Given N jobs where every job is represented by following three elements of it.
- Start Time
- Finish Time
- Profit or Value Associated (>= 0)
Find the maximum profit subset of jobs such that no two jobs in the subset overlap.
Example:
Input: Number of Jobs n = 4 Job Details {Start Time, Finish Time, Profit} Job 1: {1, 2, 50} Job 2: {3, 5, 20} Job 3: {6, 19, 100} Job 4: {2, 100, 200} Output: The maximum profit is 250. We can get the maximum profit by scheduling jobs 1 and 4. Note that there is longer schedules possible Jobs 1, 2 and 3 but the profit with this schedule is 20+50+100 which is less than 250.
A simple version of this problem is discussed here where every job has the same profit or value. The Greedy Strategy for activity selection doesn’t work here as a schedule with more jobs may have smaller profit or value.
The above problem can be solved using the following recursive solution.
1) First sort jobs according to finish time. 2) Now apply following recursive process. // Here arr[] is array of n jobs findMaximumProfit(arr[], n) { a) if (n == 1) return arr[0]; b) Return the maximum of following two profits. (i) Maximum profit by excluding current job, i.e., findMaximumProfit(arr, n-1) (ii) Maximum profit by including the current job } How to find the profit including current job? The idea is to find the latest job before the current job (in sorted array) that doesn't conflict with current job 'arr[n-1]'. Once we find such a job, we recur for all jobs till that job and add profit of current job to result. In the above example, "job 1" is the latest non-conflicting for "job 4" and "job 2" is the latest non-conflicting for "job 3".
The following is the implementation of the above naive recursive method.
// C++ program for weighted job scheduling using Naive Recursive Method #include <iostream> #include <algorithm> using namespace std;
// A job has start time, finish time and profit. struct Job
{ int start, finish, profit;
}; // A utility function that is used for sorting events // according to finish time bool jobComparator(Job s1, Job s2)
{ return (s1.finish < s2.finish);
} // Find the latest job (in sorted array) that doesn't // conflict with the job[i]. If there is no compatible job, // then it returns -1. int latestNonConflict(Job arr[], int i)
{ for ( int j=i-1; j>=0; j--)
{
if (arr[j].finish <= arr[i-1].start)
return j;
}
return -1;
} // A recursive function that returns the maximum possible // profit from given array of jobs. The array of jobs must // be sorted according to finish time. int findMaxProfitRec(Job arr[], int n)
{ // Base case
if (n == 1) return arr[n-1].profit;
// Find profit when current job is included
int inclProf = arr[n-1].profit;
int i = latestNonConflict(arr, n);
if (i != -1)
inclProf += findMaxProfitRec(arr, i+1);
// Find profit when current job is excluded
int exclProf = findMaxProfitRec(arr, n-1);
return max(inclProf, exclProf);
} // The main function that returns the maximum possible // profit from given array of jobs int findMaxProfit(Job arr[], int n)
{ // Sort jobs according to finish time
sort(arr, arr+n, jobComparator);
return findMaxProfitRec(arr, n);
} // Driver program int main()
{ Job arr[] = {{3, 10, 20}, {1, 2, 50}, {6, 19, 100}, {2, 100, 200}};
int n = sizeof (arr)/ sizeof (arr[0]);
cout << "The optimal profit is " << findMaxProfit(arr, n);
return 0;
} |
// JAVA program for weighted job scheduling using Naive Recursive Method import java.util.*;
class GFG
{ // A job has start time, finish time and profit. static class Job
{ int start, finish, profit;
Job( int start, int finish, int profit)
{
this .start = start;
this .finish = finish;
this .profit = profit;
}
} // Find the latest job (in sorted array) that doesn't // conflict with the job[i]. If there is no compatible job, // then it returns -1. static int latestNonConflict(Job arr[], int i)
{ for ( int j = i - 1 ; j >= 0 ; j--)
{
if (arr[j].finish <= arr[i - 1 ].start)
return j;
}
return - 1 ;
} // A recursive function that returns the maximum possible // profit from given array of jobs. The array of jobs must // be sorted according to finish time. static int findMaxProfitRec(Job arr[], int n)
{ // Base case
if (n == 1 ) return arr[n- 1 ].profit;
// Find profit when current job is included
int inclProf = arr[n- 1 ].profit;
int i = latestNonConflict(arr, n);
if (i != - 1 )
inclProf += findMaxProfitRec(arr, i+ 1 );
// Find profit when current job is excluded
int exclProf = findMaxProfitRec(arr, n- 1 );
return Math.max(inclProf, exclProf);
} // The main function that returns the maximum possible // profit from given array of jobs static int findMaxProfit(Job arr[], int n)
