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Activity Selection Problem | Greedy Algo-1

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  • Difficulty Level : Easy
  • Last Updated : 26 Nov, 2022
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Greedy is an algorithmic paradigm that builds up a solution piece by piece, always choosing the next piece that offers the most obvious and immediate benefit. Greedy algorithms are used for optimization problems. 

An optimization problem can be solved using Greedy if the problem has the following property: 

  • At every step, we can make a choice that looks best at the moment, and we get the optimal solution to the complete problem. 

If a Greedy Algorithm can solve a problem, then it generally becomes the best method to solve that problem as the Greedy algorithms are in general more efficient than other techniques like Dynamic Programming. But Greedy algorithms cannot always be applied. For example, the Fractional Knapsack problem can be solved using Greedy, but 0-1 Knapsack cannot be solved using Greedy.

Following are some standard algorithms that are Greedy algorithms:

1) Kruskal’s Minimum Spanning Tree (MST): 

In Kruskal’s algorithm, we create an MST by picking edges one by one. The Greedy Choice is to pick the smallest weight edge that doesn’t cause a cycle in the MST constructed so far

2) Prim’s Minimum Spanning Tree: 

In Prim’s algorithm also, we create a MST by picking edges one by one. We maintain two sets: a set of the vertices already included in MST and the set of the vertices not yet included. The Greedy Choice is to pick the smallest weight edge that connects the two sets

3) Dijkstra’s Shortest Path

Dijkstra’s algorithm is very similar to Prim’s algorithm. The shortest-path tree is built up, edge by edge. We maintain two sets: a set of the vertices already included in the tree and a set of the vertices not yet included. The Greedy Choice is to pick the edge that connects the two sets and is on the smallest weight path from the source to the set that contains not yet included vertices

4) Huffman Coding: 

Huffman Coding is a loss-less compression technique. It assigns variable-length bit codes to different characters. The Greedy Choice is to assign the least bit length code to the most frequent character.

The greedy algorithms are sometimes also used to get an approximation for Hard optimization problems. For example, Traveling Salesman Problem is an NP-Hard problem. A Greedy choice for this problem is to pick the nearest unvisited city from the current city at every step. These solutions don’t always produce the best optimal solution but can be used to get an approximately optimal solution.

Here let us see one such problem that can be solved using Greedy algorithm

Problem:

You are given n activities with their start and finish times. Select the maximum number of activities that can be performed by a single person, assuming that a person can only work on a single activity at a time. 

Examples:  

Input: start[]  =  {10, 12, 20}, finish[] =  {20, 25, 30}
Output: 0 2
Explanation: A person can perform at most two activities. The 
maximum set of activities that can be executed 
is {0, 2} [ These are indexes in start[] and finish[] ]

Input: start[]  =  {1, 3, 0, 5, 8, 5}, finish[] =  {2, 4, 6, 7, 9, 9};
Output: 0 1 3 4
Explanation: A person can perform at most four activities. The 
maximum set of activities that can be executed 
is {0, 1, 3, 4} [ These are indexes in start[] and finish[]

Recommended Practice

Approach: To solve the problem follow the below idea:

The greedy choice is to always pick the next activity whose finish time is the least among the remaining activities and the start time is more than or equal to the finish time of the previously selected activity. We can sort the activities according to their finishing time so that we always consider the next activity as the minimum finishing time activity

Follow the given steps to solve the problem:

  • Sort the activities according to their finishing time 
  • Select the first activity from the sorted array and print it 
  • Do the following for the remaining activities in the sorted array
    • If the start time of this activity is greater than or equal to the finish time of the previously selected activity then select this activity and print it

Note: In the implementation, it is assumed that the activities are already sorted according to their finish time

Below is the implementation of the above approach.

C++




// C++ program for activity selection problem.
 
