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Insert in sorted and non-overlapping interval array

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Given a set of non-overlapping intervals and a new interval, insert the interval at correct position. If the insertion results in overlapping intervals, then merge the overlapping intervals. Assume that the set of non-overlapping intervals is sorted on the basis of start time, to find correct position of insertion.

Prerequisite: Merge the intervals

Examples: 

Input: Set : [1, 3], [6, 9]
New Interval : [2, 5]
Output: [1, 5], [6, 9]
The correct position to insert new interval
[2, 5] is between the two given intervals.
The resulting set would have been
[1, 3], [2, 5], [6, 9], but the intervals
[1, 3], [2, 5] are overlapping. So, they are
merged together in one interval [1, 5].
Input: Set : [1, 2], [3, 5], [6, 7], [8, 10], [12, 16]
New Interval : [4, 9]
Output: [1, 2], [3, 10], [12, 16]
First the interval is inserted between intervals
[3, 5] and [6, 7]. Then overlapping intervals are
merged together in one interval.

Method 1:  Analyzing all Cases
Approach: 
Let the new interval to be inserted is : [a, b] 

  • Case 1 : b < (starting time of first interval in set) 
    In this case simply insert new interval at the beginning of the set.
  • Case 2 : (ending value of last interval in set) < a 
    In this case simply insert new interval at the end of the set.
  • Case 3 : a ? (starting value of first interval) and b ? (ending value of last interval) 
    In this case the new interval overlaps with all the intervals, i.e., it contains all the intervals. So the final answer is the new interval itself.
  • Case 4 : The new interval does not overlap with any interval in the set and falls between any two intervals in the set 
    In this case simply insert the interval in the correct position in the set. A sample test case for this is : 
Input: Set : [1, 2], [6, 9]
New interval : [3, 5]
Output: [1, 2], [3, 5], [6, 9]
  • Case 5 : The new interval overlaps with the interval(s) of the set. 
    In this case simply merge the new interval with overlapping intervals. To have a better understanding of how to merge overlapping intervals, refer this post : Merge Overlapping Intervals 

Example 2 of sample test cases above cover this case. 

Implementation:

C++

// C++ Code to insert a new interval in set of sorted
// intervals and merge overlapping intervals that are
// formed as a result of insertion.
#include <bits/stdc++.h>
 
using namespace std;
 
// Define the structure of interval
struct Interval
{
    int start;
    int end;
    Interval()
        : start(0), end(0)
    {
    }
    Interval(int s, int e)
        : start(s), end(e)
    {
    }
};
 
// A subroutine to check if intervals overlap or not.
bool doesOverlap(Interval a, Interval b)
{
    return (min(a.end, b.end) >= max(a.start, b.start));
}
 
// Function to insert new interval and
// merge overlapping intervals
vector<Interval> insertNewInterval
(vector<Interval>& Intervals, Interval newInterval)
{
    vector<Interval> ans;
    int n = Intervals.size();
 
    // If set is empty then simply insert
    // newInterval and return.
    if (n == 0)
    {
        ans.push_back(newInterval);
        return ans;
    }
 
 
    // Case 1 and Case 2 (new interval to be
    // inserted at corners)
    if (newInterval.end < Intervals[0].start ||
            newInterval.start > Intervals[n - 1].end)
    {
        if (newInterval.end < Intervals[0].start)
            ans.push_back(newInterval);
 
        for (int i = 0; i < n; i++)
            ans.push_back(Intervals[i]);
 
        if (newInterval.start > Intervals[n - 1].end)
            ans.push_back(newInterval);
 
        return ans;
    }
 
    // Case 3 (New interval covers all existing)
    if (newInterval.start <= Intervals[0].start &&
        newInterval.end >= Intervals[n - 1].end)
    {
        ans.push_back(newInterval);
        return ans;
    }
 
    // Case 4 and Case 5
    // These two cases need to check whether
    // intervals overlap or not. For this we
    // can use a subroutine that will perform
    // this function.
    bool overlap = true;
    for (int i = 0; i < n; i++)
    {
        overlap = doesOverlap(Intervals[i], newInterval);
        if (!overlap)
        {
            ans.push_back(Intervals[i]);
 
            // Case 4 : To check if given interval
            // lies between two intervals.
            if (i < n &&
                newInterval.start > Intervals[i].end &&
                newInterval.end < Intervals[i + 1].start)
                ans.push_back(newInterval);
 
            continue;
        }
 
        // Case 5 : Merge Overlapping Intervals.
        // Starting time of new merged interval is
        // minimum of starting time of both
        // overlapping intervals.
        Interval temp;
        temp.start = min(newInterval.start,
                         Intervals[i].start);
 
        // Traverse the set until intervals are
        // overlapping
        while (i < n && overlap)
        {
 
            // Ending time of new merged interval
            // is maximum of ending time both
            // overlapping intervals.
            temp.end = max(newInterval.end,
                           Intervals[i].end);
            if (i == n - 1)
                overlap = false;
            else
                overlap = doesOverlap(Intervals[i + 1],
                                          newInterval);
            i++;
        }
 
        i--;
        ans.push_back(temp);
    }
 
    return ans;
}
 
// Driver code
int main()
{
    vector<Interval> Intervals;
    Interval newInterval;
 
    newInterval.start = 1;
    newInterval.end = 2;
    Intervals.push_back(newInterval);
    newInterval.start = 3;
    newInterval.end = 5;
    Intervals.push_back(newInterval);
    newInterval.start = 6;
    newInterval.end = 7;
    Intervals.push_back(newInterval);
    newInterval.start = 8;
    newInterval.end = 10;
    Intervals.push_back(newInterval);
    newInterval.start = 12;
    newInterval.end = 16;
    Intervals.push_back(newInterval);
    newInterval.start = 4;
    newInterval.end = 9;
 
    vector<Interval> ans =
          insertNewInterval(Intervals, newInterval);
 
    for (int i = 0; i < ans.size(); i++)
        cout << ans[i].start << ", "
             << ans[i].end << "\n";
 
    return 0;
}

                    

Java

// java Code to insert a new interval in set of sorted
 
// intervals and merge overlapping intervals that are
// formed as a result of insertion.
import java.util.*;
// Define the structure of interval
public class Main {
    public static class Interval {
        public int start;
        public int end;
 
        public Interval()
        {
            this.start = 0;
            this.end = 0;
        }
 
        public Interval(int start, int end)
        {
            this.start = start;
            this.end = end;
        }
    }
    // A subroutine to check if intervals overlap or not.
    public static boolean doesOverlap(Interval a,
                                      Interval b)
    {
        return (Math.min(a.end, b.end)
                >= Math.max(a.start, b.start));
    }
    // Function to insert new interval and
    // merge overlapping intervals
    public static List<Interval>
    insertNewInterval(List<Interval> intervals,
                      Interval newInterval)
    {
        List<Interval> ans = new ArrayList<>();
        int n = intervals.size();
        // If set is empty then simply insert
        // newInterval and return.
        if (n == 0) {
            ans.add(newInterval);
            return ans;
        }
        // Case 1 and Case 2 (new interval to be
        // inserted at corners)
        if (newInterval.end < intervals.get(0).start
            || newInterval.start
                   > intervals.get(n - 1).end) {
            if (newInterval.end < intervals.get(0).start) {
                ans.add(newInterval);
            }
 
            for (int i = 0; i < n; i++) {
                ans.add(intervals.get(i));
            }
 
            if (newInterval.start
                > intervals.get(n - 1).end) {
                ans.add(newInterval);
            }
 
            return ans;
        }
        // Case 3 (New interval covers all existing)
 
        if (newInterval.start <= intervals.get(0).start
            && newInterval.end
                   >= intervals.get(n - 1).end) {
            ans.add(newInterval);
            return ans;
        }
 
        // Case 4 and Case 5
        // These two cases need to check whether
        // intervals overlap or not. For this we
        // can use a subroutine that will perform
        // this function.
        boolean overlap = true;
 
        for (int i = 0; i < n; i++) {
            overlap = doesOverlap(intervals.get(i),
                                  newInterval);
 
            if (!overlap) {
 
                // Case 4 : To check if given interval
                // lies between two intervals.
                ans.add(intervals.get(i));
 
                if (i < n
                    && newInterval.start
                           > intervals.get(i).end
                    && newInterval.end
                           < intervals.get(i + 1).start) {
                    ans.add(newInterval);
                }
 
                continue;
            }
            // Case 5 : Merge Overlapping Intervals.
            // Starting time of new merged interval is
            // minimum of starting time of both
            // overlapping intervals.
 
            Interval temp = new Interval();
            temp.start = Math.min(newInterval.start,
                                  intervals.get(i).start);
            // Traverse the set until intervals are
            // overlapping
            while (i < n && overlap) {
                // Ending time of new merged interval
                // is maximum of ending time both
                // overlapping intervals.
                temp.end = Math.max(newInterval.end,
                                    intervals.get(i).end);
 
                if (i == n - 1) {
                    overlap = false;
                }
                else {
                    overlap = doesOverlap(
                        intervals.get(i + 1), newInterval);
                }
 
                i++;
            }
 
            i--;
            ans.add(temp);
        }
 
        return ans;
    }
    // Driver code
    public static void main(String[] args)
    {
        List<Interval> intervals = new ArrayList<>();
        Interval newInterval = new Interval(1, 2);
        intervals.add(newInterval);
        newInterval = new Interval(3, 5);
        intervals.add(newInterval);
        newInterval = new Interval(6, 7);
        intervals.add(newInterval);
        newInterval = new Interval(8, 10);
        intervals.add(newInterval);
        newInterval = new Interval(12, 16);
        intervals.add(newInterval);
        newInterval = new Interval(4, 9);
 
        List<Interval> ans
            = insertNewInterval(intervals, newInterval);
 
        for (int i = 0; i < ans.size(); i++) {
            System.out.println(ans.get(i).start + ", "
                               + ans.get(i).end);
        }
    }
}

                    

