Make the intervals non-overlapping by assigning them to two different processors

• Last Updated : 16 Aug, 2021

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 their 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] >= firstProcessorEndTime)
firstProcessorEndTime =
max(firstProcessorEndTime, interval[i])
else if (interval[i] >= secondProcessorEndTime)
secondProcessorEndTime =
max(secondProcessorEndTime, interval[i])
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 using namespace std; // Function to assign the intervals// to two different processorsvoid assignIntervals(vector > 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 fin = ' ';    bool flag = false;     // Loop to iterate over the    // intervals with their start time    for (int i = 0; i < n; i++) {        if (interval[i] >= firstEndTime) {            firstEndTime = interval[i];            interval[i].push_back('S');        }        else if (interval[i] >= secondEndTime) {            secondEndTime = interval[i];            interval[i].push_back('F');        }        else {            flag = true;            break;        }    }     // Condition to check if there    // is a possible solution    if (flag)        cout << (-1);    else {        vector form(n, ' ');         for (int i = 0; i < n; i++) {            int indi = interval[i];            form[indi] = interval[i];        }         // form = ''.join(form)        for (int i = 0; i < form.size(); i++)            cout << form[i] << ",";    }} // Driver Codeint main(){     vector > intervals        = { { 360, 480 }, { 420, 540 }, { 600, 660 } };     // Function Call    assignIntervals(intervals, intervals.size());    return 0;}

Python3

 # Python implementation for intervals# scheduling to two processors such# that there are no overlapping intervals # Function to assign the intervals# to two different processorsdef 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)         firstEndTime = -1    secondEndTime = -1    fin = ''    flag = False         # Loop to iterate over the    # intervals with their start time    for i in range(n):        if interval[i] >= firstEndTime:            firstEndTime = interval[i]            interval[i].append('S')        elif interval[i] >= secondEndTime:            secondEndTime = interval[i]            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]            form[indi] = interval[i]        # form = ''.join(form)        print(form, ", ") # Driver Code   if __name__ == "__main__":    intervals = [[360, 480], [420, 540], [600, 660]]         # Function Call    assignIntervals(intervals, len(intervals))
Output:
['S', 'F', 'S'] ,

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

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