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Minimum number of days required to schedule all exams

  • Difficulty Level : Hard
  • Last Updated : 20 Aug, 2021
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Given a graph consisting of N nodes, where each node represents an exam and a 2D array Edges[][2] such that each pair of the exam (Edges[i][0], Edges[i][1]) denotes the edge between them, the task is to find the minimum number of days required to schedule all the exams such that no two exams connected via an edge are scheduled on the same day.

Examples:

Input: N = 5, E = 10, Edges[][] = {{0, 1}, {0, 2}, {0, 3}, {0, 4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}
Output: 5
Explanation:

In the above graph, all the nodes (representing exams) are connected to each other via a directed path. Therefore, the minimum number of days required to complete the exam is 5.



Input: N = 7, E = 12, Edges[][] = [{0, 1}, {0, 3}, {0, 4}, {0, 6}, {1, 2}, {1, 4}, {1, 6}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {4, 5}]
Output: 3

Approach: The given problem can be solved by using the concept of Graph Coloring. Although, the problem is NP complete, a good approximation is as follows.

Below is the implementation of the above approach:

C++14




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Comparator function to sort the
// vector of pairs in decreasing order
bool compare(pair<int, int> a,
             pair<int, int> b)
{
    // If the first values are the same
    if (a.first == b.first) {
        return (a.second < b.second);
    }
 
    // Otherwise
    else {
        return (a.first > b.first);
    }
}
 
// Function to add an undirected
// edge between any pair of nodes
void addEdge(vector<vector<int> >& adj,
             int u, int v)
{
    adj[u][v] = 1;
    adj[v][u] = 1;
}
 
// Function to find the minimum number
// of days to schedule all the exams
int minimumDays(int V, int Edges[][2],
                int E)
{
    // Stores the adjacency list of
    // the given graph
    vector<vector<int> > adj(
        V, vector<int>(V, 0));
 
    // Iterate over the edges
    for (int i = 0; i < E; i++) {
 
        int u = Edges[i][0];
        int v = Edges[i][1];
 
        // Add the edges between the
        // nodes u and v
        addEdge(adj, u, v);
    }
 
    // Initialize a vector of pair that
    // stores { degree, vertex }
    vector<pair<int, int> > vdegree(V);
 
    for (int i = 0; i < V; i++) {
 
        // Degree of the node
        int degree = 0;
        vdegree[i].second = i;
 
        for (int j = 0; j < V; j++) {
            if (adj[i][j] != 0) {
 
                // Increment the degree
                degree++;
            }
        }
 
        // Update the degree of the
        // current node
        vdegree[i].first = degree;
    }
 
    // Sort to arrange all vertices
    // in descending order of degree
    sort(vdegree.begin(),
         vdegree.end(), compare);
 
    // Stores the vertices according
    // to degree in descending order
    int vorder[V];
 
    for (int i = 0; i < V; i++) {
        vorder[i] = vdegree[i].second;
    }
 
    // Stores the color of the all
    // the nodes
    int color[V];
 
    for (int i = 0; i < V; i++) {
        color[i] = i + 1;
    }
 
    int colored[V];
 
    // Initialize all vertices with
    // an invalid color 0
    memset(colored, 0, sizeof(colored));
 
    // Keeps the track of number of
    // vertices colored
    int numvc = 0;
 
    // Track the different color
    // assigned
    int k = 0;
 
    for (int i = 0; i < V; i++) {
 
        // If all vertices are colored
        // then exit from the for loop
        if (numvc == V) {
            break;
        }
 
        // If vertex is already
        // colored, then continue
        if (colored[vorder[i]] != 0) {
            continue;
        }
 
        // If vertex not colored
        else {
 
            colored[vorder[i]] = color[k];
 
            // After coloring increase
            // the count of colored
            // vertex by 1
            numvc++;
 
            for (int j = 0; j < V; j++) {
 
                // If the current node
                // and its adjacent are
                // not colored
                if (colored[j] == 0
                    && adj[vorder[i]][j] == 0) {
 
                    colored[j] = color[k];
 
                    // Increment the count
                    numvc++;
                }
            }
 
