Make the intervals non-overlapping by assigning them to two different processors
Given a list of intervals interval[] where each interval contains two integers L and R, the task is to assign intervals to two different processors such that there are no overlapping intervals for each processor. To assign the interval[i] to the first processor, print “F” and to assign it to the second processor, print “S”.
Note: If there is no possible solution print -1.
Examples:
Input: interval[] = {{360, 480}, {420, 540}, {600, 660}}
Output: S, F, S
Explanation:
The intervals assigned to processors are –
Intervals of First Processor {{420, 540}}
Intervals of Second Processor {{360, 480}, {600, 660}}
As there are no overlapping intervals for each processor, it will be a valid solution.Input: interval[] = {{99, 150}, {1, 100}, {100, 301}, {2, 5}, {150, 250}}
Output: S, F, F, S, S
Explanation:
The intervals assigned to processors are –
Intervals of First Processor {{1, 100}, {100, 301}}
Intervals of Second Processor {{99, 150}, {2, 5}, {150, 250}}
As there are no overlapping intervals for each processor, it will be a valid solution.
Approach: The idea is to use Greedy algorithm to assign the intervals to the processor.
If the highest end time of the processor is less than or equal to start time of an interval, then this interval can be assigned to the processor. Otherwise, check for the another processor. If any interval cannot be assigned to any processor then there is no possible solution.
Below is the illustration of the steps of the approach:
- As in the current problem we have to print according to the order of the intervals. So to save the order of intervals, pair the intervals with their index.
- Sort the intervals by their start time. i.e. L.
- Iterate over the intervals and assign the intervals to the processors as follows:
if (interval[i][0] >= firstProcessorEndTime) answer[interval[i]] = "F" firstProcessorEndTime = max(firstProcessorEndTime, interval[i][0]) else if (interval[i][0] >= secondProcessorEndTime) answer[interval[i]] = "S" secondProcessorEndTime = max(secondProcessorEndTime, interval[i][0]) else print(-1)
Below is the implementation of the above approach:
C++
// C++ implementation for intervals // scheduling to two processors such // that there are no overlapping intervals #include <bits/stdc++.h> using namespace std; // Function to assign the intervals // to two different processors void assignIntervals(vector<vector< int > > interval, int n) { // Loop to pair the interval // with their indices for ( int i = 0; i < n; i++) interval[i].push_back(i); // sorting the interval by // their start times sort(interval.begin(), interval.end()); int firstEndTime = -1; int secondEndTime = -1; char find = ' ' ; bool flag = false ; // Loop to iterate over the // intervals with their start time for ( int i = 0; i < n; i++) { if (interval[i][0] >= firstEndTime) { firstEndTime = interval[i][1]; interval[i].push_back( 'S' ); } else if (interval[i][0] >= secondEndTime) { secondEndTime = interval[i][1]; interval[i].push_back( 'F' ); } else { flag = true ; break ; } } // Condition to check if there // is a possible solution if (flag) cout << (-1); else { vector< char > form(n, ' ' ); for ( int i = 0; i < n; i++) { int indi = interval[i][2]; form[indi] = interval[i][3]; } // form = ''.join(form) for ( int i = 0; i < form.size(); i++) cout << form[i] << "," ; } } // Driver Code int main() { vector<vector< int > > intervals = { { 360, 480 }, { 420, 540 }, { 600, 660 } }; // Function Call assignIntervals(intervals, intervals.size()); return 0; } |
Java
// Java program for the above approach import java.util.*; public class Main { public static void main(String[] args) { // Intervals to be scheduled int [][] intervals = { { 360 , 480 }, { 420 , 540 }, { 600 , 660 } }; // Function Call assignIntervals(intervals, intervals.length); } // Function to assign the intervals // to two different processors public static void assignIntervals( int [][] interval, int n) { // Loop to pair the interval // with their indices for ( int i = 0 ; i < n; i++) { interval[i] = Arrays.copyOf(interval[i], interval[i].length + 1 ); interval[i][interval[i].length - 1 ] = i; } // Sorting the interval by their start times Arrays.sort(interval, Comparator.comparingInt(a -> a[ 0 ])); int firstEndTime = - 1 ; int secondEndTime = - 1 ; boolean flag = false ; // Loop to iterate over the // intervals with their start time for ( int i = 0 ; i < n; i++) { if (interval[i][ 0 ] >= firstEndTime) { firstEndTime = interval[i][ 1 ]; interval[i] = Arrays.copyOf(interval[i], interval[i].length + 1 ); interval[i][interval[i].length - 1 ] = 'S' ; } else if (interval[i][ 0 ] >= secondEndTime) { secondEndTime = interval[i][ 1 ]; interval[i] = Arrays.copyOf(interval[i], interval[i].length + 1 ); interval[i][interval[i].length - 1 ] = 'F' ; } else { flag = true ; break ; } } // Condition to check if there // is a possible solution if (flag) { System.out.println( "-1" ); } else { List<Character> form = new ArrayList<>(Collections.nCopies(n, ' ' )); for ( int i = 0 ; i < n; i++) { int indi = interval[i][ 2 ]; form.set(indi, ( char ) interval[i][ 3 ]); } // Print the result String output = form.toString(); System.out.println(output); } } } // This code is contributed by Prince Kumar |
