Find Non-overlapping intervals among a given set of intervals
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:
- Sort the given set of intervals according to starting time.
- Traverse all the set of intervals and check whether the consecutive intervals overlaps or not.
- 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.
- 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 |
C#
using System; using System.Linq; using System.Collections.Generic; class GFG { // Interval with start time & end time class Interval { public int start, end; public 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 List<Interval> p = new List<Interval>(); // Sort the given interval according // starting time Array.Sort(arr, new Comparison< int []>( (a, b) => 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.Count; i++) { Console.WriteLine( "[" + p[i].start + ", " + p[i].end + "]" ); } } // Driver Code static void Main( string [] args) { // Given set of interval int [][] arr = { new int [] { 1, 3 }, new int [] { 2, 4 }, new int [] { 3, 5 }, new int [] { 7, 9 } }; int N = arr.Length; // Function Call FindFreeInterval(arr, N); } } // This code is contributed by phasing17 |
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> |
[5, 7]
Time Complexity: O(N*log N), where N is the number of set of intervals.
Auxiliary Space: O(N)