (let child of adj[node]) {
if (dist + 1 < d[child])
dfs(d, child, adj, dist + 1);
}
}
// Function to find the minimum distance
function minDist(N, edges) {
let ans = Number.MAX_SAFE_INTEGER;
// Creation of the adjacency matrix
let adj = new Array(N + 1).fill(0).map(() => new Array());
for (let u of edges) {
adj[u[0]].push(u[1]);
adj[u[1]].push(u[0]);
}
// Consider ith node as source
// in each iteration
for (let i = 1; i <= N; i++) {
// Distance array to store
// distance of all the nodes
// from source
let d = new Array(N + 1).fill(Number.MAX_SAFE_INTEGER);
// DFS traversal of the tree and store
// distance of all nodes from source
dfs(d, i, adj, 0);
let dist = 0;
// Find max distance from distance array
for (let j = 1; j <= N; j++)
dist = Math.max(dist, d[j]);
// If distance found is smaller than ans
// then make ans equal to distance
ans = Math.min(ans, dist);
}
// Return the minimum value
return ans;
}
// Driver Code
let N = 4;
let edges = [[1, 2], [2, 3], [2, 4]];
// Function call
document.write(minDist(N, edges));
// This code is contributed by gfgking.
</script>
Time Complexity: O(N*N)
Auxiliary Space: O(N)
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