{ // Sort jobs according to finish time
Arrays.sort(arr, new Comparator<Job>(){
public int compare(Job j1,Job j2)
{
return j1.finish-j2.finish;
}
});
return findMaxProfitRec(arr, n);
} // Driver program public static void main(String args[])
{ int m = 4 ;
Job arr[] = new Job[m];
arr[ 0 ] = new Job( 3 , 10 , 20 );
arr[ 1 ] = new Job( 1 , 2 , 50 );
arr[ 2 ] = new Job( 6 , 19 , 100 );
arr[ 3 ] = new Job( 2 , 100 , 200 );
int n =arr.length;
System.out.println( "The optimal profit is " + findMaxProfit(arr, n));
} } // This code is contributed by Debojyoti Mandal |
# Python3 program for weighted job scheduling using # Naive Recursive Method # Importing the following module to sort array # based on our custom comparison function from functools import cmp_to_key
# A job has start time, finish time and profit class Job:
def __init__( self , start, finish, profit):
self .start = start
self .finish = finish
self .profit = profit
# A utility function that is used for # sorting events according to finish time def jobComparator(s1, s2):
return s1.finish < s2.finish
# Find the latest job (in sorted array) that # doesn't conflict with the job[i]. If there # is no compatible job, then it returns -1 def latestNonConflict(arr, i):
for j in range (i - 1 , - 1 , - 1 ):
if arr[j].finish < = arr[i - 1 ].start:
return j
return - 1
# A recursive function that returns the # maximum possible profit from given # array of jobs. The array of jobs must # be sorted according to finish time def findMaxProfitRec(arr, n):
# Base case
if n = = 1 :
return arr[n - 1 ].profit
# Find profit when current job is included
inclProf = arr[n - 1 ].profit
i = latestNonConflict(arr, n)
if i ! = - 1 :
inclProf + = findMaxProfitRec(arr, i + 1 )
# Find profit when current job is excluded
exclProf = findMaxProfitRec(arr, n - 1 )
return max (inclProf, exclProf)
# The main function that returns the maximum # possible profit from given array of jobs def findMaxProfit(arr, n):
# Sort jobs according to finish time
arr = sorted (arr, key = cmp_to_key(jobComparator))
return findMaxProfitRec(arr, n)
# Driver code values = [ ( 3 , 10 , 20 ), ( 1 , 2 , 50 ),
( 6 , 19 , 100 ), ( 2 , 100 , 200 ) ]
arr = []
for i in values:
arr.append(Job(i[ 0 ], i[ 1 ], i[ 2 ]))
n = len (arr)
print ( "The optimal profit is" , findMaxProfit(arr, n))
# This code is code contributed by Kevin Joshi |
<script> // JavaScript program for weighted job scheduling using // Naive Recursive Method // A job has start time, finish time and profit class Job { constructor(start, finish, profit)
{
this .start = start
this .finish = finish
this .profit = profit
}
} // A utility function that is used for // sorting events according to finish time function jobComparator(s1, s2){
return s2.finish - s1.finish;
} // Find the latest job (in sorted array) that // doesn't conflict with the job[i]. If there // is no compatible job, then it returns -1 function latestNonConflict(arr, i){
for (let j = i - 1; j >= 0; j--)
{
if (arr[j].finish <= arr[i - 1].start)
return j
}
return -1
} // A recursive function that returns the // maximum possible profit from given // array of jobs. The array of jobs must // be sorted according to finish time function findMaxProfitRec(arr, n){
// Base case
if (n == 1)
return arr[n - 1].profit
// Find profit when current job is included
let inclProf = arr[n - 1].profit
let i = latestNonConflict(arr, n)
if (i != -1)
inclProf += findMaxProfitRec(arr, i + 1)
// Find profit when current job is excluded
let exclProf = findMaxProfitRec(arr, n - 1)
return Math.max(inclProf, exclProf)
} // The main function that returns the maximum // possible profit from given array of jobs function findMaxProfit(arr, n){
// Sort jobs according to finish time
arr.sort(jobComparator)
return findMaxProfitRec(arr, n)
} // Driver code let values = [ [3, 10, 20], [1, 2, 50], [6, 19, 100], [2, 100, 200] ]
let arr = [] for (let i of values)
arr.push( new Job(i[0], i[1], i[2]))
let n = arr.length document.write( "The optimal profit is " , findMaxProfit(arr, n), "</br>" )
// This code is code contributed by shinjanpatra </script> |
Output:
The optimal profit is 250
The above solution may contain many overlapping subproblems. For example, if lastNonConflicting() always returns the previous job, then findMaxProfitRec(arr, n-1) is called twice and the time complexity becomes O(n*2n). As another example when lastNonConflicting() returns previous to the previous job, there are two recursive calls, for n-2 and n-1. In this example case, recursion becomes the same as Fibonacci Numbers.
