Given an array of pairs Edges[][], representing edges connecting vertices in a Tree consisting of N nodes, the task is to find the minimum time required to color all the edges of a Tree based on the assumption that coloring an edge requires 1 unit of time.
Note: Multiple edges can be colored on a particular instant, but a node can be part of only one of the edges colored on a particular day.
Examples
Input: Edges[][] = ((1, 2), (3, 4), (2, 3))
Output: 2
Explanation:
Step 1: Color edges (1, 2) and (3, 4)
Step 2: Color edge (2, 3)Input: Edges[][] = ((1, 2), (1, 3), (1, 4))
Output : 3
Approach: This problem can be solved using DFS(Depth First Search). Follow the steps below to solve the problem:
- Initialize global variables, say ans as 0, to store the minimum time required to color all the edges of a Tree.
- Initialize a variable current_time as 0, to store the time required to color the current edge.
- Iterate over the children of the current node and perform the following steps:
- If the current edge is not visited, i.e the current node is not equal to the parent node:
- Increase current_time by 1.
- Check if the parent edge has been colored at the same time or not. If found to be true, then increase current_time by 1 as a node cannot be part of more than one edge which are being colored at the same time.
- Update ans as maximum of ans and current_time.
- Call the recursive function minTimeToColor for the children of the current node.
- If the current edge is not visited, i.e the current node is not equal to the parent node:
- After the end of this function, print ans.
Below is the code for the above approach.
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Stores the required answer
int ans = 0;
// Stores the graph
vector<int> edges[100000];
// Function to add edges
void Add_edge(int u, int v)
{
edges[u].push_back(v);
edges[v].push_back(u);
}
// Function to calculate the minimum time
// required to color all the edges of a tree
void minTimeToColor(int node, int parent,
int arrival_time)
{
// Starting from time = 0,
// for all the child edges
int current_time = 0;
for (auto x : edges[node]) {
// If the edge is not visited yet.
if (x != parent) {
// Time of coloring of
// the current edge
++current_time;
// If the parent edge has
// been colored at the same time
if (current_time == arrival_time)
++current_time;
// Update the maximum time
ans = max(ans, current_time);
// Recursively call the
// function to its child node
minTimeToColor(x, node, current_time);
}
}
}
// Driver Code
int main()
{
pair<int, int> A[] = { { 1, 2 },
{ 2, 3 },
{ 3, 4 } };
for (auto i : A) {
Add_edge(i.first, i.second);
}
// Function call
minTimeToColor(1, -1, 0);
// Finally, print the answer
cout << ans << "\n";
}
// Java program for the above approach
import java.util.*;
class GFG{
// Stores the required answer
static int ans = 0;
// Stores the graph
@SuppressWarnings("unchecked")
static Vector<Integer> edges[] = new Vector[100000];
// Function to add edges
static void Add_edge(int u, int v)
{
edges[u].add(v);
edges[v].add(u);
}
// Function to calculate the minimum time
// required to color all the edges of a tree
static void minTimeToColor(int node, int parent,
int arrival_time)
{
// Starting from time = 0,
// for all the child edges
int current_time = 0;
for(int x = 0; x < edges[node].size(); x++)
{
// If the edge is not visited yet.
if (edges[node].get(x) != parent)
{
// Time of coloring of
// the current edge
++current_time;
// If the parent edge has
// been colored at the same time
if (current_time == arrival_time)
++current_time;
// Update the maximum time
ans = Math.max(ans, current_time);
// Recursively call the
// function to its child node
minTimeToColor(edges[node].get(x), node,
current_time);
}
}
}
// Driver Code
public static void main(String[] args)
{
for(int i = 0; i < edges.length; i++)
edges[i] = new Vector<Integer>();
int A[][] = { { 1, 2 },
{ 2, 3 },
{ 3, 4 } };
for(int i = 0; i < 3; i++)
{
Add_edge(A[i][0], A[i][1]);
}
// Function call
minTimeToColor(1, -1, 0);
// Finally, print the answer
System.out.print(ans + "\n");
}
}
// This code is contributed by umadevi9616
# Python3 program for the above approach
# Stores the required answer
ans = 0
# Stores the graph
edges = [[] for i in range(100000)]
# Function to add edges
def Add_edge(u, v):
global edges
edges[u].append(v)
edges[v].append(u)
# Function to calculate the minimum time
# required to color all the edges of a tree
def minTimeToColor(node, parent, arrival_time):
global ans
# Starting from time = 0,
# for all the child edges
current_time = 0
for x in edges[node]:
