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# 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}}

Recommended Practice

## 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 ``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 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 stack=``new` `Stack<>();``  ` `        ``// sort the intervals in increasing order of start time``        ``Arrays.sort(arr,``new` `Comparator(){``            ``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);`

Output

` 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.

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 ``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 ``#include ``#include ` `// 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(){``            ``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[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[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);`

Output

` The Merged Intervals are: [1, 9] `

Time Complexity: O(N*log(N))
Auxiliary Space Complexity: O(1)

My Personal Notes arrow_drop_up