So this problem has both properties of Dynamic Programming, Optimal Substructure, and Overlapping Subproblems.
Like other Dynamic Programming Problems, we can solve this problem by making a table that stores solutions of subproblems.
Below is an implementation based on Dynamic Programming.
// C++ program for weighted job scheduling using Dynamic // Programming. #include <algorithm> #include <iostream> using namespace std;
// A job has start time, finish time and profit. struct Job {
int start, finish, profit;
}; // A utility function that is used for sorting events // according to finish time bool jobComparator(Job s1, Job s2)
{ return (s1.finish < s2.finish);
} // Find the latest job (in sorted array) that doesn't // conflict with the job[i] int latestNonConflict(Job arr[], int i)
{ for ( int j = i - 1; j >= 0; j--) {
if (arr[j].finish <= arr[i].start)
return j;
}
return -1;
} // The main function that returns the maximum possible // profit from given array of jobs int findMaxProfit(Job arr[], int n)
{ // Sort jobs according to finish time
sort(arr, arr + n, jobComparator);
// Create an array to store solutions of subproblems.
// table[i] stores the profit for jobs till arr[i]
// (including arr[i])
int * table = new int [n];
table[0] = arr[0].profit;
// Fill entries in M[] using recursive property
for ( int i = 1; i < n; i++) {
// Find profit including the current job
int inclProf = arr[i].profit;
int l = latestNonConflict(arr, i);
if (l != -1)
inclProf += table[l];
// Store maximum of including and excluding
table[i] = max(inclProf, table[i - 1]);
}
// Store result and free dynamic memory allocated for
// table[]
int result = table[n - 1];
delete [] table;
return result;
} // Driver program int main()
{ Job arr[] = { { 3, 10, 20 },
{ 1, 2, 50 },
{ 6, 19, 100 },
{ 2, 100, 200 } };
int n = sizeof (arr) / sizeof (arr[0]);
cout << "The optimal profit is "
<< findMaxProfit(arr, n);
return 0;
} |
// JAVA program for weighted job scheduling using Naive // Recursive Method import java.util.*;
class GFG {
// A job has start time, finish time and profit.
static class Job {
int start, finish, profit;
Job( int start, int finish, int profit)
{
this .start = start;
this .finish = finish;
this .profit = profit;
}
}
// Find the latest job (in sorted array) that doesn't
// conflict with the job[i]. If there is no compatible
// job, then it returns -1.
static int latestNonConflict(Job arr[], int i)
{
for ( int j = i - 1 ; j >= 0 ; j--) {
// finish before next is started
if (arr[j].finish <= arr[i - 1 ].start)
return j;
}
return - 1 ;
}
static int findMaxProfitDP(Job arr[], int n)
{
// Create an array to store solutions of
// subproblems. table[i] stores the profit for jobs
// till arr[i] (including arr[i])
int [] table = new int [n];
table[ 0 ] = arr[ 0 ].profit;
// Fill entries in M[] using recursive property
for ( int i = 1 ; i < n; i++) {
// Find profit including the current job
int inclProf = arr[i].profit;
int l = latestNonConflict(arr, i);
if (l != - 1 )
inclProf += table[l];
// Store maximum of including and excluding
table[i] = Math.max(inclProf, table[i - 1 ]);
}
// Store result and free dynamic memory allocated
// for table[]
int result = table[n - 1 ];
return result;
}
// The main function that returns the maximum possible
// profit from given array of jobs
static int findMaxProfit(Job arr[], int n)
{
// Sort jobs according to finish time
Arrays.sort(arr, new Comparator<Job>() {
public int compare(Job j1, Job j2)
{
return j1.finish - j2.finish;
}
});
return findMaxProfitDP(arr, n);
}
// Driver program
public static void main(String args[])
{
int m = 4 ;
Job arr[] = new Job[m];
arr[ 0 ] = new Job( 3 , 10 , 20 );
arr[ 1 ] = new Job( 1 , 2 , 50 );
arr[ 2 ] = new Job( 6 , 19 , 100 );
arr[ 3 ] = new Job( 2 , 100 , 200 );
int n = arr.length;
System.out.println( "The optimal profit is "
+ findMaxProfit(arr, n));
}
} |
# Python3 program for weighted job scheduling # using Dynamic Programming # Importing the following module to sort array # based on our custom comparison function from functools import cmp_to_key
# A job has start time, finish time and profit class Job:
def __init__( self , start, finish, profit):
self .start = start
self .finish = finish
self .profit = profit
# A utility function that is used for sorting # events according to finish time def jobComparator(s1, s2):
return s1.finish < s2.finish
# Find the latest job (in sorted array) that # doesn't conflict with the job[i]. If there # is no compatible job, then it returns -1 def latestNonConflict(arr, i):
for j in range (i - 1 , - 1 , - 1 ):
if arr[j].finish < = arr[i - 1 ].start:
return j
return - 1
# The main function that returns the maximum possible # profit from given array of jobs def findMaxProfit(arr, n):
# Sort jobs according to finish time
arr = sorted (arr, key = cmp_to_key(jobComparator))