Python

class Interval:
    def __init__(self, s=0, e=0):
        self.start = s
        self.end = e
 
def insert_and_merge_intervals(intervals, new_interval):
    merged = []
    i = 0
 
    # Add intervals that come before the new interval
    while i < len(intervals) and intervals[i].end < new_interval.start:
        merged.append(intervals[i])
        i += 1
 
    # Merge overlapping intervals
    while i < len(intervals) and intervals[i].start <= new_interval.end:
        new_interval.start = min(new_interval.start, intervals[i].start)
        new_interval.end = max(new_interval.end, intervals[i].end)
        i += 1
 
    merged.append(new_interval)
 
    # Add intervals that come after the new interval
    while i < len(intervals):
        merged.append(intervals[i])
        i += 1
 
    return merged
 
if __name__ == "__main__":
    intervals = [Interval(1, 2), Interval(3, 5), Interval(6, 7), Interval(8, 10), Interval(12, 16)]
    new_interval = Interval(4, 9)
 
    result = insert_and_merge_intervals(intervals, new_interval)
 
    for interval in result:
        print(interval.start, interval.end)

                    

C#

using System;
using System.Collections.Generic;
 
public class MainClass
{
    public class Interval
    {
        public int start;
        public int end;
 
        public Interval()
        {
            this.start = 0;
            this.end = 0;
        }
 
        public Interval(int start, int end)
        {
            this.start = start;
            this.end = end;
        }
    }
 
    public static bool DoesOverlap(Interval a, Interval b)
    {
        return Math.Min(a.end, b.end) >= Math.Max(a.start, b.start);
    }
 
    public static List<Interval> InsertNewInterval(List<Interval> intervals,
                                                   Interval newInterval)
    {
        List<Interval> ans = new List<Interval>();
        int n = intervals.Count;
 
        if (n == 0)
        {
            ans.Add(newInterval);
            return ans;
        }
 
        if (newInterval.end < intervals[0].start || newInterval.start > intervals[n - 1].end)
        {
            if (newInterval.end < intervals[0].start)
            {
                ans.Add(newInterval);
            }
 
            for (int i = 0; i < n; i++)
            {
                ans.Add(intervals[i]);
            }
 
            if (newInterval.start > intervals[n - 1].end)
            {
                ans.Add(newInterval);
            }
 
            return ans;
        }
 
        if (newInterval.start <= intervals[0].start && newInterval.end >= intervals[n - 1].end)
        {
            ans.Add(newInterval);
            return ans;
        }
 
        bool overlap = true;
 
        for (int i = 0; i < n; i++)
        {
            overlap = DoesOverlap(intervals[i], newInterval);
 
            if (!overlap)
            {
                ans.Add(intervals[i]);
 
                if (i < n && newInterval.start > intervals[i].end &&
                    newInterval.end < intervals[i + 1].start)
                {
                    ans.Add(newInterval);
                }
 
                continue;
            }
 
            Interval temp = new Interval();
            temp.start = Math.Min(newInterval.start, intervals[i].start);
 
            while (i < n && overlap)
            {
                temp.end = Math.Max(newInterval.end, intervals[i].end);
 
                if (i == n - 1)
                {
                    overlap = false;
                }
                else
                {
                    overlap = DoesOverlap(intervals[i + 1], newInterval);
                }
 
                i++;
            }
 
            i--;
            ans.Add(temp);
        }
 
        return ans;
    }
 
    public static void Main(string[] args)
    {
        List<Interval> intervals = new List<Interval>();
        Interval newInterval = new Interval(1, 2);
        intervals.Add(newInterval);
        newInterval = new Interval(3, 5);
        intervals.Add(newInterval);
        newInterval = new Interval(6, 7);
        intervals.Add(newInterval);
        newInterval = new Interval(8, 10);
        intervals.Add(newInterval);
        newInterval = new Interval(12, 16);
        intervals.Add(newInterval);
        newInterval = new Interval(4, 9);
 
        List<Interval> ans = InsertNewInterval(intervals, newInterval);
 
        foreach (Interval interval in ans)
        {
            Console.WriteLine(interval.start + ", " + interval.end);
        }
    }
}

                    

Javascript

// Javascript code addition
 
// Define the structure of interval
class Interval {
  constructor(start = 0, end = 0) {
    this.start = start;
    this.end = end;
  }
}
 
// A subroutine to check if intervals overlap or not.
function doesOverlap(a, b) {
  return Math.min(a.end, b.end) >= Math.max(a.start, b.start);
}
 
// Function to insert new interval and merge overlapping intervals
function insertNewInterval(intervals, newInterval) {
  let ans = [];
  let n = intervals.length;
 
  // If set is empty then simply insert newInterval and return.
  if (n === 0) {
    ans.push(newInterval);
    return ans;
  }
 
