Minimum number of colors required to color a graph

Given a graph with N vertices and E edges. The edges are given as U[] and V[] such that for each index i, U[i] is connected to V[i]. The task is to find the minimum number of colors needed to color the given graph.

Examples

Input: N = 5, M = 6, U[] = { 1, 2, 3, 1, 2, 3 }, V[] = { 3, 3, 4, 4, 5, 5 };
Output: 3
Explanation:
For the above graph node 1, 3, and 5 cannot have the same color. Hence the count is 3.



Approach:
We will keep two array count[] and colors[]. The array count[] will store the count of edges for each node and colors[] will store the colors of each node. Initialize count for every vertex to 0 and color for every vertex to 1.
Steps:

  1. Make a adjacency list for the given set of edges and keep the count of edges for each node
  2. Iterate over all nodes and insert the node in the queue Q which has no edge.
  3. While Q is not empty, do the following:
    • Pop the element from the queue.
    • For all the nodes connected to popped node:
      1. Decrease the count of current edge for popped node.
      2. If color of current is less than or equals to color of it’s parent(the node which was popped) then, Update the color of current node = 1 + color of popped node
      3. If count is zero then insert this node in the queue Q.
  4. The maximum element in colors[] array will give the minimum number of colors required to color the given graph.

Below is the implementation of the above approach:

C++

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// C++ program to find the minimum
// number of colors needed to color
// the graph
#include <bits/stdc++.h>
using namespace std;
  
// Function to count the minimum
// number of color required
void minimumColors(int N, int E,
                   int U[], int V[])
{
  
    // Create array of vectors
    // to make adjacency list
    vector<int> adj[N];
  
    // Intialise colors array to 1
    // and count array to 0
    vector<int> count(N, 0);
    vector<int> colors(N, 1);
  
    // Create adjacency list of
    // a graph
    for (int i = 0; i < N; i++) {
        adj[V[i] - 1].push_back(U[i] - 1);
        count[U[i] - 1]++;
    }
  
    // Declare queue Q
    queue<int> Q;
  
    // Traverse count[] and insert
    // in Q if count[i] = 0;
    for (int i = 0; i < N; i++) {
        if (count[i] == 0) {
            Q.push(i);
        }
    }
  
    // Traverse queue and update
    // the count of colors
    // adjacent node
    while (!Q.empty()) {
        int u = Q.front();
        Q.pop();
  
        // Traverse node u
        for (auto x : adj[u]) {
            count[x]--;
  
            // If count[x] = 0
            // insert in Q
            if (count[x] == 0) {
                Q.push(x);
            }
  
            // If colors of child
            // node is less than
            // parent node, update
            // the count by 1
            if (colors[x] <= colors[u]) {
                colors[x] = 1 + colors[u];
            }
        }
    }
  
    // Stores the minimumColors
    // requires to color the graph.
    int minColor = -1;
  
    // Find the maximum of colors[]
    for (int i = 0; i < N; i++) {
        minColor = max(minColor, colors[i]);
    }
  
    // Print the minimum no. of
    // colors required.
    cout << minColor << endl;
}
  
// Driver function
int main()
{
    int N = 5, E = 6;
    int U[] = { 1, 2, 3, 1, 2, 3 };
    int V[] = { 3, 3, 4, 4, 5, 5 };
  
    minimumColors(N, E, U, V);
    return 0;
}

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Java

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// Java program to find the minimum
// number of colors needed to color
// the graph
import java.util.*;
  
class GFG
{
  
// Function to count the minimum
// number of color required
static void minimumColors(int N, int E,
                int U[], int V[])
{
  
    // Create array of vectors
    // to make adjacency list
    Vector<Integer> []adj = new Vector[N];
  
    // Intialise colors array to 1
    // and count array to 0
    int []count = new int[N];
    int []colors = new int[N];
    for (int i = 0; i < N; i++) {
        adj[i] = new Vector<Integer>();
        colors[i] = 1;
    }
      
    // Create adjacency list of
    // a graph
    for (int i = 0; i < N; i++) {
        adj[V[i] - 1].add(U[i] - 1);
        count[U[i] - 1]++;
    }
  
    // Declare queue Q
    Queue<Integer> Q = new LinkedList<>();
  
    // Traverse count[] and insert
    // in Q if count[i] = 0;
    for (int i = 0; i < N; i++) {
        if (count[i] == 0) {
            Q.add(i);
        }
    }
  
    // Traverse queue and update
    // the count of colors
    // adjacent node
    while (!Q.isEmpty()) {
        int u = Q.peek();
        Q.remove();
  
        // Traverse node u
        for (int x : adj[u]) {
            count[x]--;
  
            // If count[x] = 0
            // insert in Q
            if (count[x] == 0) {
                Q.add(x);
            }
  