            // Increment k
            k++;
        }
    }
 
    // Sort the array
    sort(colored, colored + V);
 
    // Count of unique colors
    int unique_color = 1;
 
    // Count the number of unique
    // colors
    for (int i = 1; i < V; i++) {
 
        if (colored[i]
            != colored[i - 1]) {
            unique_color++;
        }
    }
 
    // Return the number of days
    // to sechedule an exam
    return unique_color;
}
 
// Driver Code
int main()
{
    int V = 7, E = 12;
    int Edges[][2]
        = { { 0, 1 }, { 0, 3 }, { 0, 4 }, { 0, 6 }, { 1, 2 }, { 1, 4 }, { 1, 6 }, { 2, 5 }, { 2, 6 }, { 3, 4 }, { 3, 5 }, { 4, 5 } };
    cout << minimumDays(V, Edges, E);
 
    return 0;
}

Python3




# Python 3 program for the above approach
 
# Comparator function to sort the
# vector of pairs in decreasing order
 
# Function to add an undirected
# edge between any pair of nodes
def addEdge(adj, u, v):
    adj[u][v] = 1
    adj[v][u] = 1
 
# Function to find the minimum number
# of days to schedule all the exams
def minimumDays(V, Edges, E):
    # Stores the adjacency list of
    # the given graph
    adj = [[0 for i in range(V)] for j in range(V)]
 
    # Iterate over the edges
    for i in range(E):
        u = Edges[i][0]
        v = Edges[i][1]
 
        # Add the edges between the
        # nodes u and v
        addEdge(adj, u, v)
 
    # Initialize a vector of pair that
    # stores [degree, vertex }
    vdegree = [[0,0] for i in range(V)]
 
    for i in range(V):
        # Degree of the node
        degree = 0
        vdegree[i][1] = i
 
        for j in range(V):
            if (adj[i][j] != 0):
                # Increment the degree
                degree += 1
 
        # Update the degree of the
        # current node
        vdegree[i][0] = degree
 
    # Sort to arrange all vertices
    # in descending order of degree
    vdegree.sort(reverse=True)
 
    # Stores the vertices according
    # to degree in descending order
    vorder = [0 for i in range(V)]
 
    for i in range(V):
        vorder[i] = vdegree[i][1]
 
    # Stores the color of the all
    # the nodes
    color = [0 for i in range(V)]
 
    for i in range(V):
        color[i] = i + 1
 
    colored = [0 for i in range(V)]
 
    # Keeps the track of number of
    # vertices colored
    numvc = 0
 
    # Track the different color
    # assigned
    k = 0
 
    for i in range(V):
        # If all vertices are colored
        # then exit from the for loop
        if (numvc == V):
            break
 
        # If vertex is already
        # colored, then continue
        if (colored[vorder[i]] != 0):
            continue
 
        # If vertex not colored
        else:
            colored[vorder[i]] = color[k]
 
            # After coloring increase
            # the count of colored
            # vertex by 1
            numvc += 1
 
            for j in range(V):
                # If the current node
                # and its adjacent are
                # not colored
                if (colored[j] == 0 and adj[vorder[i]][j] == 0):
                    colored[j] = color[k]
 
                    # Increment the count
                    numvc += 1
 
            # Increment k
            k += 1
 
    # Sort the array
    colored.sort()
 
    # Count of unique colors
    unique_color = 1
 
    # Count the number of unique
    # colors
    for i in range(1,V,1):
        if (colored[i] != colored[i - 1]):
            unique_color += 1
 
    # Return the number of days
    # to sechedule an exam
    return unique_color
 
# Driver Code
if __name__ == '__main__':
    V = 7
    E = 12
    Edges = [[0, 1 ], [0, 3 ], [0, 4 ], [0, 6 ], [1, 2 ], [1, 4 ], [1, 6 ], [2, 5 ], [2, 6 ], [3, 4 ], [3, 5 ], [4, 5 ] ]
    print(minimumDays(V, Edges, E))
     
    # This code is contributed by ipg2016107.
Output: 
3

 

Time Complexity: O(N2
Auxiliary Space: O(N)

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