// The following implementation assumes that the activities
// are already sorted according to their finish time
#include <bits/stdc++.h>
using namespace std;
 
// Prints a maximum set of activities that can be done by a
// single person, one at a time.
void printMaxActivities(int s[], int f[], int n)
{
    int i, j;
 
    cout << "Following activities are selected" << endl;
 
    // The first activity always gets selected
    i = 0;
    cout << i << " ";
 
    // Consider rest of the activities
    for (j = 1; j < n; j++) {
        // If this activity has start time greater than or
        // equal to the finish time of previously selected
        // activity, then select it
        if (s[j] >= f[i]) {
            cout << j << " ";
            i = j;
        }
    }
}
 
// Driver code
int main()
{
    int s[] = { 1, 3, 0, 5, 8, 5 };
    int f[] = { 2, 4, 6, 7, 9, 9 };
    int n = sizeof(s) / sizeof(s[0]);
 
    // Function call
    printMaxActivities(s, f, n);
    return 0;
}
// this code contributed by shivanisinghss2110

C




// C program for activity selection problem.
 
// The following implementation assumes that the activities
// are already sorted according to their finish time
#include <stdio.h>
 
// Prints a maximum set of activities that can be done by a
// single person, one at a time.
void printMaxActivities(int s[], int f[], int n)
{
    int i, j;
 
    printf("Following activities are selected\n");
 
    // The first activity always gets selected
    i = 0;
    printf("%d ", i);
 
    // Consider rest of the activities
    for (j = 1; j < n; j++) {
        // If this activity has start time greater than or
        // equal to the finish time of previously selected
        // activity, then select it
        if (s[j] >= f[i]) {
            printf("%d ", j);
            i = j;
        }
    }
}
 
// Driver code
int main()
{
    int s[] = { 1, 3, 0, 5, 8, 5 };
    int f[] = { 2, 4, 6, 7, 9, 9 };
    int n = sizeof(s) / sizeof(s[0]);
 
    // Function call
    printMaxActivities(s, f, n);
    return 0;
}

Java




// Java program for activity selection problem.
 
// The following implementation assumes that the activities
// are already sorted according to their finish time
import java.io.*;
import java.lang.*;
import java.util.*;
 
class ActivitySelection {
    // Prints a maximum set of activities that can be done
    // by a single person, one at a time.
    public static void printMaxActivities(int s[], int f[],
                                          int n)
    {
        int i, j;
 
        System.out.println(
            "Following activities are selected");
 
        // The first activity always gets selected
        i = 0;
        System.out.print(i + " ");
 
        // Consider rest of the activities
        for (j = 1; j < n; j++) {
            // If this activity has start time greater than
            // or equal to the finish time of previously
            // selected activity, then select it
            if (s[j] >= f[i]) {
                System.out.print(j + " ");
                i = j;
            }
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int s[] = { 1, 3, 0, 5, 8, 5 };
        int f[] = { 2, 4, 6, 7, 9, 9 };
        int n = s.length;
 
        // Function call
        printMaxActivities(s, f, n);
    }
}

Python3




# Python3 program for activity selection problem.
 
# The following implementation assumes that the activities
# are already sorted according to their finish time
 
# Prints a maximum set of activities that can be done
# by a single person, one at a time
def printMaxActivities(s, f):
    n = len(f)
    print("Following activities are selected")
 
    # The first activity is always selected
    i = 0
    print(i, end=' ')
 
    # Consider rest of the activities
    for j in range(1, n):
 
        # If this activity has start time greater than
        # or equal to the finish time of previously
        # selected activity, then select it
        if s[j] >= f[i]:
            print(j, end=' ')
            i = j
 
 
# Driver code
if __name__ == '__main__':
    s = [1, 3, 0, 5, 8, 5]
    f = [2, 4, 6, 7, 9, 9]
 
    # Function call
    printMaxActivities(s, f)
 
# This code is contributed by Nikhil Kumar Singh

C#




// C# program for activity selection problem.
 
