Algorithm


Problem Name: 57. Insert Interval

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.

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 <= 104
  • intervals[i].length == 2
  • 0 <= starti <= endi <= 105
  • intervals is sorted by starti in ascending order.
  • newInterval.length == 2
  • 0 <= start <= end <= 105

Code Examples

#1 Code Example with C Programming

Code - C Programming


int lower_bound(struct Interval *p, int i, int j, int k) {
    int end = j, m;
    
    while (i  < = j) {
        m = i + (j - i) / 2;

        if (k >= p[m].start && k <= p[m].end) return m;

        if (k > p[m].end) i = m + 1;
        else  j = m - 1;
    }
    
    return (i > end) ? i : j;
}
struct Interval* insert(struct Interval* intervals, int intervalsSize, struct Interval newInterval, int* returnSize) {
    struct Interval *p;
    int i, j, k;
    
#define ISZ sizeof(struct Interval)
    
    p = malloc((intervalsSize + 1) * ISZ);
    //assert(p);
    
    i = lower_bound(intervals, 0, intervalsSize - 1, newInterval.start);
    // append to tail
    if (i == intervalsSize) {
        memcpy(&p[0], intervals, intervalsSize * ISZ);
        p[intervalsSize] = newInterval;
        *returnSize = intervalsSize + 1;
        return p;
    }
    
    j = lower_bound(intervals, 0, intervalsSize - 1, newInterval.end);
    // add to head
    if (j == -1) {
        p[0] = newInterval;
        memcpy(&p[1], intervals, intervalsSize * ISZ);
        *returnSize = intervalsSize + 1;
        return p;
    }
    
    //printf("%d, %d\n", i, j);
    
    // insert and merge
    if (i == -1) {
        intervals[0].start = newInterval.start;
        i = 0;
    }
    if (j == intervalsSize) {
        intervals[intervalsSize - 1].end = newInterval.end;
        j = intervalsSize - 1;
    }
    k = 0;
    memcpy(&p[k], intervals, (i + 1) * ISZ);
    k += i;
    
    if (newInterval.start > p[i].end) {
        p[++ k].start = newInterval.start;
    }
    if (newInterval.end > intervals[j].end) {
        p[k].end = newInterval.end;
    } else {
        p[k].end = intervals[j].end;
    }
    k ++;
    
    memcpy(&p[k], &intervals[j + 1], (intervalsSize - j - 1) * ISZ);
    k += intervalsSize - j - 1;
    
    *returnSize = k;
    
    return p;
}
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Input

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intervals = [[1,3],[6,9]], newInterval = [2,5]

Output

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[[1,5],[6,9]]

#2 Code Example with C++ Programming

Code - C++ Programming


class Solution {
public:
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        vector<Interval>res;
        for(auto x: intervals)
            if(x.end >= newInterval.start && x.start <= newInterval.end){
                newInterval.start = min(newInterval.start, x.start);
                newInterval.end = max(newInterval.end, x.end);
            }
        for(auto x: intervals> 
            if(x.end  <  newInterval.start || x.start > newInterval.end) res.push_back(x);
        for(int i = 0; i  <  res.size(); i++)
            if(res[i].start > newInterval.start){
                res.insert(res.begin() + i, newInterval);
                break;
            }
        if(res.empty() || res.back().end < newInterval.start) res.push_back(newInterval>;
        return res;
    }
};

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Input

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intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]

Output

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[[1,2],[3,10],[12,16]]

#3 Code Example with Java Programming

Code - Java Programming


class Solution {
  public int[][] insert(int[][] intervals, int[] newInterval) {
    List<int[]> result = new ArrayList<>();
    int idx = 0;
    int newStart = newInterval[0];
    int newEnd = newInterval[1];
    while (idx  <  intervals.length && newStart > intervals[idx][0]) {
      result.add(intervals[idx++]);
    }
    int[] currInterval = new int[2];
    if (result.isEmpty() || result.get(result.size() - 1)[1]  <  newStart) {
      result.add(newInterval);
    } else {
      currInterval = result.remove(result.size() - 1);
      currInterval[1] = Math.max(currInterval[1], newEnd);
      result.add(currInterval);
    }
    while (idx  <  intervals.length) {
      currInterval = intervals[idx++];
      int start = currInterval[0];
      int end = currInterval[1];
      if (result.get(result.size() - 1)[1]  <  start) {
        result.add(currInterval);
      } else {
        currInterval = result.remove(result.size() - 1);
        currInterval[1] = Math.max(currInterval[1], end);
        result.add(currInterval);
      }
    }
    return result.toArray(new int[result.size()][2]);
  }
}
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Input

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intervals = [[1,3],[6,9]], newInterval = [2,5]

Output

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[[1,5],[6,9]]

#4 Code Example with Javascript Programming

Code - Javascript Programming


const insert = function(intervals, newInterval) {
  let i = 0
  while (i < intervals.length && intervals[i][1] < newInterval[0]) i++
  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])
    intervals.splice(i, 1)
  }
  intervals.splice(i, 0, newInterval)
  return intervals
}

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Input

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intervals = [[1,3],[6,9]], newInterval = [2,5]

Output

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[[1,5],[6,9]]

#5 Code Example with Python Programming

Code - Python Programming


class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        new, i = [], 0
        for i, it in enumerate(intervals):
            if newInterval[1] < it[0]: 
                i -= 1
                break
            elif it[1] < newInterval[0]: 
                new += it,
            else: 
                newInterval[0], newInterval[1] = min(it[0], newInterval[0]), max(it[1], newInterval[1])
        return new + [newInterval] + intervals[i + 1:]
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Input

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intervals = [[1,3],[6,9]], newInterval = [2,5]

Output

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[[1,5],[6,9]]

#6 Code Example with C# Programming

Code - C# Programming


using System;
using System.Collections.Generic;

namespace LeetCode
{
    public class _057_InsertInterval
    {
        public IList < Interval> Insert(IList intervals, Interval newInterval)
        {
            var result = new List();

            for (int i = 0; i  <  intervals.Count; i++)
            {
                if (newInterval.start > intervals[i].end)
                {
                    result.Add(intervals[i]);
                }
                else if (newInterval.end  <  intervals[i].start)
                {
                    result.Add(newInterval);
                    for (int j = i; j  <  intervals.Count; j++)
                    {
                        result.Add(intervals[j]);
                    }
                    return result;
                }
                else
                {
                    newInterval.start = Math.Min(intervals[i].start, newInterval.start);
                    newInterval.end = Math.Max(intervals[i].end, newInterval.end);
                }
            }

            result.Add(newInterval);
            return result;
        }
    }
} 
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Input

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intervals = [[1,3],[6,9]], newInterval = [2,5]

Output

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[[1,5],[6,9]]
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