Interval list intersections

//same as merge intervals class Solution { public int[][] intervalIntersection(int[][] firstList, int[][] secondList) { //put both the list in a single list 'list' List<int[]> list = new ArrayList<>(); for(int i =0;i<firstList.length;i++){ list.add(new int[]{firstList[i][0],firstList[i][1]}); } for(int i =0;i<secondList.length;i++){ list.add(new int[]{secondList[i][0],secondList[i][1]}); } //sort the 'list' in ascending order of the start time  Collections.sort(list,(a,b)->{ if(a[0]==b[0]) return Integer.compare(b[1],a[1]);// if the start time is same sort on the basis on ending time return Integer.compare(a[0],b[0]); //else sort on the basis of start time }); //below is same as mergeInterval problem:solved greedly int start = Integer.MAX_VALUE; int end = Integer.MIN_VALUE; List<int[]> result = new ArrayList<>(); for(int i =0;i<list.size();i++){ int intervals[] = list.get(i); //case 1 : (a,b) and (p,q) where p is between (a,b) and b is between (p,q) so intersection will be : (p,b) if(end>=intervals[0] && end<= intervals[1]){ result.add(new int[]{intervals[0],end}); start = Integer.min(start, intervals[0]); end = Integer.max(end, intervals[1]); } //case 2 : (a,b) and (p,q) and p,q both lie between (a,b) else if(end>=intervals[0] && end>=intervals[1]){ result.add(new int[]{intervals[0],intervals[1]}); } else{ start = intervals[0]; end = intervals[1]; } } int res[][] = new int[result.size()][2]; for(int i= 0;i<result.size();i++){ res[i] = result.get(i); } return res; } }