// The following implementation assumes
// that the activities are already sorted
// according to their finish time
using System;
 
class GFG {
    // Prints a maximum set of activities
    // that can be done by a single
    // person, one at a time.
    public static void printMaxActivities(int[] s, int[] f,
                                          int n)
    {
        int i, j;
 
        Console.Write(
            "Following activities are selected\n");
 
        // The first activity always gets selected
        i = 0;
        Console.Write(i + " ");
 
        // Consider rest of the activities
        for (j = 1; j < n; j++) {
            // If this activity has start time greater than
            // or equal to the finish time of previously
            // selected activity, then select it
            if (s[j] >= f[i]) {
                Console.Write(j + " ");
                i = j;
            }
        }
    }
 
    // Driver Code
    public static void Main()
    {
        int[] s = { 1, 3, 0, 5, 8, 5 };
        int[] f = { 2, 4, 6, 7, 9, 9 };
        int n = s.Length;
 
        // Function call
        printMaxActivities(s, f, n);
    }
}
 
// This code is contributed
// by ChitraNayal

PHP




<?php
// PHP program for activity selection problem.
 
// The following implementation assumes that
// the activities are already sorted according
// to their finish time
 
// Prints a maximum set of activities
// that can be done by a single
// person, one at a time.
function printMaxActivities($s, $f, $n)
{
 
    echo "Following activities are selected " . "\n";
 
    // The first activity always gets selected
    $i = 0;
    echo $i . " ";
 
    // Consider rest of the activities
    for ($j = 1; $j < $n; $j++)
    {
         
    // If this activity has start time greater
    // than or equal to the finish time of
    // previously selected activity, then select it
    if ($s[$j] >= $f[$i])
    {
        echo $j . " ";
        $i = $j;
    }
    }
}
 
// Driver Code
$s = array(1, 3, 0, 5, 8, 5);
$f = array(2, 4, 6, 7, 9, 9);
$n = sizeof($s);
 
// Function call
printMaxActivities($s, $f, $n);
 
// This code is contributed
// by Akanksha Rai
?>

Javascript




<script>
// The following implementation assumes that the activities
// are already sorted according to their finish time
 
    // Prints a maximum set of activities that can be done by a single
    // person, one at a time.
    function printMaxActivities(s,f,n)
    {
        let i, j;
        document.write("Following activities are selected : n");
         
        // The first activity always gets selected
        i = 0;
        document.write(i+" ");
         
        // Consider rest of the activities
        for (j = 1; j < n; j++)
        {
         
             // If this activity has start time greater than or
             // equal to the finish time of previously selected
             // activity, then select it
             if (s[j] >= f[i])
             {
                  document.write(j+" ");
                  i = j;
              }
        }
    }
     
    // Driver program to test above function
    let s = [1, 3, 0, 5, 8, 5]
    let f = [2, 4, 6, 7, 9, 9]
    let n = s.length;
    printMaxActivities(s, f, n);
     
    // This code is contributed by avanitrachhadiya2155
</script>

Output

Following activities are selected
0 1 3 4 

Time Complexity: O(N)
Auxiliary Space: O(1)

How does Greedy Choice work for Activities sorted according to finish time? 

Let the given set of activities be S = {1, 2, 3, …n}, and activities are sorted by finish time. The greedy choice is to always pick activity 1. How come activity 1 always provides one of the optimal solutions?

 We can prove it by showing that if there is another solution B with the first activity other than 1, then there is also a solution A of the same size as activity 1 as the first activity. Let the first activity selected by B be k, then there always exist A = {B – {k}} U {1}.

Note: The activities in B are independent and k has the smallest finishing time among all. Since k is not 1, finish(k) >= finish(1))

How to implement when given activities are not sorted? 

We create a structure/class for activities. We sort all activities by finish time (Refer sort in C++ STL). Once we have the activities sorted, we apply the same algorithm.

