LeetCode – Insert Interval

Problem:

Given a set of non-overlapping & sorted intervals, insert a new interval into the intervals (merge if necessary).

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

Analysis

Quickly summarize 3 cases. Whenever there is intersection, created a new interval.

insert-interval

Java Solution 1

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public ArrayList<Interval> insert(ArrayList<Interval> intervals, Interval newInterval) {
 
        ArrayList<Interval> result = new ArrayList<Interval>();
 
        for(Interval interval: intervals){
            if(interval.end < newInterval.start){
                result.add(interval);
            }else if(interval.start > newInterval.end){
                result.add(newInterval);
                newInterval = interval;        
            }else if(interval.end >= newInterval.start || interval.start <= newInterval.end){
                newInterval = new Interval(Math.min(interval.start, newInterval.start), Math.max(newInterval.end, interval.end));
            }
        }
 
        result.add(newInterval); 
 
        return result;
    }
}

Java Solution 2 - Binary Search

If the intervals list is an ArrayList, we can use binary search to make the best time complexity O(n).

public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
    List<Interval> result = new ArrayList<>();
 
    if (intervals.size() == 0) {
        result.add(newInterval);
        return result;
    }
 
    int p = helper(intervals, newInterval);
    result.addAll(intervals.subList(0, p));
 
    for (int i = p; i < intervals.size(); i++) {
        Interval interval = intervals.get(i);
        if (interval.end < newInterval.start) {
            result.add(interval);
        } else if (interval.start > newInterval.end) {
            result.add(newInterval);
            newInterval = interval;
        } else if (interval.end >= newInterval.start || interval.start <= newInterval.end) {
            newInterval = new Interval(Math.min(interval.start, newInterval.start), Math.max(newInterval.end, interval.end));
        }
    }
 
    result.add(newInterval);
 
    return result;
}
 
public int helper(List<Interval> intervals, Interval newInterval) {
    int low = 0;
    int high = intervals.size() - 1;
 
    while (low < high) {
        int mid = low + (high - low) / 2;
 
        if (newInterval.start <= intervals.get(mid).start) {
            high = mid;
        } else {
            low = mid + 1;
        }
    }
 
    return high == 0 ? 0 : high - 1;
}

The best time is O(log(n)) and worst case time is O(n).

Category >> Algorithms  
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  • Sreekar

    First thing that came to mind is binary search. I wonder admin considered it to be unnecessarily complex or something.

  • Matias SM

    I propose a best case O(log N) solution based on binary search. Note however that the overall algorithm can have a O(N) cost due to interval removal from the array (cost of arbitrary position removal in an array) – which could be optimized/amortized separately.


    class Interval {
    final int s;
    final int e;

    Interval(int s, int e) { this.s = s; this.e = e; }

    /** Assumes there exists an overlap */
    Interval merge(Interval o) {
    return new Interval(Math.min(s, o.s), Math.max(e, o.e));
    }
    }

    void insert(Interval i, ArrayList sortedList) {
    if (sortedList.isEmpty()) {
    sortedList.add(i);
    return;
    }

    int idxS = searchInsertIdx(i.s, sortedList);
    //we look for e+1 because we want to merge if eq too (see logic later)
    int idxE = searchInsertIdx(i.e + 1, sortedList);

    boolean replaceS = false;
    if (idxS > 0) {
    Interval prev = sortedList.get(idxS - 1);
    if (prev.e >= i.s) {
    i = i.merge(prev);
    idxS -= 1;
    replaceS = true;
    }
    }

    if (idxS < idxE) {
    Interval lastToMerge = sortedList.get(idxE - 1);
    i = i.merge(lastToMerge);
    if ((idxS + 1) < idxE) {
    removeRange(idxS + 1, idxE, sortedList);
    replaceS = true;
    }
    }

    if (replaceS) sortedList.set(idxS, i);
    else sortedList.add(idxS, i);
    }

    void removeRange(int s, int e, ArrayList sortedList) {
    sortedList.subList(s, e).clear();
    }

    /* Returns the position where an Interval starting at startValue should be inserted ignoring merges */
    int searchInsertIdx(int startValue, ArrayList sortedList) {
    if (sortedList.isEmpty()) return 0;

    int s = 0;
    int e = sortedList.size();
    while (e > s) {
    int mid = (e + s)/2;
    Interval atMid = sortedList.get(mid);

    if (atMid.s == startValue) return mid;
    else if(atMid.s < startValue) s = mid + 1;
    else e = mid - 1;
    }

    return sortedList.get(s).s < startValue? s + 1 : s;
    }

  • Jack Y

    sortmerge

  • Juan Gomez

    It’s a great solution. I think this is not necessary, though:

    || interval.start <= newInterval.end

  • Omar Edgardo Lugo S├ínchez

    just add the new interval, and run 7) Merge Intervals

  • tia

    I add some check before inserting. Sorry, attached pic twice. Don’t know how to remove it.

  • GuoJiaAgain

    public class Solution {

    public ArrayList insert(ArrayList intervals, Interval newInterval) {

    ArrayList ans = new ArrayList();

    int size = intervals.size();

    while( size > 0 ) {

    Interval i = intervals.remove(0);

    size–;

    if( i.end < newInterval.start )

    ans.add( i );

    else if ( newInterval.end < i.start ) {

    ans.add( newInterval );

    ans.add( i );

    ans.addAll( intervals );

    return ans;

    } else {

    newInterval.start = Math.min( newInterval.start, i.start );

    newInterval.end = Math.max( newInterval.end, i.end );

    }

    }

    ans.add( newInterval );

    return ans;

    }

  • Kevin

    Is there any typo in this solution. It doesn’t work.