# Tried to make the solution similar to merge interval problem using Java and min heap

• ``````public class Solution {
class Interval{
int start;
int end;

Interval(int start, int end){
this.start = start;
this.end = end;
}
}
public int findMinArrowShots(int[][] points) {
if(points == null || points.length == 0) return 0;
PriorityQueue<Interval> pq = new PriorityQueue<>(new Comparator<Interval>(){
public int compare(Interval a, Interval b){
if(a.start == b.start) return a.end - b.end;
else return a.start-b.start;
}
});
for(int i = 0;i<points.length;i++){
}
int minArrows = 1;
Interval temp = pq.poll();
int end = temp.end;
while(!pq.isEmpty()){
Interval curr = pq.poll();
if(curr.start<=end){
//shoot at a common point in both the intervals. which is min(end, curr.end)
//for example 1,6 and 2,8 ..so shoot at 6 which is covered by both.
end = Math.min(end, curr.end);
}else{

//you need a new arrow for this one buddy.
minArrows++;
end = curr.end;
}
}
return minArrows;
}
}``````

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