  // Case 1 and Case 2 (new interval to be inserted at corners)
  if (
    newInterval.end < intervals[0].start ||
    newInterval.start > intervals[n - 1].end
  ) {
    if (newInterval.end < intervals[0].start) ans.push(newInterval);
 
    for (let i = 0; i < n; i++) ans.push(intervals[i]);
 
    if (newInterval.start > intervals[n - 1].end) ans.push(newInterval);
 
    return ans;
  }
 
  // Case 3 (New interval covers all existing)
  if (
    newInterval.start <= intervals[0].start &&
    newInterval.end >= intervals[n - 1].end
  ) {
    ans.push(newInterval);
    return ans;
  }
 
  // Case 4 and Case 5
  // These two cases need to check whether intervals overlap or not.
  // For this we can use a subroutine that will perform this function.
  let overlap = true;
  for (let i = 0; i < n; i++) {
    overlap = doesOverlap(intervals[i], newInterval);
    if (!overlap) {
      ans.push(intervals[i]);
 
      // Case 4 : To check if given interval lies between two intervals.
      if (
        i < n &&
        newInterval.start > intervals[i].end &&
        newInterval.end < intervals[i + 1].start
      ) {
        ans.push(newInterval);
      }
 
      continue;
    }
 
    // Case 5 : Merge Overlapping Intervals.
    // Starting time of new merged interval is minimum of starting time of both overlapping intervals.
    let temp = new Interval(
      Math.min(newInterval.start, intervals[i].start),
      0
    );
 
    // Traverse the set until intervals are overlapping
    while (i < n && overlap) {
      // Ending time of new merged interval is maximum of ending time both overlapping intervals.
      temp.end = Math.max(newInterval.end, intervals[i].end);
      if (i === n - 1) overlap = false;
      else overlap = doesOverlap(intervals[i + 1], newInterval);
      i++;
    }
 
    i--;
    ans.push(temp);
  }
 
  return ans;
}
 
// Driver code
let intervals = [];
let newInterval = new Interval(1, 2);
intervals.push(newInterval);
newInterval = new Interval(3, 5);
intervals.push(newInterval);
newInterval = new Interval(6, 7);
intervals.push(newInterval);
newInterval = new Interval(8, 10);
intervals.push(newInterval);
newInterval = new Interval(12, 16);
intervals.push(newInterval);
newInterval = new Interval(4, 9);
 
let ans = insertNewInterval(intervals, newInterval);
 
for (let i = 0; i < ans.length; i++) {
  console.log(ans[i].start + ", " + ans[i].end);
}
 
// The code is contributed by Nidhi goel.

                    

Output
1, 2
3, 10
12, 16


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

Method 2: Using merge intervals

  1. Append the given interval to the given list of the intervals.
  2. Now it becomes the same old merge intervals problem. To have a better understanding of how to merge overlapping intervals, refer this post : Merge Overlapping Intervals 
  3. Use the merge intervals function.

Implementation:

C++

// C++ Code to insert a new interval in set of sorted
// intervals and merge overlapping intervals that are
// formed as a result of insertion.
#include <bits/stdc++.h>
 
using namespace std;
 
// Define the structure of interval
struct Interval
{
    int s, e;
};
 
// A subroutine to check if intervals overlap or not.
bool mycomp(Interval a, Interval b) { return a.s < b.s; }
 
// merge overlapping intervals
void insertNewInterval(vector<Interval>& Intervals, Interval newInterval)
{
    vector<Interval> ans;
    int n = Intervals.size();
    Intervals.push_back(newInterval); //Insert the new interval into Intervals
    sort(Intervals.begin(), Intervals.end(), mycomp);
  
    int index = 0;
    // Traverse all input Intervals
    for (int i = 1; i <=n; i++) {
        // If this is not first Interval and overlaps
        // with the previous one
        if (Intervals[index].e >= Intervals[i].s) {
            // Merge previous and current Intervals
            Intervals[index].e = max(Intervals[index].e, Intervals[i].e);
        }
        else {
            index++;
            Intervals[index] = Intervals[i];
        }
    }
   
    for (int i = 0; i <= index; i++)
        cout  << Intervals[i].s << ", " << Intervals[i].e << "\n";
}
 
// Driver code
int main()
{
    vector<Interval> Intervals;
    Interval newInterval;
 
    newInterval.s = 1;
    newInterval.e = 2;
    Intervals.push_back(newInterval);
    newInterval.s = 3;
    newInterval.e = 5;
    Intervals.push_back(newInterval);
    newInterval.s = 6;
    newInterval.e = 7;
    Intervals.push_back(newInterval);
    newInterval.s = 8;
    newInterval.e = 10;
    Intervals.push_back(newInterval);
    newInterval.s = 12;
    newInterval.e = 16;
    Intervals.push_back(newInterval);
    newInterval.s = 4;
    newInterval.e = 9;
 
    insertNewInterval(Intervals, newInterval);
 
    return 0;
}

                    

Java

// Java Code to insert a new interval in set of sorted
// intervals and merge overlapping intervals that are
// formed as a result of insertion.
 
import java.util.*;
 
public class IntervalMerge {
    // Define the structure of interval
    static class Interval {
        int s, e;
        Interval(int s, int e)
        {
            this.s = s;
            this.e = e;
        }
    }
 
