57. Insert Interval

Medium

Problem Description

You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.

Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return intervals after the insertion.

Examples

Example 1:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:
Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
Jump to Solution: Python Java C++ JavaScript C#

Python Solution


def insert(intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
    result = []
    i = 0
    n = len(intervals)
    
    # Add all intervals that end before newInterval starts
    while i < n and intervals[i][1] < newInterval[0]:
        result.append(intervals[i])
        i += 1
    
    # Merge overlapping intervals
    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
    
    result.append(newInterval)
    
    # Add remaining intervals
    while i < n:
        result.append(intervals[i])
        i += 1
    
    return result

Java Solution


class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List result = new ArrayList<>();
        int i = 0;
        int n = intervals.length;
        
        // Add all intervals that end before newInterval starts
        while (i < n && intervals[i][1] < newInterval[0]) {
            result.add(intervals[i]);
            i++;
        }
        
        // Merge overlapping intervals
        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++;
        }
        
        result.add(newInterval);
        
        // Add remaining intervals
        while (i < n) {
            result.add(intervals[i]);
            i++;
        }
        
        return result.toArray(new int[result.size()][]);
    }
}

C++ Solution


class Solution {
public:
    vector> insert(vector>& intervals, vector& newInterval) {
        vector> result;
        int i = 0;
        int n = intervals.size();
        
        // Add all intervals that end before newInterval starts
        while (i < n && intervals[i][1] < newInterval[0]) {
            result.push_back(intervals[i]);
            i++;
        }
        
        // Merge overlapping intervals
        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++;
        }
        
        result.push_back(newInterval);
        
        // Add remaining intervals
        while (i < n) {
            result.push_back(intervals[i]);
            i++;
        }
        
        return result;
    }
};

JavaScript Solution


/**
 * @param {number[][]} intervals
 * @param {number[]} newInterval
 * @return {number[][]}
 */
var insert = function(intervals, newInterval) {
    const result = [];
    let i = 0;
    const n = intervals.length;
    
    // Add all intervals that end before newInterval starts
    while (i < n && intervals[i][1] < newInterval[0]) {
        result.push(intervals[i]);
        i++;
    }
    
    // Merge overlapping intervals
    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++;
    }
    
    result.push(newInterval);
    
    // Add remaining intervals
    while (i < n) {
        result.push(intervals[i]);
        i++;
    }
    
    return result;
};

C# Solution


public class Solution {
    public int[][] Insert(int[][] intervals, int[] newInterval) {
        var result = new List();
        int i = 0;
        int n = intervals.Length;
        
        // Add all intervals that end before newInterval starts
        while (i < n && intervals[i][1] < newInterval[0]) {
            result.Add(intervals[i]);
            i++;
        }
        
        // Merge overlapping intervals
        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++;
        }
        
        result.Add(newInterval);
        
        // Add remaining intervals
        while (i < n) {
            result.Add(intervals[i]);
            i++;
        }
        
        return result.ToArray();
    }
}

Complexity Analysis

Solution Explanation

This solution uses a three-step approach to insert and merge intervals:

Key points: