Insert and Merge Interval
Last Updated : 26 Mar, 2025
Given a set of non-overlapping intervals and a new interval, the task is to insert the interval at the correct position such that after insertion, the intervals remain sorted. If the insertion results in overlapping intervals, then merge the overlapping intervals. Assume that the set of non-overlapping intervals is sorted based on start time.
Examples:
Input: intervals[][] = [[1, 3], [4, 5], [6, 7], [8, 10]], newInterval[] = [5, 6]
Output: [[1, 3], [4, 7], [8, 10]]
Explanation: The intervals [4, 5] and [6, 7] are overlapping with [5, 6]. So, they are merged into one interval [4, 7].
Input: intervals[][] = [[1, 2], [3, 5], [6, 7], [8, 10], [12, 16]], newInterval[] = [4, 9]
Output: [[1, 2], [3, 10], [12, 16]]
Explanation: The intervals [ [3, 5], [6, 7], [8, 10] ] are overlapping with [4, 9]. So, they are merged into one interval [3, 10].
[Naive Approach] Insertion and Merging - O(n*log n) Time and O(1) Space
The approach is to append the new interval to the given array of intervals and then handle the overlapping of intervals. So, we will use the same approach as Merge Overlapping Intervals to merge the overlapping intervals after insertion.
C++ #include <bits/stdc++.h> using namespace std; // Function to merge overlapping intervals vector<vector<int>> mergeOverlap(vector<vector<int>>& intervals) { // Sort intervals based on start values sort(intervals.begin(), intervals.end()); vector<vector<int>> res; // Insert the first interval into the result res.push_back(intervals[0]); for (int i = 1; i < intervals.size(); i++) { // Find the last interval in the result vector<int>& last = res.back(); vector<int>& curr = intervals[i]; // If current interval overlaps with the last interval // in the result, merge them if (curr[0] <= last[1]) last[1] = max(last[1], curr[1]); else res.push_back(curr); } return res; } vector<vector<int>> insertInterval(vector<vector<int>>& intervals, vector<int> &newInterval) { intervals.push_back(newInterval); return mergeOverlap(intervals); } int main() { vector<vector<int>> intervals = {{1, 3}, {4, 5}, {6, 7}, {8, 10}}; vector<int> newInterval = {5, 6}; vector<vector<int>> res = insertInterval(intervals, newInterval); for (vector<int> interval: res) { cout << interval[0] << " " << interval[1] << "\n"; } return 0; } import java.util.*; class GfG { // Function to merge overlapping intervals static ArrayList<int[]> mergeOverlap(int[][] intervals) { // Sort intervals based on start values Arrays.sort(intervals, Comparator.comparingInt(a -> a[0])); ArrayList<int[]> res = new ArrayList<>(); // Insert the first interval into the result res.add(intervals[0]); for (int i = 1; i < intervals.length; i++) { // Find the last interval in the result int[] last = res.get(res.size() - 1); int[] curr = intervals[i]; // If current interval overlaps with the last interval // in the result, merge them if (curr[0] <= last[1]) { last[1] = Math.max(last[1], curr[1]); } else { res.add(curr); } } return res; } static ArrayList<int[]> insertInterval(int[][] intervals, int[] newInterval) { // Create a new ArrayList to hold the intervals ArrayList<int[]> intervalList = new ArrayList<>(Arrays.asList(intervals)); intervalList.add(newInterval); return mergeOverlap(intervalList.toArray(new int[0][])); } public static void main(String[] args) { int[][] intervals = {{1, 3}, {4, 5}, {6, 7}, {8, 10}}; int[] newInterval = {5, 6}; ArrayList<int[]> res = insertInterval(intervals, newInterval); for (int[] interval : res) { System.out.println(interval[0] + " " + interval[1]); } } } def mergeOverlap(intervals): # Sort intervals based on start values intervals.sort() res = [intervals[0]] for i in range(1, len(intervals)): last = res[-1] curr = intervals[i] # If current interval overlaps with the last interval # in the result, merge them if curr[0] <= last[1]: last[1] = max(last[1], curr[1]) else: res.append(curr) return res def insertInterval(intervals, newInterval): intervals.append(newInterval) return mergeOverlap(intervals) if __name__ == "__main__": intervals = [[1, 3], [4, 5], [6, 7], [8, 10]] newInterval = [5, 6] res = insertInterval(intervals, newInterval) for interval in res: print(interval[0], interval[1])
using System; using System.Collections.Generic; using System.Linq; class GfG { // Function to merge overlapping intervals static List<int[]> mergeOverlap(int[][] intervals) { // Sort intervals based on start values Array.Sort(intervals, (a, b) => a[0].CompareTo(b[0])); List<int[]> res = new List<int[]>(); // Insert the first interval into the result res.Add(intervals[0]); for (int i = 1; i < intervals.Length; i++) { // Find the last interval in the result int[] last = res[res.Count - 1]; int[] curr = intervals[i]; // If current interval overlaps with the last interval // in the result, merge them if (curr[0] <= last[1]) last[1] = Math.Max(last[1], curr[1]); else res.Add(curr); } return res; } static List<int[]> insertInterval(int[][] intervals, int[] newInterval) { // Convert intervals to a list and add the new interval List<int[]> intervalList = intervals.ToList(); intervalList.Add(newInterval); return mergeOverlap(intervalList.ToArray()); } static void Main(string[] args) { int[][] intervals = new int[][] { new int[] {1, 3}, new int[] {4, 5}, new int[] {6, 7}, new int[] {8, 10} }; int[] newInterval = new int[] {5, 6}; List<int[]> res = insertInterval(intervals, newInterval); foreach (int[] interval in res) { Console.WriteLine(interval[0] + " " + interval[1]); } } } function mergeOverlap(intervals) { // Sort intervals based on start values intervals.sort((a, b) => a[0] - b[0]); const res = []; // Insert the first interval into the result res.push(intervals[0]); for (let i = 1; i < intervals.length; i++) { // Find the last interval in the result const last = res[res.length - 1]; const curr = intervals[i]; // If current interval overlaps with the last interval // in the result, merge them if (curr[0] <= last[1]) last[1] = Math.max(last[1], curr[1]); else res.push(curr); } return res; } function insertInterval(intervals, newInterval) { intervals.push(newInterval); return mergeOverlap(intervals); } // Driver Code const intervals = [[1, 3], [4, 5], [6, 7], [8, 10]]; const newInterval = [5, 6]; const res = insertInterval(intervals, newInterval); for (const interval of res) { console.log(interval[0] + " " + interval[1]); } [Expected Approach] Contiguous Interval Merging - O(n) Time and O(n) Space
When we add a new interval, it may overlap with some contiguous intervals in the array. The overlapping intervals can be found in a contiguous subarray because the intervals array is already sorted. To remove overlapping we find these overlapping interval's subarray and merge them with new interval, to form a single merged interval.
Now to maintain the order sorted, we first add the lower intervals, then this merged interval, and finally the remaining intervals in the result.
