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Find Non-overlapping intervals among a given set of intervals

  • Difficulty Level : Hard
  • Last Updated : 19 May, 2021
Geek Week

Given N set of time intervals, the task is to find the intervals which don’t overlap with the given set of intervals.
Examples: 
 

Input: interval arr[] = { {1, 3}, {2, 4}, {3, 5}, {7, 9} } 
Output: 
[5, 7] 
Explanation: 
The only interval which doesn’t overlaps with the other intervals is [5, 7].
Input: interval arr[] = { {1, 3}, {9, 12}, {2, 4}, {6, 8} } 
Output: 
[4, 6] 
[8, 9] 
Explanation: 
There are two intervals which don’t overlap with other intervals are [4, 6], [8, 9].
 

 

Approach: The idea is to sort the given time intervals according to starting time and if the consecutive intervals don’t overlap then the difference between them is the free interval. 
Below are the steps: 
 

  1. Sort the given set of intervals according to starting time.
  2. Traverse all the set of intervals and check whether the consecutive intervals overlaps or not.
  3. If the intervals(say interval a & interval b) doesn’t overlap then the set of pairs form by [a.end, b.start] is the non-overlapping interval.
  4. If the intervals overlaps, then check for next consecutive intervals.

Below is the implementation of the above approach: 
 



C++




// C++ program for the above approach
 
#include "bits/stdc++.h"
using namespace std;
 
// interval with start time & end time
struct interval {
    int start, end;
};
 
// Comparator function to sort the given
// interval according to time
bool compareinterval(interval i1, interval i2)
{
    return (i1.start < i2.start);
}
 
// Function that find the free interval
void findFreeinterval(interval arr[], int N)
{
 
    // If there are no set of interval
    if (N <= 0) {
        return;
    }
 
    // To store the set of free interval
    vector<pair<int, int> > P;
 
    // Sort the given interval according
    // starting time
    sort(arr, arr + N, compareinterval);
 
    // Iterate over all the interval
    for (int i = 1; i < N; i++) {
 
        // Previous interval end
        int prevEnd = arr[i - 1].end;
 
        // Current interval start
        int currStart = arr[i].start;
 
        // If ending index of previous
        // is less than starting index
        // of current, then it is free
        // interval
        if (prevEnd < currStart) {
            P.push_back({ prevEnd,
                          currStart });
        }
    }
 
    // Print the free interval
    for (auto& it : P) {
        cout << "[" << it.first << ", "
             << it.second << "]" << endl;
    }
}
 
// Driver Code
int main()
{
 
    // Given set of interval
    interval arr[] = { { 1, 3 },
                       { 2, 4 },
                       { 3, 5 },
                       { 7, 9 } };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    findFreeinterval(arr, N);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG {
     
// Interval with start time & end time
static class Interval
{
    int start, end;
 
    Interval(int start, int end)
    {
        this.start = start;
        this.end = end;
    }
}
 
// Function that find the free interval
static void findFreeinterval(int[][] arr, int N)
{
    // If there are no set of interval
    if (N <= 0)
    {
        return;
    }
 
    // To store the set of free interval
    ArrayList<Interval> p = new ArrayList<>();
 
    // Sort the given interval according
    // starting time
    Arrays.sort(arr, new Comparator<int[]>()
    {
        public int compare(int[] a, int[] b)
        {
            return a[0] - b[0];
        }
    });
 
    // Iterate over all the interval
    for (int i = 1; i < N; i++)
    {
 
        // Previous interval end
        int prevEnd = arr[i - 1][1];
 
        // Current interval start
        int currStart = arr[i][0];
 
        // If ending index of previous
        // is less than starting index
        // of current, then it is free
        // interval
        if (prevEnd < currStart)
        {
            Interval interval = new Interval(prevEnd,
                                              currStart);
            p.add(interval);
        }
    }
 
    // Print the free interval
    for (int i = 0; i < p.size(); i++)
    {
        System.out.println("[" + p.get(i).start +
                          ", " + p.get(i).end + "]");
    }
}
 
// Driver code
public static void main(String[] args)
{
 
    // Given set of interval
    int[][] arr = { { 1, 3 },
                    { 2, 4 },
                    { 3, 5 },
                    { 7, 9 } };
 
    int N = arr.length;
 
    // Function Call
    findFreeinterval(arr, N);
}
}
 
// This code is contributed by offbeat

Python3




# Python3 program for the above approach
def findFreeinterval(arr, N):
     
    # If there are no set of interval
    if N < 1:
        return
     
    # To store the set of free interval
    P = []
     
    # Sort the given interval according
    # Starting time
    arr.sort(key = lambda a:a[0])
     
    # Iterate over all the interval
    for i in range(1, N):
         
        # Previous interval end
        prevEnd = arr[i - 1][1]
         
        # Current interval start
        currStart = arr[i][0]
         
        # If Previous Interval is less
        # than current Interval then we
        # store that answer
        if prevEnd < currStart:
            P.append([prevEnd, currStart])
     
    # Print the intervals
    for i in P:
        print(i)
     
# Driver code
if __name__ == "__main__":
     
    # Given List of intervals
    arr = [ [ 1, 3 ], [ 2, 4 ],
            [ 3, 5 ], [ 7, 9 ] ]
     
    N = len(arr)
     
    # Function call
    findFreeinterval(arr, N)
     
# This code is contributed by Tokir Manva

Javascript




<script>
 
// Javascript program for the above approach
 
// Function that find the free interval
function findFreeinterval(arr, N)
{
 
    // If there are no set of interval
    if (N <= 0) {
        return;
    }
 
    // To store the set of free interval
    var P = [];
 
    // Sort the given interval according
    // starting time
    arr.sort((a,b) => a[0]-b[0])
 
    // Iterate over all the interval
    for (var i = 1; i < N; i++) {
 
        // Previous interval end
        var prevEnd = arr[i - 1][1];
 
        // Current interval start
        var currStart = arr[i][0];
 
        // If ending index of previous
        // is less than starting index
        // of current, then it is free
        // interval
        if (prevEnd < currStart) {
            P.push([prevEnd,
                          currStart]);
        }
    }
 
    // Print the free interval
    P.forEach(it => {
 
        document.write( "[" + it[0] + ", "
             + it[1] + "]");
    });
}
 
// Driver Code
 
// Given set of interval
var arr = [ [ 1, 3 ],
                   [ 2, 4 ],
                   [ 3, 5 ],
                   [ 7, 9 ] ];
var N = arr.length;
 
// Function Call
findFreeinterval(arr, N);
 
// This code is contributed by noob2000.
</script>
Output: 
[5, 7]

 

Time Complexity: O(N*log N), where N is the number of set of intervals.
 




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