    // A subroutine to check if intervals overlap or not.
    static class IntervalComparator
        implements Comparator<Interval> {
        public int compare(Interval a, Interval b)
        {
            return a.s - b.s;
        }
    }
 
    // merge overlapping intervals
    static void insertNewInterval(List<Interval> intervals,
                                  Interval newInterval)
    {
        List<Interval> ans = new ArrayList<>();
        int n = intervals.size();
        intervals.add(
            newInterval); // Insert the new interval into
                          // intervals
        Collections.sort(intervals,
                         new IntervalComparator());
 
        int index = 0;
        // Traverse all input intervals
        for (int i = 1; i <= n; i++) {
            // If this is not first Interval and overlaps
            // with the previous one
            if (intervals.get(index).e
                >= intervals.get(i).s) {
                // Merge previous and current intervals
                intervals.get(index).e
                    = Math.max(intervals.get(index).e,
                               intervals.get(i).e);
            }
            else {
                index++;
                intervals.set(index, intervals.get(i));
            }
        }
 
        for (int i = 0; i <= index; i++)
            System.out.println(intervals.get(i).s + ", "
                               + intervals.get(i).e);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        List<Interval> intervals = new ArrayList<>();
        Interval newInterval;
 
        newInterval = new Interval(1, 2);
        intervals.add(newInterval);
        newInterval = new Interval(3, 5);
        intervals.add(newInterval);
        newInterval = new Interval(6, 7);
        intervals.add(newInterval);
        newInterval = new Interval(8, 10);
        intervals.add(newInterval);
        newInterval = new Interval(12, 16);
        intervals.add(newInterval);
        newInterval = new Interval(4, 9);
 
        insertNewInterval(intervals, newInterval);
    }
}

                    

Python3

# Define the structure of interval
class Interval:
    def __init__(self, s=0, e=0):
        self.s = s
        self.e = e
 
# A subroutine to check if intervals overlap or not.
 
 
def mycomp(interval):
    return interval.s
 
# merge overlapping intervals
 
 
def insertNewInterval(Intervals, newInterval):
    ans = []
    n = len(Intervals)
    Intervals.append(newInterval)  # Insert the new interval into Intervals
    Intervals.sort(key=mycomp)
 
    index = 0
    # Traverse all input Intervals
    for i in range(1, n+1):
        # If this is not first Interval and overlaps
        # with the previous one
        if Intervals[index].e >= Intervals[i].s:
            # Merge previous and current Intervals
            Intervals[index].e = max(Intervals[index].e, Intervals[i].e)
        else:
            index += 1
            Intervals[index] = Intervals[i]
 
    for i in range(index+1):
        print(Intervals[i].s, Intervals[i].e)
 
 
Intervals = []
newInterval = Interval(1, 2)
Intervals.append(newInterval)
newInterval = Interval(3, 5)
Intervals.append(newInterval)
newInterval = Interval(6, 7)
Intervals.append(newInterval)
newInterval = Interval(8, 10)
Intervals.append(newInterval)
newInterval = Interval(12, 16)
Intervals.append(newInterval)
newInterval = Interval(4, 9)
 
insertNewInterval(Intervals, newInterval)

                    

C#

using System;
using System.Collections.Generic;
using System.Linq;
 
namespace IntervalMerge
{
    public class IntervalMergeProgram
    {
        // Define the structure of interval
        public class Interval
        {
            public int s, e;
            public Interval(int s, int e)
            {
                this.s = s;
                this.e = e;
            }
        }
 
        // A subroutine to check if intervals overlap or not.
        public class IntervalComparator : IComparer<Interval>
        {
            public int Compare(Interval a, Interval b)
            {
                return a.s - b.s;
            }
        }
 
        // merge overlapping intervals
        public static void InsertNewInterval(List<Interval> intervals, Interval newInterval)
        {
            List<Interval> ans = new List<Interval>();
            int n = intervals.Count();
            intervals.Add(newInterval); // Insert the new interval into intervals
            intervals.Sort(new IntervalComparator());
 
            int index = 0;
            // Traverse all input intervals
            for (int i = 1; i <= n; i++)
            {
                // If this is not first Interval and overlaps with the previous one
                if (intervals[index].e >= intervals[i].s)
                {
                    // Merge previous and current intervals
                    intervals[index].e = Math.Max(intervals[index].e, intervals[i].e);
                }
                else
                {
                    index++;
                    intervals[index] = intervals[i];
                }
            }
 
            for (int i = 0; i <= index; i++)
            {
                Console.WriteLine(intervals[i].s + ", " + intervals[i].e);
            }
        }
 
        // Driver code
        static void Main(string[] args)
        {
            List<Interval> intervals = new List<Interval>();
            Interval newInterval;
 
            newInterval = new Interval(1, 2);
            intervals.Add(newInterval);
            newInterval = new Interval(3, 5);
            intervals.Add(newInterval);
            newInterval = new Interval(6, 7);
            intervals.Add(newInterval);
            newInterval = new Interval(8, 10);
            intervals.Add(newInterval);
            newInterval = new Interval(12, 16);
            intervals.Add(newInterval);
            newInterval = new Interval(4, 9);
 