C++ #include <bits/stdc++.h> using namespace std; // Function to insert and merge intervals vector<vector<int>> insertInterval(vector<vector<int>>& intervals, vector<int> &newInterval) { vector<vector<int>> res; int i = 0; int n = intervals.size(); // Add all intervals that come before the new interval while (i < n && intervals[i][1] < newInterval[0]) { res.push_back(intervals[i]); i++; } // Merge all overlapping intervals with the new interval while (i < n && intervals[i][0] <= newInterval[1]) { newInterval[0] = min(newInterval[0], intervals[i][0]); newInterval[1] = max(newInterval[1], intervals[i][1]); i++; } res.push_back(newInterval); // Add all the remaining intervals while (i < n) { res.push_back(intervals[i]); i++; } return res; } int main() { vector<vector<int>> intervals = {{1, 3}, {4, 5}, {6, 7}, {8, 10}}; vector<int> newInterval = {5, 6}; vector<vector<int>> res = insertInterval(intervals, newInterval); for (vector<int> interval: res) { cout << interval[0] << " " << interval[1] << "\n"; } return 0; } import java.util.ArrayList; import java.util.Arrays; import java.util.List; class GfG { // Function to insert and merge intervals static ArrayList<int[]> insertInterval(int[][] intervals, int[] newInterval) { ArrayList<int[]> res = new ArrayList<>(); int i = 0; int n = intervals.length; // Add all intervals that come before the new interval while (i < n && intervals[i][1] < newInterval[0]) { res.add(intervals[i]); i++; } // Merge all overlapping intervals with the new interval while (i < n && intervals[i][0] <= newInterval[1]) { newInterval[0] = Math.min(newInterval[0], intervals[i][0]); newInterval[1] = Math.max(newInterval[1], intervals[i][1]); i++; } res.add(newInterval); // Add all the remaining intervals while (i < n) { res.add(intervals[i]); i++; } // Return the result as a List<int[]> return res; } public static void main(String[] args) { int[][] intervals = {{1, 3}, {4, 5}, {6, 7}, {8, 10}}; int[] newInterval = {5, 6}; ArrayList<int[]> res = insertInterval(intervals, newInterval); for (int[] interval : res) { System.out.println(interval[0] + " " + interval[1]); } } } def insertInterval(intervals, newInterval): res = [] i = 0 n = len(intervals) # Add all intervals that come before the new interval while i < n and intervals[i][1] < newInterval[0]: res.append(intervals[i]) i += 1 # Merge all overlapping intervals with the new interval while i < n and intervals[i][0] <= newInterval[1]: newInterval[0] = min(newInterval[0], intervals[i][0]) newInterval[1] = max(newInterval[1], intervals[i][1]) i += 1 res.append(newInterval) # Add all the remaining intervals while i < n: res.append(intervals[i]) i += 1 return res if __name__ == "__main__": intervals = [[1, 3], [4, 5], [6, 7], [8, 10]] newInterval = [5, 6] res = insertInterval(intervals, newInterval) for interval in res: print(interval[0], interval[1])
using System; using System.Collections.Generic; class GfG { // Function to insert and merge intervals static List<int[]> insertInterval(int[][] intervals, int[] newInterval) { List<int[]> res = new List<int[]>(); int i = 0; int n = intervals.Length; // Add all intervals that come before the new interval while (i < n && intervals[i][1] < newInterval[0]) { res.Add(intervals[i]); i++; } // Merge all overlapping intervals with the new interval while (i < n && intervals[i][0] <= newInterval[1]) { newInterval[0] = Math.Min(newInterval[0], intervals[i][0]); newInterval[1] = Math.Max(newInterval[1], intervals[i][1]); i++; } res.Add(newInterval); // Add all the remaining intervals while (i < n) { res.Add(intervals[i]); i++; } // Return result as List<int[]> return res; } static void Main(string[] args) { int[][] intervals = new int[][] { new int[] {1, 3}, new int[] {4, 5}, new int[] {6, 7}, new int[] {8, 10} }; int[] newInterval = new int[] {5, 6}; List<int[]> res = insertInterval(intervals, newInterval); foreach (int[] interval in res) { Console.WriteLine(interval[0] + " " + interval[1]); } } } function insertInterval(intervals, newInterval) { let res = []; let i = 0; const n = intervals.length; // Add all intervals that come before the new interval while (i < n && intervals[i][1] < newInterval[0]) { res.push(intervals[i]); i++; } // Merge all overlapping intervals with the new interval while (i < n && intervals[i][0] <= newInterval[1]) { newInterval[0] = Math.min(newInterval[0], intervals[i][0]); newInterval[1] = Math.max(newInterval[1], intervals[i][1]); i++; } res.push(newInterval); // Add all the remaining intervals while (i < n) { res.push(intervals[i]); i++; } return res; } // Driver Code const intervals = [[1, 3], [4, 5], [6, 7], [8, 10]]; const newInterval = [5, 6]; const res = insertInterval(intervals, newInterval); for (const interval of res) { console.log(interval[0] + " " + interval[1]); }; Similar Reads
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