Merge Overlapping Intervals
Given a set of time intervals in any order, merge all overlapping intervals into one and output the result which should have only mutually exclusive intervals.
Example:
Input: Intervals = {{1,3},{2,4},{6,8},{9,10}}
Output: {{1, 4}, {6, 8}, {9, 10}}
Explanation: Given intervals: [1,3],[2,4],[6,8],[9,10], we have only two overlapping intervals here,[1,3] and [2,4]. Therefore we will merge these two and return [1,4],[6,8], [9,10].Input: Intervals = {{6,8},{1,9},{2,4},{4,7}}
Output: {{1, 9}}
Merge Overlapping Intervals Using Nested Loop
A simple approach is to start from the first interval and compare it with all other intervals for overlapping, if it overlaps with any other interval, then remove the other interval from the list and merge the other into the first interval. Repeat the same steps for the remaining intervals after the first. This approach cannot be implemented in better than O(n^2) time.
Merge Overlapping Intervals Optimized Approach
The idea to solve this problem is, first sort the intervals according to the starting time. Once we have the sorted intervals, we can combine all intervals in a linear traversal. The idea is, in sorted array of intervals, if interval[i] doesn’t overlap with interval[i-1], then interval[i+1] cannot overlap with interval[i-1] because starting time of interval[i+1] must be greater than or equal to interval[i].
Follow the steps mentioned below to implement the approach:
- Sort the intervals based on the increasing order of starting time.
- Push the first interval into a stack.
- For each interval do the following:
- If the current interval does not overlap with the top of the stack then, push the current interval into the stack.
- If the current interval overlap with the top of the stack then, update the stack top with the ending time of the current interval.
- The end stack contains the merged intervals.
Below is the implementation of the above approach:
C++
// A C++ program for merging overlapping intervals #include <bits/stdc++.h> using namespace std; // An interval has start time and end time struct Interval { int start, end; }; // Compares two intervals according to their starting time. // This is needed for sorting the intervals using library // function std::sort(). See http://goo.gl/iGspV bool compareInterval(Interval i1, Interval i2) { return (i1.start < i2.start); } // The main function that takes a set of intervals, merges // overlapping intervals and prints the result void mergeIntervals(Interval arr[], int n) { // Test if the given set has at least one interval if (n <= 0) return ; // Create an empty stack of intervals stack<Interval> s; // sort the intervals in increasing order of start time sort(arr, arr + n, compareInterval); // push the first interval to stack s.push(arr[0]); // Start from the next interval and merge if necessary for ( int i = 1; i < n; i++) { // get interval from stack top Interval top = s.top(); // if current interval is not overlapping with stack // top, push it to the stack if (top.end < arr[i].start) s.push(arr[i]); // Otherwise update the ending time of top if ending // of current interval is more else if (top.end < arr[i].end) { top.end = arr[i].end; s.pop(); s.push(top); } } // Print contents of stack cout << "\n The Merged Intervals are: " ; while (!s.empty()) { Interval t = s.top(); cout << "[" << t.start << "," << t.end << "] " ; s.pop(); } return ; } // Driver program int main() { Interval arr[] = { { 6, 8 }, { 1, 9 }, { 2, 4 }, { 4, 7 } }; int n = sizeof (arr) / sizeof (arr[0]); mergeIntervals(arr, n); return 0; } |
Java
// A Java program for merging overlapping intervals import java.util.Arrays; import java.util.Comparator; import java.util.Stack; public class MergeOverlappingIntervals { // The main function that takes a set of intervals, merges // overlapping intervals and prints the result public static void mergeIntervals(Interval arr[]) { // Test if the given set has at least one interval if (arr.length <= 0 ) return ; // Create an empty stack of intervals Stack<Interval> stack= new Stack<>(); // sort the intervals in increasing order of start time Arrays.sort(arr, new Comparator<Interval>(){ public int compare(Interval i1,Interval i2) { return i1.start-i2.start; } }); // push the first interval to stack stack.push(arr[ 0 ]); // Start from the next interval and merge if necessary for ( int i = 1 ; i < arr.length; i++) { // get interval from stack top Interval top = stack.peek(); // if current interval is not overlapping with stack top, // push it to the stack if (top.end < arr[i].start) stack.push(arr[i]); // Otherwise update the ending time of top if ending of current // interval is more else if (top.end < arr[i].end) { top.end = arr[i].end; stack.pop(); stack.push(top); } } // Print contents of stack System.out.print( "The Merged Intervals are: " ); while (!stack.isEmpty()) { Interval t = stack.pop(); System.out.print( "[" +t.start+ "," +t.end+ "] " ); } } public static void main(String args[]) { Interval arr[]= new Interval[ 4 ]; arr[ 0 ]= new Interval( 6 , 8 ); arr[ 1 ]= new Interval( 1 , 9 ); arr[ 2 ]= new Interval( 2 , 4 ); arr[ 3 ]= new Interval( 4 , 7 ); mergeIntervals(arr); } } class Interval { int start,end; Interval( int start, int end) { this .start=start; this .end=end; } } // This code is contributed by Gaurav Tiwari |
