# 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].
```

Thoughts of This Problem

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

Java Solution

```/** * 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; } }```
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 Sánchez

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 )

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

return ans;

} else {

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

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

}

}