# Create an array to store solutions of subproblems.
# table[i] stores the profit for jobs till arr[i]
# (including arr[i])
table = [ None ] * n
table[ 0 ] = arr[ 0 ].profit
# Fill entries in M[] using recursive property
for i in range ( 1 , n):
# Find profit including the current job
inclProf = arr[i].profit
l = latestNonConflict(arr, i)
if l ! = - 1 :
inclProf + = table[l]
# Store maximum of including and excluding
table[i] = max (inclProf, table[i - 1 ])
# Store result and free dynamic memory
# allocated for table[]
result = table[n - 1 ]
return result
# Driver code values = [( 3 , 10 , 20 ), ( 1 , 2 , 50 ),
( 6 , 19 , 100 ), ( 2 , 100 , 200 )]
arr = []
for i in values:
arr.append(Job(i[ 0 ], i[ 1 ], i[ 2 ]))
n = len (arr)
print ( "The optimal profit is" , findMaxProfit(arr, n))
# This code is contributed by Kevin Joshi |
<script> // JavaScript program for weighted job scheduling using // Naive Recursive Method // A job has start time, finish time and profit class Job{ constructor(start, finish, profit){
this .start = start
this .finish = finish
this .profit = profit
}
} // A utility function that is used for // sorting events according to finish time function jobComparator(s1, s2){
return s1.finish - s2.finish
} // Find the latest job (in sorted array) that // doesn't conflict with the job[i]. If there // is no compatible job, then it returns -1 function latestNonConflict(arr, i){
for (let j=i-1;j>=0;j--){
if (arr[j].finish <= arr[i - 1].start)
return j
}
return -1
} // The main function that returns the maximum // possible profit from given array of jobs function findMaxProfit(arr, n){
// Sort jobs according to finish time
arr.sort(jobComparator)
// Create an array to store solutions of subproblems.
// table[i] stores the profit for jobs till arr[i]
// (including arr[i])
let table = new Array(n).fill( null )
table[0] = arr[0].profit
// Fill entries in M[] using recursive property
for (let i=1;i<n;i++){
// Find profit including the current job
let inclProf = arr[i].profit
let l = latestNonConflict(arr, i)
if (l != -1)
inclProf += table[l]
// Store maximum of including and excluding
table[i] = Math.max(inclProf, table[i - 1])
}
// Store result and free dynamic memory
// allocated for table[]
let result = table[n - 1]
return result
} // Driver code let values = [ [3, 10, 20], [1, 2, 50], [6, 19, 100], [2, 100, 200] ] let arr = [] for (let i of values)
arr.push( new Job(i[0], i[1], i[2]))
let n = arr.length document.write( "The optimal profit is " , findMaxProfit(arr, n), "</br>" )
// This code is code contributed by shinjanpatra </script> |
Output:
The optimal profit is 250
Time Complexity of the above Dynamic Programming Solution is O(n2). Note that the above solution can be optimized to O(nLogn) using Binary Search in latestNonConflict() instead of linear search. Thanks to Garvit for suggesting this optimization. Please refer below post for details.
Weighted Job Scheduling in O(n Log n) time
References:
http://courses.cs.washington.edu/courses/cse521/13wi/slides/06dp-sched.pdf