Below image is an illustration of the above approach: 

Approach for Solving Activity Selection ProblemBelow is the implementation of the above approach:

C++




// C++ program for activity selection problem
// when input activities may not be sorted.
#include <bits/stdc++.h>
using namespace std;
 
// A job has a start time, finish time and profit.
struct Activitiy {
    int start, finish;
};
 
// A utility function that is used for sorting
// activities according to finish time
bool activityCompare(Activitiy s1, Activitiy s2)
{
    return (s1.finish < s2.finish);
}
 
// Returns count of the maximum set of activities that can
// be done by a single person, one at a time.
void printMaxActivities(Activitiy arr[], int n)
{
    // Sort jobs according to finish time
    sort(arr, arr + n, activityCompare);
 
    cout << "Following activities are selected :\n";
 
    // The first activity always gets selected
    int i = 0;
    cout << "(" << arr[i].start << ", " << arr[i].finish
         << ")";
 
    // Consider rest of the activities
    for (int j = 1; j < n; j++) {
        // If this activity has start time greater than or
        // equal to the finish time of previously selected
        // activity, then select it
        if (arr[j].start >= arr[i].finish) {
            cout << ", (" << arr[j].start << ", "
                 << arr[j].finish << ")";
            i = j;
        }
    }
}
 
// Driver code
int main()
{
    Activitiy arr[] = { { 5, 9 }, { 1, 2 }, { 3, 4 },
                        { 0, 6 }, { 5, 7 }, { 8, 9 } };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    printMaxActivities(arr, n);
    return 0;
}

Java




// Java program for activity selection problem
// when input activities may not be sorted.
import java.io.*;
import java.util.*;
 
// A job has a start time, finish time and profit.
class Activity {
    int start, finish;
 
    // Constructor
    public Activity(int start, int finish)
    {
        this.start = start;
        this.finish = finish;
    }
}
 
// class to define user defined comparator
class Compare {
 
    // A utility function that is used for sorting
    // activities according to finish time
    static void compare(Activity arr[], int n)
    {
        Arrays.sort(arr, new Comparator<Activity>() {
            @Override
            public int compare(Activity s1, Activity s2)
            {
                return s1.finish - s2.finish;
            }
        });
    }
}
 
// Driver class
class GFG {
 
    // Returns count of the maximum set of activities that
    // can
    // be done by a single person, one at a time.
    static void printMaxActivities(Activity arr[], int n)
    {
        // Sort jobs according to finish time
        Compare obj = new Compare();
        obj.compare(arr, n);
        System.out.println(
            "Following activities are selected :");
 
        // The first activity always gets selected
        int i = 0;
        System.out.print("(" + arr[i].start + ", "
                         + arr[i].finish + ")");
 
        // Consider rest of the activities
        for (int j = 1; j < n; j++) {
 
            // If this activity has start time greater than
            // or equal to the finish time of previously
            // selected activity, then select it
            if (arr[j].start >= arr[i].finish) {
                System.out.print(", (" + arr[j].start + ", "
                                 + arr[j].finish + ")");
                i = j;
            }
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int n = 6;
        Activity arr[] = new Activity[n];
        arr[0] = new Activity(5, 9);
        arr[1] = new Activity(1, 2);
        arr[2] = new Activity(3, 4);
        arr[3] = new Activity(0, 6);
        arr[4] = new Activity(5, 7);
        arr[5] = new Activity(8, 9);
 
        // Function call
        printMaxActivities(arr, n);
    }
}
 
// This code is contributed by Dharanendra L V.

Python3




''' Python program for activity selection problem
 when input activities may not be sorted.'''
 
 
def MaxActivities(arr, n):
    selected = []
 
    # Sort jobs according to finish time
    Activity.sort(key=lambda x: x[1])
 
    # The first activity always gets selected
    i = 0
    selected.append(arr[i])
 
    for j in range(1, n):
 
        '''If this activity has start time greater than or
           equal to the finish time of previously selected
           activity, then select it'''
        if arr[j][0] >= arr[i][1]:
            selected.append(arr[j])
            i = j
    return selected
 
