# Beats 99% java. Simple with explanation.

• So the trick here is to multiply number by 10 and identify start and end with a plus one.
This way, I know which element in the array is "start" and which is "end". The rest of the
code is simply iterate through a sorted array and find a start bound and an end bound.

``````public class Solution {
public List<Interval> merge(List<Interval> intervals) {
if(intervals.size()==0) return new ArrayList<Interval>();
int[] intArr=new int[2*intervals.size()];
int scount=0,start=0;
List<Interval> result=new ArrayList<Interval>();
for(int i=0;i<2*intervals.size();i+=2){
intArr[i]=intervals.get(i/2).start*10;
intArr[i+1]=intervals.get(i/2).end*10+1;
}
Arrays.sort(intArr);
for(int i=0;i<intArr.length;i++){
if(scount==0){//initialization
start=intArr[i]/10;
scount++;
continue;
}
if(intArr[i]%10!=0){
scount--;
if(scount==0) result.add(new Interval(start,(intArr[i]-1)/10));//if start bound count is equal to end bound count
}
else{
scount++;//find more than one start bound so need to find equal number end bound count
}
}
return result;
}
}
``````

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