            InsertNewInterval(intervals, newInterval);
        }
    }
}

                    

Javascript

// Define the structure of interval
class Interval {
    constructor(s, e) {
        this.s = s;
        this.e = e;
    }
}
 
// A subroutine to check if intervals overlap or not.
function mycomp(a, b) {
    return a.s - b.s;
}
 
// merge overlapping intervals
function insertNewInterval(Intervals, newInterval) {
    let ans = [];
    let n = Intervals.length;
    Intervals.push(newInterval); //Insert the new interval into Intervals
    Intervals.sort(mycomp);
 
    let index = 0;
    // Traverse all input Intervals
    for (let i = 1; i <= n; i++) {
        // If this is not first Interval and overlaps
        // with the previous one
        if (Intervals[index].e >= Intervals[i].s) {
            // Merge previous and current Intervals
            Intervals[index].e = Math.max(Intervals[index].e, Intervals[i].e);
        }
        else {
            index++;
            Intervals[index] = Intervals[i];
        }
    }
 
    for (let i = 0; i <= index; i++) {
        console.log(Intervals[i].s + ", " + Intervals[i].e + "\n");
    }
}
 
// Driver code to test above function
let Intervals = [];
 
let newInterval = new Interval(1, 2);
Intervals.push(newInterval);
newInterval = new Interval(3, 5);
Intervals.push(newInterval);
newInterval = new Interval(6, 7);
Intervals.push(newInterval);
newInterval = new Interval(8, 10);
Intervals.push(newInterval);
newInterval = new Interval(12, 16);
Intervals.push(newInterval);
newInterval = new Interval(4, 9);
 
insertNewInterval(Intervals, newInterval);

                    

Output
1, 2
3, 10
12, 16


Time Complexity: O(nlogn) 
Auxiliary Space: O(1)

Method 3: Another Approach Using Stack:

We will be pushing pairs in the stack until it merges with the intervals or finds a suitable place for fitting it.

Below is the implementation of the above approach:

C++

// C++ program for above approach
#include <iostream>
#include <stack>
using namespace std;
 
// Program to merge interval
void mergeInterval2(pair<int, int> arr[],
                    int n, pair<int,
                               int> newPair)
     
    // Create stack of type
    // pair<int, int>
    stack< pair<int, int> > stk;
     
    // Pushing first pair
    stk.push(arr[0]);
    
    // Storing the top element
    pair<int, int> top = stk.top();
   
    // Checking is newPair.first
    // is less than top.second
    if (newPair.first < top.second)
    {
         
        // Pop the top element
        // as it will merge with the
        // previous range
        stk.pop();
       
        // Re-assigning top.first
        top.first = min(top.first,
                          newPair.first);
       
        // Re-assigning top.second
        top.second = max(top.second,
                          newPair.second);
       
        // Push the current interval
        stk.push(top);
    }
    else
    {
         
       // Push the new pair as it does
       // not intersect to first pair
       stk.push(newPair);
    }
 
    // Iterate i from 1 to n - 1
    for (int i = 1; i < n; i++)
    {
         
        // Store the top element of
        // the stack stk
        pair<int, int> top = stk.top();
       
        // Checking is arr[i].first
        // is less than top.second
        if (arr[i].first < top.second)
        {
             
           
            // Pop the top element
            // as it will merge with the
            // previous range
            stk.pop();
           
            // Re-assigning top.first
            top.first = min(top.first,
                            arr[i].first);
           
            // Re-assigning top.second
            top.second = max(top.second,
                            arr[i].second);
           
            // Push the current interval
            stk.push(top);
        }
         
        else
        {
             
            // Push the new pair as it does
            // not intersect to first pair
            stk.push(arr[i]);
        }
    }
     
    // Storing the final intervals
    stack< pair<int,int> > finalIntervals;
   
    // Popping the stack elements
    while (stk.empty() != true)
    {
        pair<int, int> ele =
                       stk.top();
        stk.pop();
        
        // Push ele in finalIntervals
        finalIntervals.push(ele);  
    }
   
     
    // Displaying the final result
    while (finalIntervals.empty() != true)
    {
        pair<int, int> ele =
                       finalIntervals.top();
        finalIntervals.pop();
       
        cout << ele.first << ", "
             << ele.second << endl;
    }
}
 
// Driver Code
int main()
{
 
    pair<int, int> arr2[] = {
        { 1, 2 }, { 3, 5 }, { 6, 7 },
                 { 8, 10 }, { 12, 16 }
    };
    pair<int, int> newPair{ 4, 9 };
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
 
    // Function Call
    mergeInterval2(arr2, n2, newPair);
 
    return 0;
}

                    

Java

import java.util.*;
 
public class Main {
 
  // Program to merge interval
  public static void mergeInterval2(Pair[] arr, int n,
                                    Pair newPair)
  {
 
    // Create stack of type Pair
    Stack<Pair> stk = new Stack<Pair>();
 
    // Pushing first pair
    stk.push(arr[0]);
 
    // Storing the top element
    Pair top = stk.peek();
 
    // Checking is newPair.first
    // is less than top.second
    if (newPair.first < top.second) {
 
