Maximize jobs that can be completed under given constraint

07 Apr, 2022
Given an integer N denoting number of jobs and a matrix ranges[] consisting of a range [start day, end day] for each job within which it needs to be completed, the task is to find the maximum possible jobs that can be completed.Examples:  
Input: N = 5, Ranges = {{1, 5}, {1, 5}, {1, 5}, {2, 3}, {2, 3}} Output: 5 Explanation: Job 1 on day 1, Job 4 on day 2, Job 5 on day 3, Job 2 on day 4, Job 3 on day 5
Input: N=6, Ranges = {{1, 3}, {1, 3}, {2, 3}, {2, 3}, {1, 4}, {2, 5}} Output: 5  
 
Approach: The above problem can be solved using a Priority Queue. Follow the steps below to solve the problems:  
Find the minimum and maximum day in the range of jobs.
Sort all jobs in increasing order of start day.
Iterate from the minimum to maximum day, and for every ith day, select the job having least end day which can be completed on that day.
In order to perform the above step, maintain a Min Heap, and on every ith day, insert the jobs that can be completed on that day, into the Min Heap sorted by end day. If any job can completed on the ith day, consider the one with the lowest end day and increase the count of jobs completed.
Repeat this process for all the days and finally print the count of jobs completed.
Below is an implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ Program to implement the// above approach #include <bits/stdc++.h>using namespace std; // Function to find maximum// number of jobsint find_maximum_jobs(    int N,    vector<pair<int, int> > ranges){    // Min Heap    priority_queue<int, vector<int>,                   greater<int> >        queue;     // Sort ranges by start day    sort(ranges.begin(), ranges.end());     // Stores the minimum and maximum    // day in the ranges    int min_day = ranges[0].first;    int max_day = 0;     for (int i = 0; i < N; i++)        max_day            = max(max_day,                  ranges[i].second);     int index = 0, count_jobs = 0;     // Iterating from min_day to max_day    for (int i = min_day; i <= max_day; i++) {         // Insert the end day of the jobs        // which can be completed on        // i-th day in a priority queue        while (index < ranges.size()               && ranges[index].first <= i) {             queue.push(ranges[index].second);            index++;        }         // Pop all jobs whose end day        // is less than current day        while (!queue.empty()               && queue.top() < i)            queue.pop();         // If queue is empty, no job        // can be completed on        // the i-th day        if (queue.empty())            continue;         // Increment the count of        // jobs completed        count_jobs++;         // Pop the job with        // least end day        queue.pop();    }     // Return the jobs    // on the last day    return count_jobs;} // Driver Codeint main(){     int N = 5;    vector<pair<int, int> > ranges;    ranges.push_back({ 1, 5 });    ranges.push_back({ 1, 5 });    ranges.push_back({ 1, 5 });    ranges.push_back({ 2, 3 });    ranges.push_back({ 2, 3 });     cout << find_maximum_jobs(N, ranges);}
// Java Program to implement the// above approachimport java.io.*;import java.util.*; class GFG {   // Function to find maximum  // number of jobs  static int find_maximum_jobs(int N, ArrayList<ArrayList<Integer>> ranges)  {    // Min Heap    ArrayList<Integer> queue = new ArrayList<Integer>();    // Sort ranges by start day    Collections.sort(ranges, new Comparator<ArrayList<Integer>>() {         @Override      public int compare(ArrayList<Integer> o1, ArrayList<Integer> o2) {        return o1.get(0).compareTo(o2.get(0));      }                  });     // Stores the minimum and maximum    // day in the ranges    int min_day = ranges.get(0).get(0);    int max_day = 0;    for (int i = 0; i < N; i++)      max_day = Math.max(max_day, ranges.get(i).get(1));    int index = 0, count_jobs = 0;     // Iterating from min_day to max_day    for (int i = min_day; i <= max_day; i++)    {      // Insert the end day of the jobs      // which can be completed on      // i-th day in a priority queue      while (index < ranges.size() && ranges.get(index).get(0) <= i)      {        queue.add(ranges.get(index).get(1));        index++;      }      Collections.sort(queue);       // Pop all jobs whose end day      // is less than current day      while (queue.size() > 0 && queue.get(0) < i)        queue.remove(0);       // If queue is empty, no job      // can be completed on      // the i-th day      if (queue.size() == 0)        continue;      // Increment the count of      // jobs completed      count_jobs++;       // Pop the job with      // least end day      queue.remove(0);    }    // Return the jobs    // on the last day    return count_jobs;   }   // Driver code  public static void main (String[] args) {     int N = 5;    ArrayList<ArrayList<Integer>> ranges = new ArrayList<ArrayList<Integer>>();    ranges.add(new ArrayList<Integer>(Arrays.asList(1, 5)));    ranges.add(new ArrayList<Integer>(Arrays.asList(1, 5)));    ranges.add(new ArrayList<Integer>(Arrays.asList(1, 5)));    ranges.add(new ArrayList<Integer>(Arrays.asList(2, 3)));    ranges.add(new ArrayList<Integer>(Arrays.asList(2, 3)));    System.out.println(find_maximum_jobs(N, ranges));  }}// This code is contributed by avanitrachhadiya2155