Python3
# Python implementation for intervals # scheduling to two processors such # that there are no overlapping intervals # Function to assign the intervals # to two different processors def assignIntervals(interval, n): # Loop to pair the interval # with their indices for i in range (n): interval[i].append(i) # sorting the interval by # their startb times interval.sort(key = lambda x: x[ 0 ]) firstEndTime = - 1 secondEndTime = - 1 find = '' flag = False # Loop to iterate over the # intervals with their start time for i in range (n): if interval[i][ 0 ] > = firstEndTime: firstEndTime = interval[i][ 1 ] interval[i].append( 'S' ) elif interval[i][ 0 ] > = secondEndTime: secondEndTime = interval[i][ 1 ] interval[i].append( 'F' ) else : flag = True break # Condition to check if there # is a possible solution if flag: print ( - 1 ) else : form = [''] * n for i in range (n): indi = interval[i][ 2 ] form[indi] = interval[i][ 3 ] # form = ''.join(form) print (form, ", " ) # Driver Code if __name__ = = "__main__" : intervals = [[ 360 , 480 ], [ 420 , 540 ], [ 600 , 660 ]] # Function Call assignIntervals(intervals, len (intervals)) |
C#
using System; using System.Collections.Generic; using System.Linq; class MainClass { public static void Main( string [] args) { // intervals to be scheduled int [][] intervals = { new int [] {360, 480}, new int [] {420, 540}, new int [] {600, 660} }; // Function Call AssignIntervals(intervals, intervals.Length); } // Function to assign the intervals // to two different processors public static void AssignIntervals( int [][] interval, int n) { // Loop to pair the interval // with their indices for ( int i = 0; i < n; i++) { interval[i] = interval[i].Concat( new int [] {i}).ToArray(); } // sorting the interval by // their start times interval = interval.OrderBy(x => x[0]).ToArray(); int firstEndTime = -1; int secondEndTime = -1; bool flag = false ; // Loop to iterate over the // intervals with their start time for ( int i = 0; i < n; i++) { if (interval[i][0] >= firstEndTime) { firstEndTime = interval[i][1]; interval[i] = interval[i].Concat( new int [] { 'S' }).ToArray(); } else if (interval[i][0] >= secondEndTime) { secondEndTime = interval[i][1]; interval[i] = interval[i].Concat( new int [] { 'F' }).ToArray(); } else { flag = true ; break ; } } // Condition to check if there // is a possible solution if (flag) { Console.WriteLine( "-1" ); } else { List< char > form = new List< char >( new char [n]); for ( int i = 0; i < n; i++) { int indi = interval[i][2]; form[indi] = ( char )interval[i][3]; } string output = string .Join( "," , form); Console.WriteLine( "[" +output+ "]" ); } } } |
Javascript
// JavaScript implementation for intervals // scheduling to two processors such // that there are no overlapping intervals // Function to assign the intervals // to two different processors function assignIntervals(interval, n) { // Loop to pair the interval // with their indices for (let i = 0; i < n; i++) { interval[i].push(i); } // sorting the interval by // their startb times interval.sort((a, b) => a[0] - b[0]); let firstEndTime = -1; let secondEndTime = -1; let find = '' ; let flag = false ; // Loop to iterate over the // intervals with their start time for (let i = 0; i < n; i++) { if (interval[i][0] >= firstEndTime) { firstEndTime = interval[i][1]; interval[i].push( 'S' ); } else if (interval[i][0] >= secondEndTime) { secondEndTime = interval[i][1]; interval[i].push( 'F' ); } else { flag = true ; break ; } } // Condition to check if there // is a possible solution if (flag) { console.log(-1); } else { let form = Array(n).fill( '' ); for (let i = 0; i < n; i++) { let indi = interval[i][2]; form[indi] = interval[i][3]; } // form = form.join(''); console.log(form, ", " ); } } // Driver Code let intervals = [[360, 480], [420, 540], [600, 660]]; // Function Call assignIntervals(intervals, intervals.length); // This code is contributed by phasing17. |
S,F,S,
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
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