C#
// A C# program for merging overlapping intervals using System; using System.Collections; using System.Collections.Generic; public class MergeOverlappingIntervals { // sort the intervals in increasing order of start time class sortHelper : IComparer { int IComparer.Compare( object a, object b) { Interval first = (Interval)a; Interval second = (Interval)b; if (first.start == second.start) { return first.end - second.end; } return first.start - second.start; } } // The main function that takes a set of intervals, merges // overlapping intervals and prints the result public static void mergeIntervals(Interval []arr) { // Test if the given set has at least one interval if (arr.Length <= 0) return ; Array.Sort(arr, new sortHelper()); // Create an empty stack of intervals Stack stack = new Stack(); // Push the first interval to stack stack.Push(arr[0]); // Start from the next interval and merge if necessary for ( int i = 1 ; i < arr.Length; i++) { // get interval from stack top Interval top = (Interval)stack.Peek(); // if current interval is not overlapping with stack top, // Push it to the stack if (top.end < arr[i].start) stack.Push(arr[i]); // Otherwise update the ending time of top if ending of current // interval is more else if (top.end < arr[i].end) { top.end = arr[i].end; stack.Pop(); stack.Push(top); } } // Print contents of stack Console.Write( "The Merged Intervals are: " ); while (stack.Count != 0) { Interval t = (Interval)stack.Pop(); Console.Write( "[" + t.start + "," + t.end + "] " ); } } // Driver code public static void Main() { Interval []arr = new Interval[4]; arr[0] = new Interval(6, 8); arr[1] = new Interval(1, 9); arr[2] = new Interval(2, 4); arr[3] = new Interval(4, 7); mergeIntervals(arr); } } public class Interval { public int start,end; public Interval( int start, int end) { this .start = start; this .end = end; } } // This code is contributed by rutvik_56. |
Python3
def mergeIntervals(intervals): # Sort the array on the basis of start values of intervals. intervals.sort() stack = [] # insert first interval into stack stack.append(intervals[ 0 ]) for i in intervals[ 1 :]: # Check for overlapping interval, # if interval overlap if stack[ - 1 ][ 0 ] < = i[ 0 ] < = stack[ - 1 ][ - 1 ]: stack[ - 1 ][ - 1 ] = max (stack[ - 1 ][ - 1 ], i[ - 1 ]) else : stack.append(i) print ( "The Merged Intervals are :" , end = " " ) for i in range ( len (stack)): print (stack[i], end = " " ) arr = [[ 6 , 8 ], [ 1 , 9 ], [ 2 , 4 ], [ 4 , 7 ]] mergeIntervals(arr) |
Javascript
// A JavaScript program for merging overlapping intervals // An interval has start time and end time class Interval { constructor(start, end) { this .start = start; this .end = end; } } // Compares two intervals according to their starting time. function compareInterval(i1, i2) { return (i1.start < i2.start); } // The main function that takes a set of intervals, merges // overlapping intervals and prints the result function mergeIntervals(arr) { // Test if the given set has at least one interval if (arr.length <= 0) return ; // Create an empty stack of intervals let s = []; // sort the intervals in increasing order of start time arr.sort(compareInterval); // push the first interval to stack s.push(arr[0]); // Start from the next interval and merge if necessary for (let i = 1; i < arr.length; i++) { // get interval from stack top let top = s[s.length-1]; // if current interval is not overlapping with stack // top, push it to the stack if (top.end < arr[i].start) s.push(arr[i]); // Otherwise update the ending time of top if ending // of current interval is more else if (top.end < arr[i].end) { top.end = arr[i].end; s.pop(); s.push(top); } } // Print contents of stack console.log( "The Merged Intervals are: " ); while (s.length > 0) { let t = s.pop(); console.log( "[" + t.start + "," + t.end + "] " ); } return ; } // Driver program let arr = [ new Interval(6,8), new Interval(1,9), new Interval(2,4), new Interval(4,7)]; mergeIntervals(arr); |
The Merged Intervals are: [1,9]
Time complexity: O(N*log(N))
Auxiliary Space: O(N)
Merge Overlapping Intervals Space Optimized Approach
The above solution requires O(n) extra space for the stack. We can avoid the use of extra space by doing merge operations in place. Below are detailed steps.