      // Pop the top element
      // as it will merge with the
      // previous range
      stk.pop();
 
      // Re-assigning top.first
      top.first = Math.min(top.first, newPair.first);
 
      // Re-assigning top.second
      top.second
        = Math.max(top.second, newPair.second);
 
      // Push the current interval
      stk.push(top);
    }
    else {
 
      // Push the new pair as it does
      // not intersect to first pair
      stk.push(newPair);
    }
 
    // Iterate i from 1 to n - 1
    for (int i = 1; i < n; i++) {
 
      // Store the top element of
      // the stack stk
      Pair topElement = stk.peek();
 
      // Checking is arr[i].first
      // is less than top.second
      if (arr[i].first < topElement.second) {
 
        // Pop the top element
        // as it will merge with the
        // previous range
        stk.pop();
 
        // Re-assigning top.first
        topElement.first = Math.min(
          topElement.first, arr[i].first);
 
        // Re-assigning top.second
        topElement.second = Math.max(
          topElement.second, arr[i].second);
 
        // Push the current interval
        stk.push(topElement);
      }
      else {
 
        // Push the new pair as it does
        // not intersect to first pair
        stk.push(arr[i]);
      }
    }
 
    // Storing the final intervals
    Stack<Pair> finalIntervals = new Stack<Pair>();
 
    // Popping the stack elements
    while (stk.empty() != true) {
      Pair ele = stk.pop();
 
      // Push ele in finalIntervals
      finalIntervals.push(ele);
    }
 
    // Displaying the final result
    while (finalIntervals.empty() != true) {
      Pair ele = finalIntervals.pop();
      System.out.println(ele.first + ", "
                         + ele.second);
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    Pair[] arr2 = { new Pair(1, 2), new Pair(3, 5),
                   new Pair(6, 7), new Pair(8, 10),
                   new Pair(12, 16) };
    Pair newPair = new Pair(4, 9);
    int n2 = arr2.length;
 
    // Function Call
    mergeInterval2(arr2, n2, newPair);
  }
 
  // Pair class
  static class Pair {
    int first, second;
 
    public Pair(int first, int second)
    {
      this.first = first;
      this.second = second;
    }
  }
}

                    

Python3

# Python3 program for above approach
 
# Program to merge interval
def mergeInterval2(arr, n, newPair) :
     
    # Create stack of type
    # pair<int, int>
    stk = []
     
    # Pushing first pair
    stk.append(arr[0])
     
    # Storing the top element
    top = stk[len(stk) - 1]
     
    # Checking is newPair.first
    # is less than top.second
    if (newPair[0] < top[1]) :
         
        # Pop the top element
        # as it will merge with the
        # previous range
        stk.pop()
         
        # Re-assigning top.first
        top[0] = min(top[0], newPair[0])
         
        # Re-assigning top.second
        top[1] = max(top[1], newPair[1])
         
        # Push the current interval
        stk.append(top)
 
    else :
        # Push the new pair as it does
        # not intersect to first pair
        stk.append(newPair)
 
    # Iterate i from 1 to n - 1
    for i in range(1, n) :
         
        # Store the top element of
        # the stack stk
        top = stk[len(stk) - 1]
         
        # Checking is arr[i].first
        # is less than top.second
        if (arr[i][0] < top[1]) :
             
            # Pop the top element
            # as it will merge with the
            # previous range
            stk.pop()
             
            # Re-assigning top.first
            top[0] = min(top[0], arr[i][0])
             
            # Re-assigning top.second
            top[1] = max(top[1], arr[i][1])
             
            # Push the current interval
            stk.append(top)
         
        else :
             
            # Push the new pair as it does
            # not intersect to first pair
            stk.append(arr[i])
     
    # Storing the final intervals
    finalIntervals = []
     
    # Popping the stack elements
    while (len(stk) > 0) :
     
        ele = stk[len(stk) - 1]
        stk.pop()
         
        # Push ele in finalIntervals
        finalIntervals.append(ele)
     
    # Displaying the final result
    while (len(finalIntervals) > 0) :
     
        ele = finalIntervals[len(finalIntervals) - 1]
        finalIntervals.pop()
         
        print(ele[0] , end = ", ")
        print(ele[1])
 
arr2 = [ [ 1, 2 ], [ 3, 5 ], [ 6, 7 ], [ 8, 10 ], [ 12, 16 ] ]
 
newPair = [ 4, 9 ]
n2 = len(arr2)
 
# Function Call
mergeInterval2(arr2, n2, newPair)
 
# This code is contributed by divyesh072019

                    

C#

// C# program for above approach
using System;
using System.Collections;
 
class GFG{
     
// Function to merge interval
static void mergeInterval2(Tuple<int, int>[] arr,
                    int n, Tuple<int, int> newPair)
     
    // Create stack of type
    // pair<int, int>
    Stack stk = new Stack();
      
    // Pushing first pair
    stk.Push(arr[0]);
     
    // Storing the top element
    Tuple<int,
          int> top = (Tuple<int,
                            int>)stk.Peek();
    