# If the edge is not visited yet.
if (x != parent):
# Time of coloring of
# the current edge
current_time += 1
# If the parent edge has
# been colored at the same time
if (current_time == arrival_time):
current_time += 1
# Update the maximum time
ans = max(ans, current_time)
# Recursively call the
# function to its child node
minTimeToColor(x, node, current_time)
# Driver Code
if __name__ == '__main__':
A = [ [ 1, 2 ],
[ 2, 3 ],
[ 3, 4 ] ]
for i in A:
Add_edge(i[0], i[1])
# Function call
minTimeToColor(1, -1, 0)
# Finally, print the answer
print(ans)
# This code is contributed by mohit kumar 29
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Stores the required answer
static int ans = 0;
// Stores the graph
static List<List<int>> edges = new List<List<int>>();
// Function to add edges
static void Add_edge(int u, int v)
{
edges[u].Add(v);
edges[v].Add(u);
}
// Function to calculate the minimum time
// required to color all the edges of a tree
static void minTimeToColor(int node, int parent,
int arrival_time)
{
// Starting from time = 0,
// for all the child edges
int current_time = 0;
for(int x = 0; x < edges[node].Count; x++) {
// If the edge is not visited yet.
if (edges[node][x] != parent) {
// Time of coloring of
// the current edge
++current_time;
// If the parent edge has
// been colored at the same time
if (current_time == arrival_time)
++current_time;
// Update the maximum time
ans = Math.Max(ans, current_time);
// Recursively call the
// function to its child node
minTimeToColor(edges[node][x], node, current_time);
}
}
}
// Driver code
static void Main() {
for(int i = 0; i < 100000; i++)
{
edges.Add(new List<int>());
}
int[,] A = { { 1, 2 }, { 2, 3 }, { 3, 4 } };
for(int i = 0; i < 3; i++)
{
Add_edge(A[i,0], A[i,1]);
}
// Function call
minTimeToColor(1, -1, 0);
// Finally, print the answer
Console.WriteLine(ans);
}
}
// This code is contributed by divyeshrabadiya07.
<script>
// JavaScript program for the above approach
// Stores the required answer
let ans = 0;
// Stores the graph
let edges=new Array(100000);
for(let i=0;i<100000;i++)
edges[i]=[];
// Function to add edges
function Add_edge(u,v)
{
edges[u].push(v);
edges[v].push(u);
}
// Function to calculate the minimum time
// required to color all the edges of a tree
function minTimeToColor(node,parent,arrival_time)
{
// Starting from time = 0,
// for all the child edges
let current_time = 0;
for (let x=0;x<edges[node].length;x++) {
// If the edge is not visited yet.
if (edges[node][x] != parent) {
// Time of coloring of
// the current edge
++current_time;
// If the parent edge has
// been colored at the same time
if (current_time == arrival_time)
++current_time;
// Update the maximum time
ans = Math.max(ans, current_time);
// Recursively call the
// function to its child node
minTimeToColor(edges[node][x], node, current_time);
}
}
}
// Driver Code
let A=[[ 1, 2 ],[ 2, 3 ],[ 3, 4 ] ];
for(let i=0;i<A.length;i++)
{
Add_edge(A[i][0],A[i][1]);
}
// Function call
minTimeToColor(1, -1, 0);
// Finally, print the answer
document.write(ans);
// This code is contributed by patel2127
</script>
Output:
2
Time Complexity: O(N)
Auxiliary Space: O(N)