 
# Driver code
if __name__ == '__main__':
    Activity = [[5, 9], [1, 2], [3, 4], [0, 6], [5, 7], [8, 9]]
    n = len(Activity)
 
    # Function call
    selected = MaxActivities(Activity, n)
    print("Following activities are selected :")
    print(selected[0], end = "");
    for i in range (1, len(selected)):
        print(",", end = " ")
        print(selected[i], end = "")
 
# This code is contributed by kshitijjainm

Javascript




<script>
/* JavaScript program for activity selection problem
 when input activities may not be sorted.*/
function MaxActivities(arr, n){
    let selected = [];
     
    // Sort jobs according to finish time
       Activity = Activity.sort(function(a,b) {
    return a[1] - b[1];
    });
     
    // The first activity always gets selected
    let i = 0
    selected.push(arr[i]);
 
    for(let j=1;j<n;j++){
      /*If this activity has start time greater than or
         equal to the finish time of previously selected
         activity, then select it*/
      if( arr[j][0] >= arr[i][1]){
          selected.push(arr[j]);
          i = j;
      }
    }
    return selected;
}
// Driver code
Activity = [[5, 9], [1, 2], [3, 4], [0, 6],[5, 7], [8, 9]];
n = Activity.length;
selected = MaxActivities(Activity, n);
document.write("Following activities are selected : <br>")
console.log(selected)
for(let i = 0;i<selected.length;i++)
    document.write("("+selected[i]+"), ")
</script>

Output

Following activities are selected :
(1, 2), (3, 4), (5, 7), (8, 9)

Time Complexity: O(N log N), If input activities may not be sorted. It takes O(n) time when it is given that input activities are always sorted.
Auxiliary Space: O(1)

Activity Selection Problem using Priority-Queue:

We can use Min-Heap to get the activity with minimum finish time. Min-Heap can be implemented using priority-queue

Follow the given steps to solve the problem:

  • Create a priority queue (Min-Heap) and push the activities into it.
  • Push the top of the priority queue into the answer vector and set the variable start to the start time of the first activity and end to the finish time of the activity
  • While priority is not empty do the following:
    • Take the top of the priority queue and check
    • If the start time of this activity is greater than or equal to the finish time of the last chosen activity then push this activity into the answer vector
    • Else ignore it
  • Print the activities chosen, stored in the answer vector

Below is the implementation of the above approach:

CPP




// C++ program for activity selection problem
// when input activities may not be sorted.
#include <bits/stdc++.h>
using namespace std;
 
void SelectActivities(vector<int> s, vector<int> f)
{
    // Vector to store results.
    vector<pair<int, int> > ans;
 
    // Minimum Priority Queue to sort activities in
    // ascending order of finishing time (f[i]).
 
    priority_queue<pair<int, int>, vector<pair<int, int> >,
                   greater<pair<int, int> > >
        p;
 
    for (int i = 0; i < s.size(); i++) {
        // Pushing elements in priority queue where the key
        // is f[i]
        p.push(make_pair(f[i], s[i]));
    }
 
    auto it = p.top();
    int start = it.second;
    int end = it.first;
    p.pop();
    ans.push_back(make_pair(start, end));
 
    while (!p.empty()) {
        auto itr = p.top();
        p.pop();
        if (itr.second >= end) {
            start = itr.second;
            end = itr.first;
            ans.push_back(make_pair(start, end));
        }
    }
    cout << "Following Activities should be selected. "
         << endl
         << endl;
 
    for (auto itr = ans.begin(); itr != ans.end(); itr++) {
        cout << "Activity started at " << (*itr).first
             << " and ends at " << (*itr).second << endl;
    }
}
 
// Driver code
int main()
{
    vector<int> s = { 1, 3, 0, 5, 8, 5 };
    vector<int> f = { 2, 4, 6, 7, 9, 9 };
 
    // Function call
    SelectActivities(s, f);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
    // Pair class
    static class Pair {
 
        int first;
        int second;
 
