In this Leetcode Insert Interval problem solution, we have given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary). You may assume that the intervals were initially sorted according to their start times.

Leetcode Insert Interval problem solution


Problem solution in Python.

class Solution(object):
    def insert(self, intervals, newInterval):
        # O(n)
        left, right, s, e = [], [], newInterval.start, newInterval.end
        for i in intervals:
            if i.end < s: left.append(i)
            elif i.start > e: right.append(i)
            else: s, e = min(s, i.start), max(e, i.end)
        return left + [Interval(s, e)] + right
        
        # O(nlogn)
        res = []
        for i in sorted(intervals + [newInterval], key = lambda x: x.start):
            if not res: res.append(i)
            else:
                if i.start <= res[-1].end: res[-1].end = max(res[-1].end, i.end)
                else: res.append(i)
        return res



Problem solution in Java.

public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
    List<Interval> ans = new ArrayList<>();
    int i = 0, start = 0, end = 0;
    for (Interval e : intervals) {
        if (e.end < newInterval.start)
            ans.add(e);
        else if (end == 0 && e.end >= newInterval.start){ 
            if (e.start > newInterval.end) {
                ans.add(newInterval);
                ans.add(e);
                end = 1; start = 1;
            } else { 
                newInterval.start = Math.min(e.start, newInterval.start);
                newInterval.end = Math.max(e.end, newInterval.end);
                end = 1;
            }       
        } else if (start == 0 && e.start <= newInterval.end) { 
            newInterval.end = Math.max(e.end, newInterval.end);
        } else {
            if (start == 0) {
                ans.add(newInterval);
                start = 1;
            }
            ans.add(e);
        }
    }
    if (end == 0 || start == 0) {
        ans.add(newInterval);
        return ans;
    }          
    return ans;
}


Problem solution in C++.

class Solution {
public:
    vector<Interval> insert(vector<Interval>& intervals, Interval t) {
        vector<Interval> result; 
        int i=0;
        bool t_added=false;
        for(i=0; i<intervals.size(); ++i) {
            if(intervals[i].end<t.start)
                result.push_back(intervals[i]);
            else if(intervals[i].start > t.end){
                result.push_back(t);
                t_added=true;
                break;
            }
            else {
                t.start=min(t.start, intervals[i].start);
                t.end=max(t.end, intervals[i].end);
            }
        }
        if(!t_added) result.push_back(t);
        for(; i<intervals.size(); ++i){
            result.push_back(intervals[i]);
        }
        return result;
    }
};