# Python3 Program to implement the# above approach # Function to find maximum# number of jobsdef find_maximum_jobs(N, ranges) :     # Min Heap    queue = []         # Sort ranges by start day    ranges.sort()         # Stores the minimum and maximum    # day in the ranges    min_day = ranges[0][0]    max_day = 0    for i in range(N) :      max_day = max(max_day, ranges[i][1])    index, count_jobs = 0, 0         # Iterating from min_day to max_day    for i in range(min_day, max_day + 1) :           # Insert the end day of the jobs      # which can be completed on      # i-th day in a priority queue      while (index < len(ranges) and ranges[index][0] <= i) :               queue.append(ranges[index][1])        index += 1             queue.sort()           # Pop all jobs whose end day      # is less than current day      while (len(queue) > 0 and queue[0] < i) :        queue.pop(0)           # If queue is empty, no job      # can be completed on      # the i-th day      if (len(queue) == 0) :        continue           # Increment the count of      # jobs completed      count_jobs += 1           # Pop the job with      # least end day      queue.pop(0)         # Return the jobs    # on the last day    return count_jobs         # Driver codeN = 5ranges = []ranges.append((1, 5))ranges.append((1, 5))ranges.append((1, 5))ranges.append((2, 3))ranges.append((2, 3)) print(find_maximum_jobs(N, ranges)) # This code is contributed by divyeshrabadiya07.
// C# Program to implement the// above approachusing System;using System.Collections.Generic;class GFG {   // Function to find maximum  // number of jobs  static int find_maximum_jobs(int N, List<Tuple<int, int>> ranges)  {    // Min Heap    List<int> queue = new List<int>();     // Sort ranges by start day    ranges.Sort();     // Stores the minimum and maximum    // day in the ranges    int min_day = ranges[0].Item1;    int max_day = 0;    for (int i = 0; i < N; i++)      max_day = Math.Max(max_day, ranges[i].Item2);    int index = 0, count_jobs = 0;     // Iterating from min_day to max_day    for (int i = min_day; i <= max_day; i++)    {       // Insert the end day of the jobs      // which can be completed on      // i-th day in a priority queue      while (index < ranges.Count && ranges[index].Item1 <= i)      {        queue.Add(ranges[index].Item2);        index++;      }      queue.Sort();       // Pop all jobs whose end day      // is less than current day      while (queue.Count > 0 && queue[0] < i)        queue.RemoveAt(0);       // If queue is empty, no job      // can be completed on      // the i-th day      if (queue.Count == 0)        continue;       // Increment the count of      // jobs completed      count_jobs++;       // Pop the job with      // least end day      queue.RemoveAt(0);    }     // Return the jobs    // on the last day    return count_jobs;  }   // Driver code  static void Main()  {    int N = 5;    List<Tuple<int, int>> ranges = new List<Tuple<int, int>>();    ranges.Add(new Tuple<int,int>(1, 5));    ranges.Add(new Tuple<int,int>(1, 5));    ranges.Add(new Tuple<int,int>(1, 5));    ranges.Add(new Tuple<int,int>(2, 3));    ranges.Add(new Tuple<int,int>(2, 3));     Console.Write(find_maximum_jobs(N, ranges));  }} // This code is contributed by divyesh072019.
<script> // JavaScript program for the above approach // Function to find maximum// number of jobsfunction find_maximum_jobs(N, ranges){     // Min Heap    let queue = []         // Sort ranges by start day    ranges.sort()         // Stores the minimum && maximum    // day in the ranges    let min_day = ranges[0][0]    let max_day = 0    for(let i = 0;i<N ; i++){      max_day = Math.max(max_day, ranges[i][1])    }    let index = 0, count_jobs = 0         // Iterating from min_day to max_day    for(let i = min_day;i< max_day + 1;i++) {           // Insert the end day of the jobs      // which can be completed on      // i-th day in a priority queue      while (index < ranges.length && ranges[index][0] <= i){               queue.push(ranges[index][1])        index += 1      }             queue.sort()           // Pop all jobs whose end day      // is less than current day      while (queue.length > 0 && queue[0] < i)        queue.shift()           // If queue is empty, no job      // can be completed on      // the i-th day      if (queue.length == 0)        continue           // Increment the count of      // jobs completed      count_jobs += 1           // Pop the job with      // least end day      queue.shift()    }         // Return the jobs    // on the last day    return count_jobs}     // Driver codelet N = 5let ranges = []ranges.push([1, 5])ranges.push([1, 5])ranges.push([1, 5])ranges.push([2, 3])ranges.push([2, 3]) document.write(find_maximum_jobs(N, ranges)) // This code is contributed by shinjanpatra </script>
5
 
Time Complexity: O(Xlog(N)), where X is the difference between maximum and minimum day and N is the number of jobs. Auxiliary Space: O(N2) 
divyesh072019
divyeshrabadiya07
avanitrachhadiya2155
sumitgumber28
shinjanpatra
cpp-priority-queue
interview-preparation
min-heap
Algorithms
Competitive Programming
Greedy
Heap
Sorting
Greedy
Sorting
Heap
Algorithms
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.