0057 - Insert Interval (Medium)

Problem Link

https://leetcode.com/problems/insert-interval/

Problem Statement

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 intervalsafter the insertion.

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].

Constraints:

• 0 <= intervals.length <= 10^4
• intervals[i].length == 2
• 0 <= start_i <= endi <= 10^5
• intervals is sorted by start_i in ascending order.
• newInterval.length == 2
• 0 <= start <= end <= 10^5

Approach 1: One Pass

C++

Written by @wingkwong

class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        vector<vector<int>> ans;
        int n = intervals.size();
        for (int i = 0; i < n; i++) {
            if (intervals[i][1] < newInterval[0]) {
                // current interval starts first & not covered by newInterval, add intervals[i] to ans
                // [curInterval]
                //                      [newInterval]
                ans.push_back(intervals[i]);
            } else if (intervals[i][0] > newInterval[1]) {
                // newInterval starts first and not covered by current interval
                // add newInterval to ans and set newInterval = curInterval
                //                       [curInterval]
                // [newInterval]
                ans.push_back(newInterval);
                newInterval = intervals[i];
            } else if (intervals[i][1] >= newInterval[0] || intervals[i][0] <= newInterval[1]) {
                // they are overlapped, merge them
                // [curInterval]
                //        [newInterval]
                // or
                // [newInterval]
                //        [curInterval]
                newInterval[0] = min(newInterval[0], intervals[i][0]);
                newInterval[1] = max(newInterval[1], intervals[i][1]);
            }
        }
        // add the last interval
        ans.push_back(newInterval);
        return ans;
    }
};

Java

Written by @wingkwong

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List<int[]> ans = new ArrayList<>();
        int n = intervals.length;
        for (int i = 0; i < n; i++) {
            int[] curInterval = intervals[i];
            if (intervals[i][1] < newInterval[0]) {
                // current interval starts first & not covered by newInterval, add intervals[i] to ans
                // [curInterval]
                //                      [newInterval]
                ans.add(intervals[i]);
            } else if (intervals[i][0] > newInterval[1]) {
                // newInterval starts first and not covered by current interval
                // add newInterval to ans and set newInterval = curInterval
                //                       [curInterval]
                // [newInterval]
                ans.add(newInterval);
                newInterval = intervals[i];
            } else if (intervals[i][1] >= newInterval[0] || intervals[i][0] <= newInterval[1]) {
                // they are overlapped, merge them
                // [curInterval]
                //        [newInterval]
                // or
                // [newInterval]
                //        [curInterval]
                newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
                newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
            }
        }
        // add the last interval
        ans.add(newInterval);
        return ans.toArray(new int[ans.size()][]);
    }
}

Python

Written by @wingkwong

class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        ans = []
        n = len(intervals)
        for i in range(n):
            if intervals[i][1] < newInterval[0]:
                # current interval starts first & not covered by newInterval, add intervals[i] to ans
                # [curInterval]
                #                      [newInterval]
                ans.append(intervals[i])
            elif intervals[i][0] > newInterval[1]:
                # newInterval starts first and not covered by current interval
                # Add newInterval to ans and set newInterval = curInterval
                #                       [curInterval]
                # [newInterval]
                ans.append(newInterval)
                newInterval = intervals[i]
            elif intervals[i][1] >= newInterval[0] or intervals[i][0] <= newInterval[1]:
                # They are overlapped, merge them
                # [curInterval]
                #        [newInterval]
                # or
                # [newInterval]
                #        [curInterval]
                newInterval[0] = min(newInterval[0], intervals[i][0])
                newInterval[1] = max(newInterval[1], intervals[i][1])
        # add the last interval
        ans.append(newInterval)
        return ans

Rust

Written by @wingkwong

use std::cmp::min;
use std::cmp::max;

impl Solution {
    pub fn insert(intervals: Vec<Vec<i32>>, new_interval: Vec<i32>) -> Vec<Vec<i32>> {
        let mut ans = vec![];
        let n = intervals.len();
        let mut new_interval = new_interval;
        for i in 0 .. n {
            if intervals[i][1] < new_interval[0] {
                // current interval starts first & not covered by newInterval, add intervals[i] to ans
                // [curInterval]
                //                      [newInterval]
                ans.push(intervals[i].clone());
            } else if intervals[i][0] > new_interval[1] {
                // NewInterval starts first and not covered by current interval
                // Add newInterval to ans and set newInterval = curInterval
                //                       [curInterval]
                // [newInterval]
                ans.push(new_interval.clone());
                new_interval = intervals[i].clone();
            } else if intervals[i][1] >= new_interval[0] || intervals[i][0] <= new_interval[1] {
                // they are overlapped, merge them
                // [curInterval]
                //        [newInterval]
                // or
                // [newInterval]
                //        [curInterval]
                new_interval[0] = min(new_interval[0], intervals[i][0]);
                new_interval[1] = max(new_interval[1], intervals[i][1]);
            }
        }
        // add the last interval
        ans.push(new_interval.clone());
        ans
    }
}

JavaScript

Written by @radojicic23

/**
 * @param {number[][]} intervals
 * @param {number[]} newInterval
 * @return {number[][]}
 */
var insert = function (intervals, newInterval) {
  let res = [];
  for (i = 0; i < intervals.length; i++) {
    if (newInterval[1] < intervals[i][0]) {
      res.push(newInterval);
      return res.concat(intervals.slice(i, intervals.length));
    } else if (newInterval[0] > intervals[i][1]) {
      res.push(intervals[i]);
    } else {
      newInterval[0] = [Math.min(newInterval[0], intervals[i][0])];
      newInterval[1] = [Math.max(newInterval[1], intervals[i][1])];
    }
  }
  res.push(newInterval);
  return res;
};