    // Checking is newPair.first
    // is less than top.second
    if (newPair.Item1 < top.Item2)
    {
         
        // Pop the top element
        // as it will merge with the
        // previous range
        stk.Pop();
        
        // Re-assigning top.first and top.second
        top = new Tuple<int, int>(Math.Min(top.Item1,
                                           newPair.Item1),
                                  Math.Max(top.Item2,
                                           newPair.Item2));
        
        // Push the current interval
        stk.Push(top);
    }
    else
    {
         
        // Push the new pair as it does
        // not intersect to first pair
        stk.Push(newPair);
    }
  
    // Iterate i from 1 to n - 1
    for(int i = 1; i < n; i++)
    {
         
        // Store the top element of
        // the stack stk
        Tuple<int,
              int> Top = (Tuple<int,
                                int>)stk.Peek();
        
        // Checking is arr[i].first
        // is less than top.second
        if (arr[i].Item1 < Top.Item2)
        {
             
            // Pop the top element
            // as it will merge with the
            // previous range
            stk.Pop();
            
            // Re-assigning top.first and top.second
            Top = new Tuple<int, int>(Math.Min(Top.Item1,
                                               arr[i].Item1),
                                      Math.Max(Top.Item2,
                                               arr[i].Item2));
            
            // Push the current interval
            stk.Push(Top);
        }
        else
        {
             
            // Push the new pair as it does
            // not intersect to first pair
            stk.Push(arr[i]);
        }
    }
      
    // Storing the final intervals
    Stack finalIntervals = new Stack();
    
    // Popping the stack elements
    while (stk.Count != 0)
    {
        Tuple<int,
              int> ele = (Tuple<int,
                                int>)stk.Peek();
        stk.Pop();
         
        // Push ele in finalIntervals
        finalIntervals.Push(ele);  
    }
     
    // Displaying the final result
    while (finalIntervals.Count != 0)
    {
        Tuple<int,
              int> ele = (Tuple<int,
                                int>)finalIntervals.Peek();
                                 
        finalIntervals.Pop();
        
        Console.WriteLine(ele.Item1 + ", " + ele.Item2);
    }
}
 
// Driver Code
static void Main()
{
    Tuple<int, int>[] arr2 =
    {
        Tuple.Create(1, 2),
        Tuple.Create(3, 5),
        Tuple.Create(6, 7),
        Tuple.Create(8, 10),
        Tuple.Create(12, 16),
    };
     
    Tuple<int,
          int> newPair = new Tuple<int,
                                   int>(4, 9);
    int n2 = arr2.Length;
     
    // Function Call
    mergeInterval2(arr2, n2, newPair);
}
}
 
// This code is contributed by divyeshrabadiya07

                    

Javascript

// Program to merge interval
function mergeInterval2(arr, n, newPair) {
    // Create stack of type pair<int, int>
    const stk = [];
     
    // Pushing first pair
    stk.push(arr[0]);
     
    // Storing the top element
    let top = stk[stk.length - 1];
 
    // Checking is newPair.first is less than top.second
    if (newPair.first < top.second) {
        // Pop the top element as it will merge with the previous range
        stk.pop();
     
        // Re-assigning top.first
        top.first = Math.min(top.first, newPair.first);
     
        // Re-assigning top.second
        top.second = Math.max(top.second, newPair.second);
     
        // Push the current interval
        stk.push(top);
    } else {
        // Push the new pair as it does not intersect with the first pair
        stk.push(newPair);
    }
 
    // Iterate i from 1 to n - 1
    for (let i = 1; i < n; i++) {
        // Store the top element of the stack stk
        top = stk[stk.length - 1];
     
        // Checking is arr[i].first is less than top.second
        if (arr[i].first < top.second) {
            // Pop the top element as it will merge with the previous range
            stk.pop();
         
            // Re-assigning top.first
            top.first = Math.min(top.first, arr[i].first);
         
            // Re-assigning top.second
            top.second = Math.max(top.second, arr[i].second);
         
            // Push the current interval
            stk.push(top);
        } else {
            // Push the new pair as it does not intersect with the first pair
            stk.push(arr[i]);
        }
    }
     
    // Storing the final intervals
    const finalIntervals = [];
 
    // Popping the stack elements
    while (stk.length > 0) {
        const ele = stk.pop();
         
        // Push ele in finalIntervals
        finalIntervals.push(ele);
    }
 
    // Displaying the final result
    while (finalIntervals.length > 0) {
        const ele = finalIntervals.pop();
        console.log(`${ele.first}, ${ele.second}`);
    }
}
 
// Driver Code
(function main() {
    const arr2 = [
        { first: 1, second: 2 },
        { first: 3, second: 5 },
        { first: 6, second: 7 },
        { first: 8, second: 10 },
        { first: 12, second: 16 }
    ];
    const newPair = { first: 4, second: 9 };
    const n2 = arr2.length;
 
    // Function Call
    mergeInterval2(arr2, n2, newPair);
})();

                    

Output
1, 2
3, 10
12, 16


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



Last Updated : 14 Oct, 2023
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