            // If colors of child
            // node is less than
            // parent node, update
            // the count by 1
            if (colors[x] <= colors[u]) {
                colors[x] = 1 + colors[u];
            }
        }
    }
  
    // Stores the minimumColors
    // requires to color the graph.
    int minColor = -1;
  
    // Find the maximum of colors[]
    for (int i = 0; i < N; i++) {
        minColor = Math.max(minColor, colors[i]);
    }
  
    // Print the minimum no. of
    // colors required.
    System.out.print(minColor +"\n");
}
  
// Driver function
public static void main(String[] args)
{
    int N = 5, E = 6;
    int U[] = { 1, 2, 3, 1, 2, 3 };
    int V[] = { 3, 3, 4, 4, 5, 5 };
  
    minimumColors(N, E, U, V);
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 program to find the minimum
# number of colors needed to color
# the graph
from collections import deque
  
# Function to count the minimum
# number of color required
def minimumColors(N, E, U, V):
  
    # Create array of vectors
    # to make adjacency list
    adj = [[] for i in range(N)]
  
    # Intialise colors array to 1
    # and count array to 0
    count = [0]*N
    colors = [1]*(N)
  
    # Create adjacency list of
    # a graph
    for i in range(N):
        adj[V[i] - 1].append(U[i] - 1)
        count[U[i] - 1] += 1
  
    # Declare queue Q
    Q = deque()
  
    # Traverse count[] and insert
    # in Q if count[i] = 0
    for i in range(N):
        if (count[i] == 0):
            Q.append(i)
  
    # Traverse queue and update
    # the count of colors
    # adjacent node
    while len(Q) > 0:
        u = Q.popleft()
  
        # Traverse node u
        for x in adj[u]:
            count[x] -= 1
  
            # If count[x] = 0
            # insert in Q
            if (count[x] == 0):
                Q.append(x)
  
            # If colors of child
            # node is less than
            # parent node, update
            # the count by 1
            if (colors[x] <= colors[u]):
                colors[x] = 1 + colors[u]
  
    # Stores the minimumColors
    # requires to color the graph.
    minColor = -1
  
    # Find the maximum of colors[]
    for i in range(N):
        minColor = max(minColor, colors[i])
  
    # Print the minimum no. of
    # colors required.
    print(minColor)
  
# Driver function
N = 5
E = 6
U = [1, 2, 3, 1, 2, 3]
V = [3, 3, 4, 4, 5, 5]
  
minimumColors(N, E, U, V)
  
# This code is contributed by mohit kumar 29

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C#

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// C# program to find the minimum
// number of colors needed to color
// the graph
using System;
using System.Collections.Generic;
  
class GFG
{
   
// Function to count the minimum
// number of color required
static void minimumColors(int N, int E,
                int []U, int []V)
{
   
    // Create array of vectors
    // to make adjacency list
    List<int> []adj = new List<int>[N];
   
    // Intialise colors array to 1
    // and count array to 0
    int []count = new int[N];
    int []colors = new int[N];
    for (int i = 0; i < N; i++) {
        adj[i] = new List<int>();
        colors[i] = 1;
    }
       
    // Create adjacency list of
    // a graph
    for (int i = 0; i < N; i++) {
        adj[V[i] - 1].Add(U[i] - 1);
        count[U[i] - 1]++;
    }
   
    // Declare queue Q
    List<int> Q = new List<int>();
   
    // Traverse []count and insert
    // in Q if count[i] = 0;
    for (int i = 0; i < N; i++) {
        if (count[i] == 0) {
            Q.Add(i);
        }
    }
   
    // Traverse queue and update
    // the count of colors
    // adjacent node
    while (Q.Count!=0) {
        int u = Q[0];
        Q.RemoveAt(0);
   
        // Traverse node u
        foreach (int x in adj[u]) {
            count[x]--;
   
            // If count[x] = 0
            // insert in Q
            if (count[x] == 0) {
                Q.Add(x);
            }
   
            // If colors of child
            // node is less than
            // parent node, update
            // the count by 1
            if (colors[x] <= colors[u]) {
                colors[x] = 1 + colors[u];
            }
        }
    }
   
    // Stores the minimumColors
    // requires to color the graph.
    int minColor = -1;
   
    // Find the maximum of colors[]
    for (int i = 0; i < N; i++) {
        minColor = Math.Max(minColor, colors[i]);
    }
   
    // Print the minimum no. of
    // colors required.
    Console.Write(minColor +"\n");
}
   
// Driver function
public static void Main(String[] args)
{
    int N = 5, E = 6;
    int []U = { 1, 2, 3, 1, 2, 3 };
    int []V = { 3, 3, 4, 4, 5, 5 };
   
    minimumColors(N, E, U, V);
}
}
  
// This code is contributed by 29AjayKumar

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Output:

3

Time Complexity: O(N+E), where N = number of vertices and E = Number of edges

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