        Pair(int first, int second)
        {
            this.first = first;
            this.second = second;
        }
    }
 
    static void SelectActivities(int s[], int f[])
    {
 
        // Vector to store results.
        ArrayList<Pair> ans = new ArrayList<>();
 
        // Minimum Priority Queue to sort activities in
        // ascending order of finishing time (f[i]).
        PriorityQueue<Pair> p = new PriorityQueue<>(
            (p1, p2) -> p1.first - p2.first);
 
        for (int i = 0; i < s.length; i++) {
            // Pushing elements in priority queue where the
            // key is f[i]
            p.add(new Pair(f[i], s[i]));
        }
 
        Pair it = p.poll();
        int start = it.second;
        int end = it.first;
        ans.add(new Pair(start, end));
 
        while (!p.isEmpty()) {
            Pair itr = p.poll();
            if (itr.second >= end) {
                start = itr.second;
                end = itr.first;
                ans.add(new Pair(start, end));
            }
        }
        System.out.println(
            "Following Activities should be selected. \n");
 
        for (Pair itr : ans) {
            System.out.println(
                "Activity started at " + itr.first
                + " and ends at " + itr.second);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int s[] = { 1, 3, 0, 5, 8, 5 };
        int f[] = { 2, 4, 6, 7, 9, 9 };
 
        // Function call
        SelectActivities(s, f);
    }
}
 
// This code is contributed by Kingash.

Python3




# Python3 program for activity selection problem
# when input activities may not be sorted.
from heapq import heappop, heappush
 
# Function to select activites
 
 
def SelectActivities(s, f):
    ans = []
    p = []
 
    # Pushing elements in the list
    for i, j in zip(s, f):
        heappush(p, (j, i))
 
    it = heappop(p)
    start = it[1]
    end = it[0]
    ans.append(it)
 
    # Sorting process
    while p:
        it = heappop(p)
        if it[1] >= end:
            start = it[1]
            end = it[0]
            ans.append(it)
 
    print("Following Activities should be selected.\n")
    for f, s in ans:
        print(f"Activity started at {s} and ends at {f}")
 
 
# Driver code
if __name__ == "__main__":
    s = [1, 3, 0, 5, 8, 5]
    finish = [2, 4, 6, 7, 9, 9]
 
    # Function call
    SelectActivities(s, finish)
 
# This code is contributed by kraanzu.

Javascript




<script>
// javascript program for the above approach
 
 // Pair class
class Pair
{
    constructor(first,second)
    {
        this.first = first;
          this.second = second;
    }
}
 
function SelectActivities(s,f)
{
    // Vector to store results.
    let ans = [];
   
    // Minimum Priority Queue to sort activities in
    // ascending order of finishing time (f[i]).
    let p = [];
   
    for (let i = 0; i < s.length; i++) {
      // Pushing elements in priority queue where the
      // key is f[i]
      p.push(new Pair(f[i], s[i]));
    }
    p.sort(function(a,b){return a.first-b.first;});
   
    let it = p.shift();
    let start = it.second;
    let end = it.first;
    ans.push(new Pair(start, end));
   
    while (p.length!=0) {
      let itr = p.shift();
      if (itr.second >= end) {
        start = itr.second;
        end = itr.first;
        ans.push(new Pair(start, end));
      }
    }
    document.write(
      "Following Activities should be selected. <br>");
   
    for(let itr of ans.values()) {
      document.write(
        "Activity started at: " + itr.first
        + " and ends at  " + itr.second+"<br>");
    }
}
 
// Driver Code
let s=[1, 3, 0, 5, 8, 5 ];
let f=[2, 4, 6, 7, 9, 9 ];
// Function call
SelectActivities(s, f);
 
 
// This code is contributed by rag2127
</script>

Output

Following Activities should be selected. 

Activity started at: 1 and ends at  2
Activity started at: 3 and ends at  4
Activity started at: 5 and ends at  7
Activity started at: 8 and ends at  9

Time Complexity: O(N * log N)
Auxiliary Space: O(N)


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