Insert Interval

Problem Statement

Given a list of non-overlapping intervals sorted by their start time, insert a given interval at the correct position and merge all necessary intervals to produce a list that has only mutually exclusive intervals.

Example 1

Input: intervals: [[1, 3], [5, 7], [8, 12]], newInterval: [4, 6]
Output: [[1, 3], [4, 7], [8, 12]]
Explanation: After insertion, since [4, 6] overlaps with [5, 7], we merged them into one [4, 7].

Example 2

Input: intervals: [[1, 3], [5, 7], [8, 12]], newInterval: [4, 10]
Output: [[1, 3], [4, 12]]
Explanation: After insertion, since [4, 10] overlaps with [5, 7] & [8, 12], we merged them into [4, 12].

Example 3

Input: intervals: [[2, 3],[5, 7]], newInterval: [1, 4]
Output: [[1, 4], [5, 7]]
Explanation: After insertion, since [1, 4] overlaps with [2, 3], we merged them into one [1, 4].

Solution

If the given list was not sorted, we could have simply appended the new interval to it and used the merge() function from Merge Intervals. But since the given list is sorted, we should try to come up with a solution better than O(N * logN).

When inserting a new interval in a sorted list, we need to first find the correct index where the new interval can be placed. In other words, we need to skip all the intervals which end before the start of the new interval. So we can iterate through the given sorted listed of intervals and skip all the intervals with the following condition:

intervals[i].end < newInterval.start

The diagram above clearly shows the merging approach. To handle all four merging scenarios, we need to do something like this:

c.start = min(a.start, b.start)
c.end = max(a.end, b.end)

Our overall algorithm will look like this:

1. Skip all intervals which end before the start of the new interval, i.e., skip all intervals with the following condition:

intervals[i].end < newInterval.start

2. Let’s call the last interval ‘b’ that does not satisfy the above condition. If ‘b’ overlaps with the new interval (a) (i.e. b.start <= a.end), we need to merge them into a new interval ‘c’:

c.start = min(a.start, b.start)
c.end = max(a.end, b.end)

3. We will repeat the above two steps to merge ‘c’ with the next overlapping interval.

Java Code

import java.util.*;

class Innoskrit {
    public static int[][] insert(int[][] intervals, int[] newInterval) {
        
        int i = 0;
        List<int[]> ans = new ArrayList<>();
        while(i < intervals.length && intervals[i][1] < newInterval[0]) {
            ans.add(intervals[i]);
            i += 1;
        }
        
        while(i < intervals.length && intervals[i][0] <= newInterval[1]) {
            newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
            newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
            i += 1;
        }
        ans.add(newInterval);
        
        while(i < intervals.length) {
            ans.add(intervals[i]);
            i += 1;
        }
        
        return ans.toArray(new int[ans.size()][2]);
    }

    public static void main(String[] args) {
        int intervals[][] = new int[][]{{1, 3}, {5, 7}, {8, 12}};
        int newInterval[] = new int[]{4, 6};
        for (int[] interval : Innoskrit.insert(intervals, newInterval))
            System.out.print("[" + interval[0] + "," + interval[1] + "] ");
        System.out.println();
    }
    
}

C++ Code

#include <bits/stdc++.h>
using namespace std;

class Innoskrit {
public:
    static vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        int i = 0;
        vector<vector<int>> ans;
        
        while(i < intervals.size() && intervals[i][1] < newInterval[0]) {
            ans.push_back(intervals[i]);
            i += 1;
        }

        while(i < intervals.size() && intervals[i][0] <= newInterval[1]) {
            newInterval[0] = min(newInterval[0], intervals[i][0]);
            newInterval[1] = max(newInterval[1], intervals[i][1]);
            i += 1;
        }
        ans.push_back(newInterval);
        
        while(i < intervals.size()) {
            ans.push_back(intervals[i]);
            i += 1;
        }
        
        return ans; 
    }
};

int main() {
    vector<vector<int>> intervals = {{1, 3}, {5, 7}, {8, 12}};
    vector<int> newInterval = {4, 6};
    for (auto interval : Innoskrit::insert(intervals, newInterval)) {
        cout << "[" << interval[0] << "," << interval[1] << "] ";
    }
    cout << endl;
} 

Python Code

class Innoskrit:
    def insert(self, intervals, newInterval):
        
        i = 0
        ans = []
        
        while i < len(intervals) and intervals[i][1] < newInterval[0]:
            ans.append(intervals[i])
            i += 1
            
        while i < len(intervals) and intervals[i][0] <= newInterval[1]:
            newInterval[0] = min(newInterval[0], intervals[i][0])
            newInterval[1] = max(newInterval[1], intervals[i][1])
            i += 1
        ans.append(newInterval)
        
        while i < len(intervals):
            ans.append(intervals[i])
            i += 1
            
        return ans

if __name__ == "__main__":
    intervals = [[1, 3], [5, 7], [8, 12]]
    newInterval = [4, 6]
    for interval in Innoskrit().insert(intervals, newInterval):
        print(interval, end = " ")
    print()

Output

[1,3] [4,7] [8,12]

Time Complexity

As we are iterating through all the intervals only once, the time complexity of the above algorithm is O(N), where ‘N’ is the total number of intervals.

Space Complexity

The space complexity of the above algorithm will be O(N) as we need to return a list containing all the merged intervals.