Follow the steps mentioned below to implement the approach:
- Sort all intervals in increasing order of start time.
- Traverse sorted intervals starting from the first interval,
- Do the following for every interval.
- If the current interval is not the first interval and it overlaps with the previous interval,
then merge it with the previous interval. Keep doing it while the interval overlaps with the previous one. - Otherwise, Add the current interval to the output list of intervals.
- If the current interval is not the first interval and it overlaps with the previous interval,
Below is the implementation of the above approach:
C++
// C++ program to merge overlapping Intervals in // O(n Log n) time and O(1) extra space. #include <bits/stdc++.h> using namespace std; // An Interval struct Interval { int s, e; }; // Function used in sort bool mycomp(Interval a, Interval b) { return a.s < b.s; } void mergeIntervals(Interval arr[], int n) { // Sort Intervals in increasing order of // start time sort(arr, arr + n, mycomp); int index = 0; // Stores index of last element // in output array (modified arr[]) // Traverse all input Intervals for ( int i = 1; i < n; i++) { // If this is not first Interval and overlaps // with the previous one if (arr[index].e >= arr[i].s) { // Merge previous and current Intervals arr[index].e = max(arr[index].e, arr[i].e); } else { index++; arr[index] = arr[i]; } } // Now arr[0..index-1] stores the merged Intervals cout << "\n The Merged Intervals are: " ; for ( int i = 0; i <= index; i++) cout << "[" << arr[i].s << ", " << arr[i].e << "] " ; } // Driver program int main() { Interval arr[] = { { 6, 8 }, { 1, 9 }, { 2, 4 }, { 4, 7 } }; int n = sizeof (arr) / sizeof (arr[0]); mergeIntervals(arr, n); return 0; } // This code is contributed by Aditya Kumar (adityakumar129) |
C
// C program to merge overlapping Intervals in // O(n Log n) time and O(1) extra space. #include <stdbool.h> #include <stdio.h> #include <stdlib.h> // An Interval typedef struct Interval { int s, e; } Interval; // Function used in sort int mycomp( const void * a, const void * b) { Interval* data_1 = (Interval*)a; Interval* data_2 = (Interval*)b; return (data_1->s - data_2->s); } // Find maximum between two numbers. int max( int num1, int num2) { return (num1 > num2) ? num1 : num2; } void mergeIntervals(Interval arr[], int n) { // Sort Intervals in increasing order of // start time qsort (arr, n, sizeof (Interval), mycomp); int index = 0; // Stores index of last element // in output array (modified arr[]) // Traverse all input Intervals for ( int i = 1; i < n; i++) { // If this is not first Interval and overlaps // with the previous one if (arr[index].e >= arr[i].s) { // Merge previous and current Intervals arr[index].e = max(arr[index].e, arr[i].e); } else { index++; arr[index] = arr[i]; } } // Now arr[0..index-1] stores the merged Intervals printf ( "\n The Merged Intervals are: " ); for ( int i = 0; i <= index; i++) printf ( "[%d, %d]" , arr[i].s, arr[i].e); } // Driver program int main() { Interval arr[] = { { 6, 8 }, { 1, 9 }, { 2, 4 }, { 4, 7 } }; int n = sizeof (arr) / sizeof (arr[0]); mergeIntervals(arr, n); return 0; } // This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java program to merge overlapping Intervals in // O(n Log n) time and O(1) extra space import java.util.Arrays; import java.util.Comparator; // An Interval class Interval { int start,end; Interval( int start, int end) { this .start=start; this .end=end; } } public class MergeOverlappingIntervals { // Function that takes a set of intervals, merges // overlapping intervals and prints the result public static void mergeIntervals(Interval arr[]) { // Sort Intervals in increasing order of // start time Arrays.sort(arr, new Comparator<Interval>(){ public int compare(Interval i1,Interval i2) { return i1.start - i2.start; } }); int index = 0 ; // Stores index of last element // in output array (modified arr[]) // Traverse all input Intervals for ( int i= 1 ; i<arr.length; i++) { // If this is not first Interval and overlaps // with the previous one if (arr[index].end >= arr[i].start) { // Merge previous and current Intervals arr[index].end = Math.max(arr[index].end, arr[i].end); } else { index++; arr[index] = arr[i]; } } // Now arr[0..index-1] stores the merged Intervals System.out.print( "The Merged Intervals are: " ); for ( int i = 0 ; i <= index; i++) { System.out.print( "[" + arr[i].start + "," + arr[i].end + "]" ); } } // Driver Code public static void main(String args[]) { Interval arr[]= new Interval[ 4 ]; arr[ 0 ]= new Interval( 6 , 8 ); arr[ 1 ]= new Interval( 1 , 9 ); arr[ 2 ]= new Interval( 2 , 4 ); arr[ 3 ]= new Interval( 4 , 7 ); mergeIntervals(arr); } } // This code is contributed by Gaurav Tiwari // This code was fixed by Subham Mukhopadhyay |
Python3
# Python program to merge overlapping Intervals in # O(n Log n) time and O(1) extra space def mergeIntervals(arr): # Sorting based on the increasing order # of the start intervals arr.sort(key = lambda x: x[ 0 ]) # Stores index of last element # in output array (modified arr[]) index = 0 # Traverse all input Intervals starting from # second interval for i in range ( 1 , len (arr)): # If this is not first Interval and overlaps # with the previous one, Merge previous and # current Intervals if (arr[index][ 1 ] > = arr[i][ 0 ]): arr[index][ 1 ] = max (arr[index][ 1 ], arr[i][ 1 ]) else : index = index + 1 arr[index] = arr[i] print ( "The Merged Intervals are :" , end = " " ) for i in range (index + 1 ): print (arr[i], end = " " ) # Driver code arr = [[ 6 , 8 ], [ 1 , 9 ], [ 2 , 4 ], [ 4 , 7 ]] mergeIntervals(arr) |
C#
using System; using System.Linq; class Interval { public int start,end; public Interval( int start, int end) { this .start=start; this .end=end; } } class MergeOverlappingIntervals { // Function that takes a set of intervals, merges // overlapping intervals and prints the result public static void MergeIntervals(Interval[] arr) { // Sort Intervals in increasing order of // start time arr = arr.OrderBy(x => x.start).ToArray(); int index = 0; // Stores index of last element // in output array (modified arr[]) // Traverse all input Intervals for ( int i=1; i<arr.Length; i++) { // If this is not first Interval and overlaps // with the previous one if (arr[index].end >= arr[i].start) { // Merge previous and current Intervals arr[index].end = Math.Max(arr[index].end, arr[i].end); } else { index++; arr[index] = arr[i]; } } // Now arr[0..index-1] stores the merged Intervals Console.WriteLine( "The Merged Intervals are: " ); for ( int i = 0; i <= index; i++) { Console.WriteLine( "[" + arr[i].start + "," + arr[i].end + "]" ); } } // Driver Code public static void Main( string [] args) { Interval[] arr = new Interval[4]; arr[0]= new Interval(6,8); arr[1]= new Interval(1,9); arr[2]= new Interval(2,4); arr[3]= new Interval(4,7); MergeIntervals(arr); } } |
Javascript
// A JavaScript program for merging overlapping intervals // An interval has start time and end time class Interval { constructor(s, e) { this .s = s; this .e = e; } } // Function used in sort function mycomp(a, b) { return a.s < b.s; } // The main function that takes a set of intervals, merges // overlapping intervals and prints the result function mergeIntervals(arr) { // Sort Intervals in increasing order of // start time arr.sort(mycomp); let index = 0; // Stores index of last element // in output array (modified arr[]) // Traverse all input Intervals for (let i = 1; i < arr.length; i++) { // If this is not first Interval and overlaps // with the previous one if (arr[index].e >= arr[i].s) { // Merge previous and current Intervals arr[index].e = Math.max(arr[index].e, arr[i].e); } else { index++; arr[index] = arr[i]; } } // Now arr[0..index-1] stores the merged Intervals console.log( "\n The Merged Intervals are: " ); for (let i = 0; i <= index; i++) console.log( "[" + arr[i].s + ", " + arr[i].e + "] " ); } // Driver program let arr = [ new Interval(6,8), new Interval(1,9), new Interval(2,4), new Interval(4,7)]; mergeIntervals(arr); |
The Merged Intervals are: [1, 9]
Time Complexity: O(N*log(N))
Auxiliary Space Complexity: O(1)
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